寒假-第一周-几何-(点线关系的问题)

寒假第一周：几何

POJ 2318 TOYS
POJ 2398 Toy Storage(叉积判断左右+二分)
POJ 3304 Segments(思维)
POJ 1269 Intersecting Lines
POJ 1556 The Doors(判断线段相交+最短路径)
POJ 1066 Treasure Hunt
POJ 1696 Space Ant(极角排序)
POJ 3347 Kadj Squares(扩大数据避免小数)
POJ 2826 An Easy Problem?!(大力出奇迹?)
POJ 3449 Geometric Shapes(非常恶心的读入)
POJ 1584 A Round Peg in a Ground Hole(凸包+点线圆的相对关系)
POJ 2074 Line of Sight(平面视线问题)

POJ 2318 TOYS

题意： 给定一个矩形区域，再给 n n n 个隔板（保证隔板相互之间交叉），现在有 m m m 个玩具（用二维坐标表示，保证不会超出边界），要求 n n n 个挡板划分出的 n + 1 n + 1 n+1 个区域中各有多少个玩具。

思路：简单处理一下，用叉积来判断该点在直线的左边还是右边。

注意：在用叉积判断左右的时候要先确保，所有直线的起始点和终点的方向大致一样，例如都是下往上，否则判断结果会不同

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}
struct Point {double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const{return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const { //小在前return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}Point operator*(const double& k) const {return Point(k * x, k * y);}Point operator/(const double& k) const {return Point(x / k, y / k);}double operator^(const Point& b) const { //叉积return x * b.y - y * b.x;}double operator*(const Point& b) const {return x * b.x + y * b.y;}double len() {  //长度return sqrt(x * x + y * y);
//        return hypot(x, y);}double len2() {  //长度的平方return x * x + y * y;}double dis(Point p) { //两点距离return hypot(x - p.x, y - p.y);  //hypot是计算斜边长用的
//        return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));}double dis2(Point p) { //两点距离的平方return pow(x - p.x, 2.0) + pow(y - p.y, 2.0);}double rad(Point a, Point b) { //atan2(y, x),表示(x,y)在的角度(-pi,pi];Point p = *this;return fabs(atan2(fabs((a - p) ^ (b - p)), (a - p) * (b - p)));//角P的夹角}Point trunc(double r) { //转换为长度为r的向量，零向量除外double l = len();if (!dcmp(l)) return *this;r /= l;return Point(x * r, y * r);}Point rotleft() { //逆时针旋转90度return Point(-y, x);}Point rotright() { //顺时针旋转90度return Point(y, -x);}Point Rotate(Point p, double angle) {Point v = (*this) - p;double c = cos(angle), s = sin(angle);return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);}
};
struct Line {Point s, e;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}bool operator==(const Line& v) const {return (s == v.s) && (e == v.e);}Line(Point p, double angle) { //需要保证angle在[0, pi)s = p;if (dcmp(angle - pi / 2.0) == 0) {e = (s + Point(0, 1));}else {e = (s + Point(1, tan(angle)));}}Line(double a, double b, double c) { //ax + by + c = 0 一般式if (dcmp(a) == 0) {s = Point(0, -c / b);e = Point(1, -c / b);}else if (dcmp(b == 0)) {s = Point(-c / a, 0);e = Point(-c / a, 1);}else {s = Point(0, -c / b);e = Point(1, (-c - a) / b);}}void input() {s.input(); e.input();}void adjust() {if (e < s) swap(s, e);}double length() {return s.dis(e);}double angle() { //斜率,保证在[0, pi)double k = atan2(e.y - s.y, e.x - s.x);if (dcmp(k) < 0) k += pi;if (dcmp(k - pi) == 0) k -= pi;return k;}int relation(Point p) {int c = dcmp((p - s) ^ (e - s));if (c < 0) return 1; //在直线左侧else if (c > 0) return 2; //在直线右侧else return 3; //在直线上}bool pointonseg(Point p) { //点在线段上的判断(<=0表示包括端点)return dcmp((p - s) ^ (e - s)) == 0 && dcmp((p - s) * (p - e)) <= 0;}bool parallel(Line v) { //判断两向量平行return dcmp((e - s) ^ (v.e - v.s)) == 0;}double disPointToLine(Point p) { //点到直线的距离return fabs((p - s) ^ (e - s) / length());}double disPointToSeg(Point p) {  //点到线段的距离if (dcmp((p - s) * (e - s)) < 0 || dcmp((p - e) * (s - e)) < 0)return min(p.dis(s), p.dis(e));return disPointToLine(p);}
};
Line li[5005];
Point p[5005];
int main() {int n, m;while (scanf("%d", &n)) {if (n == 0) break;Point a, b;scanf("%d", &m);a.input(); b.input(); //a为左上角，b为右下角li[0].s = Point(a.x, b.y);li[0].e = Point(a.x, a.y);for (int i = 1; i <= n; i++) {int u, v;scanf("%d%d", &u, &v);li[i].e = Point(u, a.y);li[i].s = Point(v, b.y);}li[n + 1].s = Point(b.x, b.y);li[n + 1].e = Point(b.x, a.y);for (int i = 1; i <= m; i++) {p[i].input();}vector<int> pos(n + 2, 0);for (int i = 1; i <= m; i++) {for (int j = 0; j <= n; j++) {int d1 = li[j].relation(p[i]);int d2 = li[j + 1].relation(p[i]);if (d1 != d2) {pos[j]++;break;}}}for (int i = 0; i <= n; i++) {printf("%d: %d\n", i, pos[i]);}printf("\n");}return 0;
}

POJ 2398 Toy Storage(叉积判断左右+二分)

题意： 和上题差不多，但是是挡板是乱序输入的，还有OJ上的样例是错的。。。。。

思路： 因为和上题大致一样，换了一种思路，二分，利用叉积来做判断，需要注意的点和上题一致，也是要将线段的起点和重点的方向统一，不然无法用叉积判断是否在一边

代码:

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}
struct Point {double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const{return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const { //小在前return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}Point operator*(const double& k) const {return Point(k * x, k * y);}Point operator/(const double& k) const {return Point(x / k, y / k);}double operator^(const Point& b) const { //叉积return x * b.y - y * b.x;}double operator*(const Point& b) const {return x * b.x + y * b.y;}double len() {  //长度return sqrt(x * x + y * y);
//        return hypot(x, y);}double len2() {  //长度的平方return x * x + y * y;}double dis(Point p) { //两点距离return hypot(x - p.x, y - p.y);  //hypot是计算斜边长用的
//        return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));}
};
struct Line {Point s, e;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}bool operator==(const Line& v) const {return (s == v.s) && (e == v.e);}int relation(Point p) {int c = dcmp((p - s) ^ (e - s));if (c < 0) return 1; //在直线左侧else if (c > 0) return 2; //在直线右侧else return 3; //在直线上}
};
Line li[5005];
Point p[5005];
bool cmp(const Line& v1, const Line& v2) {if (v1.s.x == v2.s.x) return v1.e.x < v2.e.x;return v1.s.x < v2.s.x;
}
int main() {int n, m;while (scanf("%d", &n) && n) {scanf("%d", &m);Point p1, p2;p1.input(); p2.input(); //p1是左上角，p2是右下角for (int i = 0; i < n; i++) {int u, v;scanf("%d%d", &u, &v);li[i] = Line(Point(v, p2.y), Point(u, p1.y));}sort(li, li + n, cmp);
//        li[n] = Line(Point(p2.x, p2.y), Point(p2.x, p1.y));
//        printf("-------\n");
//        for (int i = 0; i <= n; i++) {//            printf("%.2f %.2f\n", li[i].s.x, li[i].e.x);
//        }
//        printf("-------\n");vector<int> pos(n + 1, 0);for (int i = 1; i <= m; i++) {p[i].input();int l = 0, r = n;while (r > l) {int mid = (l + r) >> 1;if (((p[i] - li[mid].s) ^ (li[mid].e - li[mid].s)) < 0)r = mid;else l = mid + 1;}pos[r]++;}
//        for (int i = 0; i <= n; i++) {//            printf("%d: %d\n", i, pos[i]);
//        }vector<int> ans(m + 1);for (int i = 0; i <= n; i++) {if (pos[i]) ans[pos[i]]++;}printf("Box\n");for (int i = 1; i <= m; i++) {if (ans[i])printf("%d: %d\n", i, ans[i]);}}return 0;
}

POJ 3304 Segments(思维)

题意：

题目大意：给n条线段，求是否存在一条直线，将所有线段投影到上面，使得所有投影至少交于一点。

思路： 这题是看题解的，刚开始想到与端点有关，但是接下来就没有思路了

这里有个很好的转换：

存在一条直线，将所有的线段投影到该直线上后，至少重叠于一个点<==>过重合点的直线，做该点的垂线，必然会与所有的直线相交。

所以枚举所有的端点，两两构成直线，来判断是否与所有线段相交即可。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}
struct Point {double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const{return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const { //小在前return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}Point operator*(const double& k) const {return Point(k * x, k * y);}Point operator/(const double& k) const {return Point(x / k, y / k);}double operator^(const Point& b) const { //叉积return x * b.y - y * b.x;}double operator*(const Point& b) const {return x * b.x + y * b.y;}double len() {  //长度return sqrt(x * x + y * y);
//        return hypot(x, y);}double len2() {  //长度的平方return x * x + y * y;}double dis(Point p) { //两点距离return hypot(x - p.x, y - p.y);  //hypot是计算斜边长用的
//        return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));}double dis2(Point p) { //两点距离的平方return pow(x - p.x, 2.0) + pow(y - p.y, 2.0);}double rad(Point a, Point b) { //atan2(y, x),表示(x,y)在的角度(-pi,pi];Point p = *this;return fabs(atan2(fabs((a - p) ^ (b - p)), (a - p) * (b - p)));//角P的夹角}Point trunc(double r) { //转换为长度为r的向量，零向量除外double l = len();if (!dcmp(l)) return *this;r /= l;return Point(x * r, y * r);}Point rotleft() { //逆时针旋转90度return Point(-y, x);}Point rotright() { //顺时针旋转90度return Point(y, -x);}Point Rotate(Point p, double angle) {Point v = (*this) - p;double c = cos(angle), s = sin(angle);return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);}
};
struct Line {Point s, e;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}bool operator==(const Line& v) const {return (s == v.s) && (e == v.e);}Line(Point p, double angle) { //需要保证angle在[0, pi)s = p;if (dcmp(angle - pi / 2.0) == 0) {e = (s + Point(0, 1));}else {e = (s + Point(1, tan(angle)));}}Line(double a, double b, double c) { //ax + by + c = 0 一般式if (dcmp(a) == 0) {s = Point(0, -c / b);e = Point(1, -c / b);}else if (dcmp(b == 0)) {s = Point(-c / a, 0);e = Point(-c / a, 1);}else {s = Point(0, -c / b);e = Point(1, (-c - a) / b);}}void input() {s.input(); e.input();}void adjust() {if (e < s) swap(s, e);}double length() {return s.dis(e);}double angle() { //斜率,保证在[0, pi)double k = atan2(e.y - s.y, e.x - s.x);if (dcmp(k) < 0) k += pi;if (dcmp(k - pi) == 0) k -= pi;return k;}int relation(Point p) {int c = dcmp((p - s) ^ (e - s));if (c < 0) return 1; //在直线左侧else if (c > 0) return 2; //在直线右侧else return 3; //在直线上}bool pointonseg(Point p) { //点在线段上的判断(<=0表示包括端点)return dcmp((p - s) ^ (e - s)) == 0 && dcmp((p - s) * (p - e)) <= 0;}bool parallel(Line v) { //判断两向量平行return dcmp((e - s) ^ (v.e - v.s)) == 0;}double disPointToLine(Point p) { //点到直线的距离return fabs((p - s) ^ (e - s) / length());}double disPointToSeg(Point p) {  //点到线段的距离if (dcmp((p - s) * (e - s)) < 0 || dcmp((p - e) * (s - e)) < 0)return min(p.dis(s), p.dis(e));return disPointToLine(p);}Point lineprog(Point p) {  //点在直线上的投影点return s + ( ((e - s) * ((e - s) * (p - s))) / ((e - s).len2()) );}Point symmetrypoint(Point p) {  //点关于直线的对称点Point q = lineprog(p);return Point(2 * q.x - p.x, 2 * q.y - p.y);}int segcrossseg(Line v) { //两线段相交，2为规范相交，1为非规范相交int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));int d3 = dcmp((v.e - v.s) ^ (s - v.s));int d4 = dcmp((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;return (d1 == 0 && dcmp((v.s - s) * (v.s - e)) <= 0) ||(d2 == 0 && dcmp((v.e - s) * (v.e - e)) <= 0) ||(d3 == 0 && dcmp((s - v.s) * (s - v.e)) <= 0) ||(d4 == 0 && dcmp((e - v.s) * (e - v.e)) <= 0);}int linecrossseg(Line v) {//直线与线段，2为规范相交，1为非规范相交int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));if ((d1 ^ d2) == -2) return 2;return (d1 == 0 || d2 == 0);}Point crossPoint(Line v) { //两直线的交点(保证不重合或平行)double a1 = (v.e - v.s) ^ (s - v.s);double a2 = (v.e - v.s) ^ (e - v.s);return Point((s.x * a2 - e.x * a1) / (a2 - a1), (s.y * a2 - e.y * a1) / (a2 - a1));}double disSegToSeg(Line v) { //线段到线段的距离,相交就是0return min(min(disPointToSeg(v.s), disPointToSeg(v.e)),min(v.disPointToSeg(s), v.disPointToSeg(e)));}
};int main() {int t;scanf("%d", &t);while(t--) {int n;scanf("%d", &n);vector<Line> li(n + 1);vector<Point> p(n * 2 + 1);for (int i = 0; i < n; i++) {li[i].s.input();li[i].e.input();p[i * 2 + 0] = li[i].s;p[i * 2 + 1] = li[i].e;}int np = 2 * n;bool f = false;for (int i = 0; i < np && !f; i++) {for (int j = i + 1; j < np && !f; j++) {if (p[i] == p[j]) continue;Line tmp = Line(p[i], p[j]);int cnt = 0;for (int k = 0; k < n; k++) {int isok = tmp.linecrossseg(li[k]);if (isok > 0) {cnt++;}}if (cnt == n) f = true;}}if (f) printf("Yes!\n");else printf("No!\n");}return 0;
}

POJ 1269 Intersecting Lines

题意： 判断平行，还是重合，还是相交

思路： 是否相交用叉积判断，是否重合用点到直线的距离判断

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}
struct Point {double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const{return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const { //小在前return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}Point operator*(const double& k) const {return Point(k * x, k * y);}Point operator/(const double& k) const {return Point(x / k, y / k);}double operator^(const Point& b) const { //叉积return x * b.y - y * b.x;}double operator*(const Point& b) const {return x * b.x + y * b.y;}double len() {  //长度return sqrt(x * x + y * y);
//        return hypot(x, y);}double len2() {  //长度的平方return x * x + y * y;}double dis(Point p) { //两点距离return hypot(x - p.x, y - p.y);  //hypot是计算斜边长用的
//        return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));}double dis2(Point p) { //两点距离的平方return pow(x - p.x, 2.0) + pow(y - p.y, 2.0);}double rad(Point a, Point b) { //atan2(y, x),表示(x,y)在的角度(-pi,pi];Point p = *this;return fabs(atan2(fabs((a - p) ^ (b - p)), (a - p) * (b - p)));//角P的夹角}Point trunc(double r) { //转换为长度为r的向量，零向量除外double l = len();if (!dcmp(l)) return *this;r /= l;return Point(x * r, y * r);}Point rotleft() { //逆时针旋转90度return Point(-y, x);}Point rotright() { //顺时针旋转90度return Point(y, -x);}Point Rotate(Point p, double angle) {Point v = (*this) - p;double c = cos(angle), s = sin(angle);return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);}
};
struct Line {Point s, e;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}bool operator==(const Line& v) const {return (s == v.s) && (e == v.e);}Line(Point p, double angle) { //需要保证angle在[0, pi)s = p;if (dcmp(angle - pi / 2.0) == 0) {e = (s + Point(0, 1));}else {e = (s + Point(1, tan(angle)));}}Line(double a, double b, double c) { //ax + by + c = 0 一般式if (dcmp(a) == 0) {s = Point(0, -c / b);e = Point(1, -c / b);}else if (dcmp(b == 0)) {s = Point(-c / a, 0);e = Point(-c / a, 1);}else {s = Point(0, -c / b);e = Point(1, (-c - a) / b);}}void input() {s.input(); e.input();}void adjust() {if (e < s) swap(s, e);}double length() {return s.dis(e);}double angle() { //斜率,保证在[0, pi)double k = atan2(e.y - s.y, e.x - s.x);if (dcmp(k) < 0) k += pi;if (dcmp(k - pi) == 0) k -= pi;return k;}int relation(Point p) {int c = dcmp((p - s) ^ (e - s));if (c < 0) return 1; //在直线左侧else if (c > 0) return 2; //在直线右侧else return 3; //在直线上}bool pointonseg(Point p) { //点在线段上的判断(<=0表示包括端点)return dcmp((p - s) ^ (e - s)) == 0 && dcmp((p - s) * (p - e)) <= 0;}bool parallel(Line v) { //判断两向量平行return dcmp((e - s) ^ (v.e - v.s)) == 0;}double disPointToLine(Point p) { //点到直线的距离return fabs((p - s) ^ (e - s) / length());}double disPointToSeg(Point p) {  //点到线段的距离if (dcmp((p - s) * (e - s)) < 0 || dcmp((p - e) * (s - e)) < 0)return min(p.dis(s), p.dis(e));return disPointToLine(p);}Point lineprog(Point p) {  //点在直线上的投影点return s + ( ((e - s) * ((e - s) * (p - s))) / ((e - s).len2()) );}Point symmetrypoint(Point p) {  //点关于直线的对称点Point q = lineprog(p);return Point(2 * q.x - p.x, 2 * q.y - p.y);}int segcrossseg(Line v) { //两线段相交，2为规范相交，1为非规范相交int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));int d3 = dcmp((v.e - v.s) ^ (s - v.s));int d4 = dcmp((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;return (d1 == 0 && dcmp((v.s - s) * (v.s - e)) <= 0) ||(d2 == 0 && dcmp((v.e - s) * (v.e - e)) <= 0) ||(d3 == 0 && dcmp((s - v.s) * (s - v.e)) <= 0) ||(d4 == 0 && dcmp((e - v.s) * (e - v.e)) <= 0);}int linecrossseg(Line v) {//直线与线段，2为规范相交，1为非规范相交int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));if ((d1 ^ d2) == -2) return 2;return (d1 == 0 || d2 == 0);}Point crossPoint(Line v) { //两直线的交点(保证不重合或平行)double a1 = (v.e - v.s) ^ (s - v.s);double a2 = (v.e - v.s) ^ (e - v.s);return Point((s.x * a2 - e.x * a1) / (a2 - a1), (s.y * a2 - e.y * a1) / (a2 - a1));}double disSegToSeg(Line v) { //线段到线段的距离,相交就是0return min(min(disPointToSeg(v.s), disPointToSeg(v.e)),min(v.disPointToSeg(s), v.disPointToSeg(e)));}
};int main() {int n;scanf("%d", &n);printf("INTERSECTING LINES OUTPUT\n");for (int i = 1; i <= n; i++) {Point a, b, c, d;a.input();b.input();c.input();d.input();Line ab = Line(a, b);Line cd = Line(c, d);if (ab.parallel(cd)) {int d = ab.disPointToLine(c);if (d > 0) {printf("NONE\n");}else {printf("LINE\n");}}else {Point res = ab.crossPoint(cd);printf("POINT %.2f %.2f\n", res.x, res.y);}}printf("END OF OUTPUT\n");return 0;
}

POJ 1556 The Doors(判断线段相交+最短路径)

题意： 给定一个 $10 \times 10 $ 大小的矩形，起点为 ( 0 , 5 ) (0, 5) (0,5)，终点为 ( 10 , 5 ) (10, 5) (10,5) 中间竖直方向最多有 18 18 18 堵墙壁，每堵墙壁都有两个门。求起点到终点的最短路径。

思路： 一开始是打算按照端点直接 d f s dfs dfs 爆搜，但是每堵墙不一定一定是经过端点才是最优的，因此直接 p a s s pass pass 。然后打算从当前点往能到达的最远点走，但是不行

写了半天，结果是个图论：

首先，枚举所有端点（包括起点和终点）连成的线段，若这条线段没有经过任何墙壁，就将这两个点建边，边权值为两点之间的距离，建完图后跑一边 d i j dij dij 求单源最短路就可以了。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}
struct Point {double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const{return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const { //小在前return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}Point operator*(const double& k) const {return Point(k * x, k * y);}Point operator/(const double& k) const {return Point(x / k, y / k);}double operator^(const Point& b) const { //叉积return x * b.y - y * b.x;}double operator*(const Point& b) const {return x * b.x + y * b.y;}double len() {  //长度return sqrt(x * x + y * y);
//        return hypot(x, y);}double len2() {  //长度的平方return x * x + y * y;}double dis(Point p) { //两点距离return hypot(x - p.x, y - p.y);  //hypot是计算斜边长用的
//        return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));}double dis2(Point p) { //两点距离的平方return pow(x - p.x, 2.0) + pow(y - p.y, 2.0);}double rad(Point a, Point b) { //atan2(y, x),表示(x,y)在的角度(-pi,pi];Point p = *this;return fabs(atan2(fabs((a - p) ^ (b - p)), (a - p) * (b - p)));//角P的夹角}Point trunc(double r) { //转换为长度为r的向量，零向量除外double l = len();if (!dcmp(l)) return *this;r /= l;return Point(x * r, y * r);}Point rotleft() { //逆时针旋转90度return Point(-y, x);}Point rotright() { //顺时针旋转90度return Point(y, -x);}Point Rotate(Point p, double angle) {Point v = (*this) - p;double c = cos(angle), s = sin(angle);return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);}
};
struct Line {Point s, e;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}bool operator==(const Line& v) const {return (s == v.s) && (e == v.e);}void input() {s.input(); e.input();}int relation(Point p) {int c = dcmp((p - s) ^ (e - s));if (c < 0) return 1; //在直线左侧else if (c > 0) return 2; //在直线右侧else return 3; //在直线上}bool pointonseg(Point p) { //点在线段上的判断(<=0表示包括端点)return dcmp((p - s) ^ (e - s)) == 0 && dcmp((p - s) * (p - e)) <= 0;}bool parallel(Line v) { //判断两向量平行return dcmp((e - s) ^ (v.e - v.s)) == 0;}double disPointToLine(Point p) { //点到直线的距离return fabs((p - s) ^ (e - s) / length());}double disPointToSeg(Point p) {  //点到线段的距离if (dcmp((p - s) * (e - s)) < 0 || dcmp((p - e) * (s - e)) < 0)return min(p.dis(s), p.dis(e));return disPointToLine(p);}Point lineprog(Point p) {  //点在直线上的投影点return s + ( ((e - s) * ((e - s) * (p - s))) / ((e - s).len2()) );}Point symmetrypoint(Point p) {  //点关于直线的对称点Point q = lineprog(p);return Point(2 * q.x - p.x, 2 * q.y - p.y);}int segcrossseg(Line v) { //两线段相交，2为规范相交，1为非规范相交int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));int d3 = dcmp((v.e - v.s) ^ (s - v.s));int d4 = dcmp((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;return (d1 == 0 && dcmp((v.s - s) * (v.s - e)) <= 0) ||(d2 == 0 && dcmp((v.e - s) * (v.e - e)) <= 0) ||(d3 == 0 && dcmp((s - v.s) * (s - v.e)) <= 0) ||(d4 == 0 && dcmp((e - v.s) * (e - v.e)) <= 0);}
};
vector<Line> g[100];
vector<Point> p;
double ans;
int n;
int st, ed;
typedef pair<double, int> P;
struct DIJ{struct edge{int u, v;double cost;int next;}es[10010];int head[100];double d[100];int V, tot;void init(int _n) {V = _n; tot = 0;for (int i = 0; i <= _n; i++) {head[i] = -1;}}void add_edge(int u, int v, double cost) {es[tot].u = u;es[tot].v = v;es[tot].cost = cost;es[tot].next = head[u];head[u] = tot++;}void dijstra(int s) {priority_queue<P, vector<P>, greater<P> > pq;for (int i = 0; i <= V; i++) d[i] = 1e9;d[s] = 0;pq.push(P(0, s));while (pq.size()) {P p = pq.top(); pq.pop();int v = p.second;if (d[v] < p.first) continue;for (int i = head[v]; i != -1; i = es[i].next) {int from = es[i].u;int to = es[i].v;double cost = es[i].cost;if (d[to] > d[from] + cost) {d[to] = d[from] + cost;pq.push(P(d[to], to));}}}}
}G;int main() {while (scanf("%d", &n)) {if (n == -1) break;if (n == 0) {printf("10.00\n");continue;}p.clear();for (int i = 0; i < n; i++) {g[i].clear();}p.push_back(Point(0, 5));for (int i = 0; i < n; i++) {double x;scanf("%lf", &x);vector<double> y;y.push_back(0);for (int j = 0; j < 4; j++) {double yy;scanf("%lf", &yy);p.push_back(Point(x, yy));y.push_back(yy);}y.push_back(10.0);for (int j = 0; j < (int)y.size(); j += 2) {g[i].push_back(Line(Point(x, y[j]), Point(x, y[j + 1])));}}p.push_back(Point(10, 5));st = 0;ed = p.size() - 1;
//        for (int i = 0; i < n; i++) {//            printf("%d: \n", i);
//            for (int j = 0; j < (int)g[i].size(); j++) {//                g[i][j].s.output();
//                g[i][j].e.output();
//            }
//        }G.init(ed);for (int i = 0; i <= ed; i++) {for (int j = i + 1; j <= ed; j++) {Line tmp = Line(p[i], p[j]);bool f = true;for (int k = 0; k < n && f; k++) {for (int c = 0; c < (int)g[k].size() && f; c++) {int isok = tmp.segcrossseg(g[k][c]);if (isok == 2) f = false;}}if (f) {G.add_edge(i, j, p[i].dis(p[j]));G.add_edge(j, i, p[i].dis(p[j]));}}}G.dijstra(0);printf("%.2f\n", G.d[ed]);}return 0;
}

POJ 1066 Treasure Hunt

题意： 一个 100 × 100 100 \times 100 100×100 的矩形中，有 n n n 面交错的墙。其中有一个宝藏点，求从矩形外走到宝藏处需要凿开的最少墙数。（只能在墙的中间凿洞）

思路：

开始理解的墙中间是题目给的那 n n n 块墙，然后先建图在求最短路做，但是一直 w a wa wa。后来看了题解，才知道，这个中间是两块区域间的墙，不是整块 … \dots … 这样只需要枚举起点，然后与宝藏点连成线段，再找到所有相交的墙壁面数，其中最小的就是答案。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}
struct Point {double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const{return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const { //小在前return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}Point operator*(const double& k) const {return Point(k * x, k * y);}Point operator/(const double& k) const {return Point(x / k, y / k);}double operator^(const Point& b) const { //叉积return x * b.y - y * b.x;}double operator*(const Point& b) const {return x * b.x + y * b.y;}double len() {  //长度return sqrt(x * x + y * y);
//        return hypot(x, y);}double len2() {  //长度的平方return x * x + y * y;}double dis(Point p) { //两点距离return hypot(x - p.x, y - p.y);  //hypot是计算斜边长用的
//        return sqrt((x - p.x) * (x - p.x) + (y - p.y) * (y - p.y));}double dis2(Point p) { //两点距离的平方return pow(x - p.x, 2.0) + pow(y - p.y, 2.0);}
};
struct Line {Point s, e;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}bool operator==(const Line& v) const {return (s == v.s) && (e == v.e);}void input() {s.input(); e.input();}int segcrossseg(Line v) { //两线段相交，2为规范相交，1为非规范相交int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));int d3 = dcmp((v.e - v.s) ^ (s - v.s));int d4 = dcmp((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;return (d1 == 0 && dcmp((v.s - s) * (v.s - e)) <= 0) ||(d2 == 0 && dcmp((v.e - s) * (v.e - e)) <= 0) ||(d3 == 0 && dcmp((s - v.s) * (s - v.e)) <= 0) ||(d4 == 0 && dcmp((e - v.s) * (e - v.e)) <= 0);}int linecrossseg(Line v) {//直线与线段，2为规范相交，1为非规范相交int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));if ((d1 ^ d2) == -2) return 2;return (d1 == 0 || d2 == 0);}
};
int n;
vector<int> x1(105, 0), x2(105, 0), yy(105, 0), y2(105, 0);
vector<Line> li;
vector<Point> in, out;
vector<Point> all;
Point treasure;
int main() {scanf("%d", &n);for (int i = 0; i < n; i++) {Point a, b;a.input(); b.input();if (dcmp(a.x - 100) == 0) y2[a.y]++;else if (dcmp(a.x - 0) == 0) yy[a.y]++;else if (dcmp(a.y - 100) == 0) x2[a.x]++;else if (dcmp(a.y - 0) == 0) x1[a.x]++;if (dcmp(b.x - 100) == 0) y2[b.y]++;else if (dcmp(b.x - 0) == 0) yy[b.y]++;else if (dcmp(b.y - 100) == 0) x2[b.x]++;else if (dcmp(b.y - 0) == 0) x1[b.x]++;li.push_back(Line(a, b));in.push_back((a + b) / 2);}li.push_back(Line(Point(0, 0), Point(0, 100)));li.push_back(Line(Point(0, 100), Point(100, 100)));li.push_back(Line(Point(100, 100), Point(100, 0)));li.push_back(Line(Point(0, 0), Point(100, 0)));treasure.input();all.push_back(treasure);int cnt1 = in.size();for (int i = 0; i < cnt1; i++)all.push_back(in[i]);x1[0]++; x1[100]++;x2[0]++; x2[100]++;yy[0]++; yy[100]++;y2[0]++; y2[100]++;int last = 0;for (int i = 1; i <= 100; i++) {if (x1[i] > 0) {double pos = (last + i) / 2.0;out.push_back(Point(pos, 0));last = i;}}last = 0;for (int i = 1; i <= 100; i++) {if (x2[i] > 0) {double pos = (last + i) / 2.0;out.push_back(Point(pos, 100.0));last = i;}}last = 0;for (int i = 1; i <= 100; i++) {if (yy[i] > 0) {double pos = (last + i) / 2.0;out.push_back(Point(0, pos));last = i;}}last = 0;for (int i = 1; i <= 100; i++) {if (y2[i] > 0) {double pos = (last + i) / 2.0;out.push_back(Point(100.0, pos));last = i;}}int cnt2 = out.size();for (int i = 0; i < (int)out.size(); i++) {all.push_back(out[i]);}int ans = INF;for (int i = 0; i < cnt2; i++) {Line tmp = Line(treasure, out[i]);int cnt = 0;for(int j = 0; j < (int)li.size(); j++) {int k = tmp.segcrossseg(li[j]);if (k > 0) cnt++;}ans = min(ans, cnt);}printf("Number of doors = %d\n", ans);return 0;
}

POJ 1696 Space Ant(极角排序)

题意： 给定 n n n 个点，要求你找出一条路径经过最多的点，路径不能交叉，路径始终只能是当前前进方向的左边。

思路： 不难发现，无论怎么给定点，都能够走完所有的点，因为可以画出一个凸包，凸包内的点始终在凸包前进方向的左侧，这样逆时针一圈一圈地走就能得到答案。

用极角排序，这里用叉积来作极角排序，具体实现看代码

每次选定上一个点作为基准对剩下的点进行极角排序，然后剩下的点中的第一个点就是下一个路径点。一直排序直到走完所有点。

注意题目要求从 y y y 坐标最小的点出发。。。。。。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}
struct Point{double x, y;int id;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const{return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const {return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const{return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const{return Point(x + b.x, y + b.y);}Point operator*(const double& k) const{return Point(x * k, y * k);}double operator*(const Point& b) const {return x * b.x + y * b.y;}Point operator/(const double& k) const{return Point(x / k, y / k);}double operator^(const Point& b) const{return x * b.y - y * b.x;}double dis(Point p) {return hypot(x - p.x, y - p.y);
//        return sqrt((*this - p) * (*this - p));}
};
Point p[105];
int s;
bool cmp(Point a, Point b) { //极角排序double tmp = (a - p[s]) ^ (b - p[s]);if (dcmp(tmp) == 0)return a.dis(p[s]) < b.dis(p[s]);else if (dcmp(tmp) < 0) //点b偏下return false;elsereturn true;
}
int main() {int t;scanf("%d", &t);while (t--) {int n;scanf("%d", &n);vector<Point> last;for (int i = 0, id; i < n; i++) {scanf("%d", &id);p[i + 1].input(), p[i + 1].id = id;last.push_back(p[i + 1]);if (i > 0 && (last[i].y < last[0].y || (last[i].y == last[0].y && last[i].x < last[0].x)))swap(last[0], last[i]);}
//        sort(last.begin(), last.end());vector<int> ans;do {if ((int)last.size() == 0) break;ans.push_back(last[0].id);s = last[0].id;last.erase(last.begin());sort(last.begin(), last.end(), cmp);} while (true);printf("%d", ans.size());for (int i = 0; i < (int)ans.size(); i++)printf(" %d", ans[i]);printf("\n");}return 0;
}

POJ 3347 Kadj Squares(扩大数据避免小数)

题意： 有若干个边长为整数的正方形，要将他们倾斜 45 45 45度放在第一象限中，底要接触 x x x 轴，并且尽可能的紧凑，求从上方观察，能看到哪些正方形。

思路： 先将每个正方形的边长放大 2 \sqrt2 2 倍，再投影下来就是整数了，避免的浮点数的精度问题。然后将所有正方形的区间范围表示出来，再对重合的部分根据两个正方形的边长大小比较进行修改。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;struct node{int l, r, len;node() {}node(int _l, int _r, int _len) {l = _l; r = _r;len = _len;}
};
int main() {int n;while (scanf("%d", &n) && n) {vector<node> a(n + 1);for (int i = 1; i <= n; i++) {a[i].l = a[i].r = 0;scanf("%d", &a[i].len);}for (int i = 1; i <= n; i++) {for (int j = 1; j < i; j++) { //找到左端点a[i].l = max(a[i].l, a[j].r - abs(a[i].len - a[j].len));}a[i].r = a[i].l + 2 * a[i].len; //确定所在区间}for (int i = 1; i <= n; i++) {for (int j = 1; j < i; j++) {if (a[j].r > a[i].l) { //前面正方形的右边界比后面正方形的左边界要大if (a[j].len > a[i].len) { //前面的正方形更大a[i].l = a[j].r;}else if (a[j].len < a[i].len) {a[j].r = a[i].l;}}}}int c = 0;for (int i = 1; i <= n; i++) {if (a[i].l < a[i].r) {if (c++ > 0) printf(" ");printf("%d", i);}}printf("\n");}return 0;
}

POJ 2826 An Easy Problem?!(大力出奇迹?)

题意： 就给两块木板，看能接到多少面积的雨水，要注意这里的雨是垂直下的

思路： 大力出奇迹，最吐血的是答案加一个 e p s eps eps 才能过。。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#include<stdlib.h>
#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-6;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}struct Point{double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const {return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}
//    bool operator<(const Point& b) const {//        return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;
//    }Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}double operator*(const Point& b) const {return x * b.x + y * b.y;}double operator^(const Point& b) const {return x * b.y - y * b.x;}double dis(Point b) {return sqrt((*this - b) * (*this - b));}
};struct Line{Point s, e;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}bool operator==(const Line& v) const {return (s == v.s) && (e == v.e);}void adjust() { //保证从下往上if (s.y > e.y) swap(s, e);}bool parallel(Line v) {return dcmp((e - s) ^ (v.e -v.s)) == 0;}int segcrossseg(Line v) {int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));int d3 = dcmp((v.e - v.s) ^ (s - v.s));int d4 = dcmp((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;return (d1 == 0 && dcmp((v.s - s) * (v.s - e)) <= 0) ||(d2 == 0 && dcmp((v.e - s) * (v.e - e)) <= 0) ||(d3 == 0 && dcmp((s - v.s) * (s - v.e)) <= 0) ||(d4 == 0 && dcmp((e - v.s) * (e - v.e)) <= 0);}int linecrossseg(Line v) {int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));if ((d1 ^ d2) == -2) return 2;return (d1 == 0 || d2 == 0);}Point crossPoint(Line v) {double a1 = (v.e - v.s) ^ (s - v.s);double a2 = (v.e - v.s) ^ (e - v.s);return Point((s.x * a2 - e.x * a1) / (a2 - a1), (s.y * a2 - e.y * a1) / (a2 - a1));}
};
int main() {int t;scanf("%d", &t);while (t--) {Point a, b, c, d;a.input(); b.input();c.input(); d.input();Line l1 = Line(a, b); l1.adjust();Line l2 = Line(c, d); l2.adjust();Line lx = Line(Point(0, 0), Point(1, 0));Line ly = Line(Point(0, 0), Point(0, 1));if (l1.parallel(lx) || l2.parallel(lx)) { //存在水平printf("0.00\n");continue;}
//        printf("2\n");if (l1.parallel(l2)) { //平行线printf("0.00\n");continue;}
//        printf("3\n");int tmp = l1.segcrossseg(l2); //判断相交情况if (tmp == 0) {  //不相交printf("0.00\n");continue;}
//        printf("4\n");Point p = l1.crossPoint(l2);Line lp = Line(p, Point(p.x, p.y + 1));int k = lp.linecrossseg(Line(l1.e, l2.e)); //判断是否在两边if (p == l1.e || p == l2.e) { //交点为某一条线段的偏上的点printf("0.00\n");continue;}
//        printf("5\n");if (k == 0) { //同一边if (((l1.e - p) ^ (l2.e - p)) > 0 && dcmp(l2.e.x - l1.e.x) >= 0) { //l2刚好遮住l1printf("0.00\n");continue;}if (((l1.e - p) ^ (l2.e - p)) < 0 && dcmp(l1.e.x - l2.e.x) >= 0) { //l1刚好遮住l2printf("0.00\n");continue;}}
//        printf("6\n");double ans = 0;Line lpp1 = Line(l1.e, Point(l1.e.x + 1, l1.e.y));Line lpp2 = Line(l2.e, Point(l2.e.x + 1, l2.e.y));int s1 = lpp1.linecrossseg(l2);int s2 = lpp2.linecrossseg(l1);if (s1 > 0) {Point o = l2.crossPoint(lpp1);ans = ((o - p) ^ (l1.e - p)) / 2.0;ans = fabs(ans);}else {Point o = l1.crossPoint(lpp2);ans = ((o - p) ^ (l2.e - p)) / 2.0;ans = fabs(ans);}printf("%.2f\n", ans + eps);}return 0;
}/*
6259 2664 8292 9080
1224 2972 9097 9680
5928.89
*/

POJ 3449 Geometric Shapes(非常恶心的读入)

题意： 给定 n n n 若干个平面封闭图形，判断相交的情况，边碰到也算相交。

有个知识点，已知正方形对角线的两顶点，求另外两个顶点

将对角线左旋 45 45 45度和右旋 45 45 45 度，再分别求投影点即可

思路： emmm，写的时候仔细一点，写好读入就基本OK了。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<string>
#include<algorithm>
#define fi first
#define se second
//#include<stdlib.h>
//#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const double eps = 1e-7;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}struct Point{double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const {return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const {return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}double operator*(const Point& b) const {return x * b.x + y * b.y;}double operator^(const Point& b) const {return x * b.y - y * b.x;}Point operator/(const double& k) const {return Point(x / k, y / k);}Point operator*(const double& k) const {return Point(x * k, y * k);}double dis(Point p) {return sqrt((*this - p) * (*this - p));}double len() {return sqrt(x * x + y * y);}double len2() {return x * x + y * y;}Point Ratate(Point p, double angle) {Point v = (*this) - p;double c = cos(angle), s = sin(angle);return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);}
};struct Line{Point s, e;int id;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}void output() {printf("%d:\n", id);s.output(); e.output();}void adjust() {if (s.y > e.y) swap(s, e);}double length() {return s.dis(e);}bool parallel(Line v) {return dcmp((e - s) ^ (v.e - v.s)) == 0;}Point lineprog(Point p) {return s + ( ((e - s) * ((e - s) * (p - s))) / ((e - s).len2()));}double disPointToLine(Point p) {return fabs((p - s) ^ (e - s) / length());}double disPointToSeg(Point p) {if (dcmp((p - s) * (e - s)) < 0 || dcmp((p - e) * (s - e)) < 0)return min(p.dis(s), p.dis(e));return disPointToLine(p);}int segcrossseg(Line v) {int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));int d3 = dcmp((v.e - v.s) ^ (s - v.s));int d4 = dcmp((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;return (d1 == 0 && dcmp((v.s - s) * (v.s - e)) <= 0) ||(d2 == 0 && dcmp((v.e - s) * (v.e - e)) <= 0) ||(d3 == 0 && dcmp((s - v.s) * (s - v.e)) <= 0) ||(d4 == 0 && dcmp((e - v.s) * (e - v.e)) <= 0);}Point crossPoint(Line v) {double a1 = (v.e - v.s) ^ (s - v.s);double a2 = (v.e - v.s) ^ (e - v.s);return Point((s.x * a2 - e.x * a1) / (a2 - a1), (s.y * a2 - e.y * a1) / (a2 - a1));}double dissegtoseg(Line v) {return min(min(disPointToSeg(v.s), disPointToSeg(v.e)),min(v.disPointToSeg(s), v.disPointToSeg(e)));}
};vector<vector<int>> g(30, vector<int>(30));
vector<Line> li;
vector<int> cl(30, 0);
vector<int> vis(30, 0);void add_edge(int u, int v) {if (!g[u][v]) {cl[u]++;cl[v]++;}g[u][v] = g[v][u] = 1;
}void getsquare(Point a, Point b, Point& b1, Point& b2) {Point c = b.Ratate(a, -pi / 4.0);Point d = b.Ratate(a, pi / 4.0);Line ac = Line(a, c);Line ad = Line(a, d);b1 = ac.lineprog(b);b2 = ad.lineprog(b);
}void getRectangle(Point a, Point b, Point c, Point& d) {//保证第二个点为直角点Point o = (a + c) / 2;d = Point(2 * o.x - b.x, 2 * o.y - b.y);
}Point change(string str) {int num[2] = {0};int c = 0;int f = 1;for (int i = 0; i < str.size(); i++) {if (str[i] >= '0' && str[i] <= '9') {num[c] = num[c] * 10 + str[i] - '0';}else if (str[i] == ',') {num[c] *= f;f = 1;c++;}else if (str[i] == '-') f = -1;else if (str[i] == ')') {num[c] *= f;}}return Point(num[0], num[1]);
}void show(vector<Point> p) {for (int i = 0; i < (int)p.size(); i++)p[i].output();
}void solve() {for (int i = 0; i < 26; i++) {if (vis[i]) {if (cl[i] == 0) {printf("%c has no intersections\n", 'A' + i);}else if (cl[i] == 1) {printf("%c intersects with ", 'A' + i);for (int j = 0; j < 26; j++) {if (g[i][j])printf("%c\n", 'A' + j);}}else if (cl[i] == 2) {printf("%c intersects with ", 'A' + i);int f = 0;for (int j = 0; j < 26; j++) {if (g[i][j]) {if (f > 0) printf(" and ");printf("%c", 'A' + j);if (f > 0) printf("\n");f++;}}}else {printf("%c intersects with ", 'A' + i);int f = 1;for (int j = 0; j < 26; j++) {if (g[i][j]) {if (f == cl[i]) {printf("and %c\n", 'A' + j);}else {printf("%c, ", 'A' + j);}f++;}}}}}
}int main() {string s;while (cin >> s) {if (s == ".") break;else if (s == "-") {//            for (int i = 0; i < (int)li.size(); i++) {//                li[i].output();
//            }for (int i = 0; i < (int)li.size(); i++) {for (int j = i + 1; j < (int)li.size(); j++) {if (li[i].id == li[j].id) continue;int s = li[i].segcrossseg(li[j]);if (s > 0) add_edge(li[i].id, li[j].id);}}solve();printf("\n");for (int i = 0; i < 30; i++) vis[i] = 0;for (int i = 0; i < 30; i++) {vis[i] = 0;cl[i] = 0;for (int j = 0; j < 30; j++) {g[i][j] = 0;}}li.clear();}else {int id = s[0] - 'A';vis[id] = 1;string kind;cin >> kind;string pt;if (kind == "square") {vector<Point> p(4);cin >> pt; p[0] = change(pt);cin >> pt; p[2] = change(pt);getsquare(p[0], p[2], p[1], p[3]);
//                show(p);for (int i = 0; i < 4; i++) {Line line = Line(p[i], p[(i + 1) % 4]);line.id = id;li.push_back(line);}}else if (kind == "rectangle") {vector<Point> p(4);for (int i = 0; i < 3; i++) {cin >> pt;p[i] = change(pt);}getRectangle(p[0], p[1], p[2], p[3]);
//                show(p);for (int i = 0; i < 4; i++) {Line line = Line(p[i], p[(i + 1) % 4]);line.id = id;li.push_back(line);}}else if (kind == "line") {vector<Point> p(2);for (int i = 0; i < 2; i++) {cin >> pt;p[i] = change(pt);}
//                show(p);Line line = Line(p[0], p[1]);line.id = id;li.push_back(line);}else if (kind == "triangle") {vector<Point> p(3);for (int i = 0; i < 3; i++) {cin >> pt;p[i] = change(pt);}
//                show(p);for (int i = 0; i < 3; i++) {Line line = Line(p[i], p[(i + 1) % 3]);line.id = id;li.push_back(line);}}else if (kind == "polygon") {int cnt;cin >> cnt;vector<Point> p(cnt);for (int i = 0; i < cnt; i++) {cin >> pt;p[i] = change(pt);}
//                show(p);for (int i = 0; i < cnt; i++) {Line line = Line(p[i], p[(i + 1) % cnt]);line.id = id;li.push_back(line);}}}}return 0;
}

POJ 1584 A Round Peg in a Ground Hole(凸包+点线圆的相对关系)

题意： 给定一个多边形，判断是否为凸多边形，以及圆是否在该凸多边形内部

思路： 直接先对所给定的点求一个凸包，看凸包的节点数 (注意：这里的凸包要求不是严格的，可以存在点的角度为180)，或者对每一对相邻的边进行叉积**(叉积注意方向要一致)** ，如果有正有负就说明不是凸包。接下来先判断圆心是否在凸包内，可以用面积法求。最后，因为圆心已经在凸包内，只需要求一下圆心到各线段（可以当做直线来求，以为保证了圆心已经在凸包内了）的距离与半径的关系即可

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<string>
#include<algorithm>
#define fi first
#define se second
//#include<stdlib.h>
//#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const int N = 1e3 + 10;const double eps = 1e-7;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}struct Point{double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const {return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const {return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}double operator*(const Point& b) const {return x * b.x + y * b.y;}double operator^(const Point& b) const {return x * b.y - y * b.x;}Point operator/(const double& k) const {return Point(x / k, y / k);}Point operator*(const double& k) const {return Point(x * k, y * k);}double dis(Point p) {return sqrt((*this - p) * (*this - p));}double len() {return sqrt(x * x + y * y);}double len2() {return x * x + y * y;}Point Ratate(Point p, double angle) {Point v = (*this) - p;double c = cos(angle), s = sin(angle);return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);}
}p[N], ch[N];struct Line{Point s, e;int id;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}void output() {printf("%d:\n", id);s.output(); e.output();}void adjust() {if (s.y > e.y) swap(s, e);}double length() {return s.dis(e);}bool parallel(Line v) {return dcmp((e - s) ^ (v.e - v.s)) == 0;}Point lineprog(Point p) {return s + ( ((e - s) * ((e - s) * (p - s))) / ((e - s).len2()));}double disPointToLine(Point p) {return fabs((p - s) ^ (e - s) / length());}double disPointToSeg(Point p) {  //点到线段的距离if (dcmp((p - s) * (e - s)) < 0 || dcmp((p - e) * (s - e)) < 0)return min(p.dis(s), p.dis(e));return disPointToLine(p);}int segcrossseg(Line v) {int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));int d3 = dcmp((v.e - v.s) ^ (s - v.s));int d4 = dcmp((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;return (d1 == 0 && dcmp((v.s - s) * (v.s - e)) <= 0) ||(d2 == 0 && dcmp((v.e - s) * (v.e - e)) <= 0) ||(d3 == 0 && dcmp((s - v.s) * (s - v.e)) <= 0) ||(d4 == 0 && dcmp((e - v.s) * (e - v.e)) <= 0);}Point crossPoint(Line v) {double a1 = (v.e - v.s) ^ (s - v.s);double a2 = (v.e - v.s) ^ (e - v.s);return Point((s.x * a2 - e.x * a1) / (a2 - a1), (s.y * a2 - e.y * a1) / (a2 - a1));}double dissegtoseg(Line v) {return min(min(disPointToSeg(v.s), disPointToSeg(v.e)),min(v.disPointToSeg(s), v.disPointToSeg(e)));}
};struct Circle{Point o;double r;Circle() {}Circle(Point _o, double _r) {o = _o;r = _r;}
};int andrew(Point p[], Point ch[], int n) {sort(p + 1, p + n + 1);int top = 0;for (int i = 1; i <= n; i++) {while ((top > 1) & (((ch[top] - ch[top - 1]) ^ (p[i] - ch[top])) < 0))--top;ch[++top] = p[i];}int tmp = top;for (int i = n - 1; i; --i) {while ((top > tmp) && (((ch[top] - ch[top - 1]) ^ (p[i] - ch[top])) < 0))--top;ch[++top] = p[i];}if (n > 1) top--;return top;
}bool is_in(Point ch[], int n, Point o) {double S = 0;for (int i = 2; i < n; i++) {double tmp = (ch[i - 1] - ch[0]) ^ (ch[i] - ch[0]);tmp = fabs(tmp);S += tmp;}double s = 0;for (int i = 1; i <= n; i++) {double tmp = (ch[i - 1] - o) ^ (ch[i] - o);tmp = fabs(tmp);s += tmp;}
//    printf("%.2f %.2f\n", S, s);return dcmp(S - s) == 0;
}bool is_ok(Point ch[], int n, Circle cir) {for (int i = 1; i <= n; i++) {Line li = Line(ch[i], ch[i - 1]);double dis = li.disPointToLine(cir.o);if (dcmp(cir.r - dis) > 0) return false;}return true;
}
int main() {int n;while (scanf("%d", &n)) {if (n < 3) break;Circle cir;scanf("%lf%lf%lf", &cir.r, &cir.o.x, &cir.o.y);for (int i = 1; i <= n; i++) {p[i].input();}int cnt = andrew(p, ch, n);ch[0] = ch[cnt];if (cnt < n) {printf("HOLE IS ILL-FORMED\n");continue;}
//        for (int i = 0; i < cnt; i++) {//            ch[i].output();
//        }if (is_in(ch, cnt, cir.o) && is_ok(ch, cnt, cir)) {printf("PEG WILL FIT\n");}else {printf("PEG WILL NOT FIT\n");}}return 0;
}/*
6 1 2 2
1 3
1 2
1 1
3 1
4 2
2 4
*/

POJ 2074 Line of Sight(平面视线问题)

题意： 给一条马路，一座房子，以及若干障碍物（所有的东西都是平行于 x x x 轴的）要求在马路上能看到整个房子的最长路段

思路： 先将在房子旁边和房子上面的障碍物都过滤掉，在剩下的障碍物中去掉末尾相同但起始点更小的障碍物，然后按末尾点为关键字进行升序排序，排序后对相邻的两个障碍物，进行计算，求在马路中能看到完整房子的路段长度。对开始和结束的端点在进行特判，最后取最大值就是答案。

代码：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cmath>
#include<map>
#include<cstring>
#include<string>
#include<algorithm>
#define fi first
#define se second
//#include<stdlib.h>
//#include <time.h>
//srand((unsigned)time(NULL));
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
using namespace std;const int N = 1e3 + 10;const double eps = 1e-7;
const double pi = acos(-1.0);
int dcmp(double x) {if (fabs(x) < eps) return 0;return x > 0 ? 1 : -1;
}struct Point{double x, y;Point() {}Point(double _x, double _y) {x = _x; y = _y;}void input() {scanf("%lf%lf", &x, &y);}void output() {printf("%.2f %.2f\n", x, y);}bool operator==(const Point& b) const {return dcmp(x - b.x) == 0 && dcmp(y - b.y) == 0;}bool operator<(const Point& b) const {return dcmp(x - b.x) == 0 ? dcmp(y - b.y) < 0 : x < b.x;}Point operator-(const Point& b) const {return Point(x - b.x, y - b.y);}Point operator+(const Point& b) const {return Point(x + b.x, y + b.y);}double operator*(const Point& b) const {return x * b.x + y * b.y;}double operator^(const Point& b) const {return x * b.y - y * b.x;}Point operator/(const double& k) const {return Point(x / k, y / k);}Point operator*(const double& k) const {return Point(x * k, y * k);}double dis(Point p) {return sqrt((*this - p) * (*this - p));}double len() {return sqrt(x * x + y * y);}double len2() {return x * x + y * y;}Point Ratate(Point p, double angle) {Point v = (*this) - p;double c = cos(angle), s = sin(angle);return Point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);}
};struct Line{Point s, e;int id;Line() {}Line(Point _s, Point _e) {s = _s; e = _e;}void output() {printf("%d:\n", id);s.output(); e.output();}void adjust() {if (s.y > e.y) swap(s, e);}double length() {return s.dis(e);}bool parallel(Line v) {return dcmp((e - s) ^ (v.e - v.s)) == 0;}Point lineprog(Point p) {return s + ( ((e - s) * ((e - s) * (p - s))) / ((e - s).len2()));}double disPointToLine(Point p) {return fabs((p - s) ^ (e - s) / length());}double disPointToSeg(Point p) {  //点到线段的距离if (dcmp((p - s) * (e - s)) < 0 || dcmp((p - e) * (s - e)) < 0)return min(p.dis(s), p.dis(e));return disPointToLine(p);}int segcrossseg(Line v) {int d1 = dcmp((e - s) ^ (v.s - s));int d2 = dcmp((e - s) ^ (v.e - s));int d3 = dcmp((v.e - v.s) ^ (s - v.s));int d4 = dcmp((v.e - v.s) ^ (e - v.s));if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;return (d1 == 0 && dcmp((v.s - s) * (v.s - e)) <= 0) ||(d2 == 0 && dcmp((v.e - s) * (v.e - e)) <= 0) ||(d3 == 0 && dcmp((s - v.s) * (s - v.e)) <= 0) ||(d4 == 0 && dcmp((e - v.s) * (e - v.e)) <= 0);}Point crossPoint(Line v) {double a1 = (v.e - v.s) ^ (s - v.s);double a2 = (v.e - v.s) ^ (e - v.s);return Point((s.x * a2 - e.x * a1) / (a2 - a1), (s.y * a2 - e.y * a1) / (a2 - a1));}double dissegtoseg(Line v) {return min(min(disPointToSeg(v.s), disPointToSeg(v.e)),min(v.disPointToSeg(s), v.disPointToSeg(e)));}
};bool cmp(Line v1, Line v2) {if (v1.e.x != v2.e.x) return v1.e.x < v2.e.x;return v1.s.x > v2.s.x;
}bool cmp1(Line v1, Line v2) { //最左上if (v1.s.x == v2.s.x) return v1.s.y > v2.s.y;return v1.s.x < v2.s.x;
}
bool cmp2(Line v1, Line v2) { //最右上if (v1.e.x == v2.e.x) return v1.e.y > v2.e.y;return v1.e.x > v2.e.x;
}double getdis(Line line, Line house, Line a, Line b) {Point p1 = a.e; Point p2 = b.s;Line l1 = Line(house.s, p1);Line l2 = Line(house.e, p2);Point c1 = l1.crossPoint(line);Point c2 = l2.crossPoint(line);
//     c1.output();
//     c2.output();double l = c1.x, r = c2.x;l = max(l, line.s.x); l = min(l, line.e.x);r = max(r, line.s.x); r = min(r, line.e.x);if (dcmp(l - r) >= 0) return 0;return r - l;
}int main() {double x1, x2, y;while (scanf("%lf%lf%lf", &x1, &x2, &y)) {if (x1 == 0 && x2 == 0 && y == 0) break;Line house = Line(Point(x1, y), Point(x2, y));scanf("%lf%lf%lf", &x1, &x2, &y);Line line = Line(Point(x1, y), Point(x2, y));int n;scanf("%d", &n);vector<Line> li;for (int i = 1; i <= n; i++) {scanf("%lf%lf%lf", &x1, &x2, &y);if (dcmp(y - house.e.y) >= 0) continue;li.push_back(Line(Point(x1, y), Point(x2, y)));}int cnt = li.size();if (cnt == 0) {printf("%.2f\n", line.e.x - line.s.x);continue;}vector<Line> nl;sort(li.begin(), li.end(), cmp);nl.push_back(li[0]);for (int i = 1; i < (int)li.size(); i++) {if (dcmp(li[i].e.x - li[i - 1].e.x) == 0) continue;nl.push_back(li[i]);}cnt = nl.size();double ans = 0;for (int i = 1; i < cnt; i++) {double tmp = getdis(line, house, nl[i - 1], nl[i]);
//            printf("%.2f\n", tmp);ans = max(ans, tmp);}//最左端特判sort(nl.begin(), nl.end(), cmp1);Line lp = Line(house.e, nl[0].s);Point cross = lp.crossPoint(line);ans = max(ans, cross.x - line.s.x);
//        printf("最左端: %.2f\n", cross.x - line.s.x);//最右端特判sort(nl.begin(), nl.end(), cmp2);lp = Line(house.s, nl[0].e);cross = lp.crossPoint(line);ans = max(ans, line.e.x - cross.x);
//        printf("最右端: %.2f\n", line.e.x - cross.x);if (dcmp(ans) == 0) printf("No View\n");else printf("%.2f\n", ans);}return 0;
}