2021牛客寒假算法基础集训营1-E-三棱锥之刻-(计算几何)
2024-06-13 05:44:20
题目链接：点击进入
题目
思路
求相交内表面的面积，分三种情况：
1、圆与三棱锥没交上；
2、圆与三棱锥完全交上（就是三棱锥的内表面全都可以染到）
3、圆与三棱锥部分相交
剩下的就是计算了，算公式，贴模板（我偷了刘爷的模板，不得不说真香）
代码
#include<iostream>
#include<cstdio>
#include<cmath>
#include<map>
#include<cstring>
#include<vector>
#include<algorithm>
#define inf 0x3f3f3f3f
//#define int long long
using namespace std;
typedef long long ll;
//const int maxn=1e7+10;
const int mod=1e9+7;
//const double eps=1e-3;
const double pai=acos(-1);
// `计算几何模板`
const double eps = 1e-8;
const double pi = acos(-1.0);
const int maxp = 1010;
//`Compares a double to zero`
int sgn(double x){if(fabs(x) < eps)return 0;if(x < 0)return -1;else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}
/** Point* Point() - Empty constructor* Point(double _x,double _y) - constructor* input() - double input* output() - %.2f output* operator == - compares x and y* operator < - compares first by x, then by y* operator - - return new Point after subtracting curresponging x and y* operator ^ - cross product of 2d points* operator * - dot product* len() - gives length from origin* len2() - gives square of length from origin* distance(Point p) - gives distance from p* operator + Point b - returns new Point after adding curresponging x and y* operator * double k - returns new Point after multiplieing x and y by k* operator / double k - returns new Point after divideing x and y by k* rad(Point a,Point b)- returns the angle of Point a and Point b from this Point* trunc(double r) - return Point that if truncated the distance from center to r* rotleft() - returns 90 degree ccw rotated point* rotright() - returns 90 degree cw rotated point* rotate(Point p,double angle) - returns Point after rotateing the Point centering at p by angle radian ccw*/
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 == (Point b)const{return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;}bool operator < (Point b)const{return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;}Point operator -(const Point &b)const{return Point(x-b.x,y-b.y);}//叉积double operator ^(const Point &b)const{return x*b.y - y*b.x;}//点积double operator *(const Point &b)const{return x*b.x + y*b.y;}//返回长度double len(){return hypot(x,y);//库函数}//返回长度的平方double len2(){return x*x + y*y;}//返回两点的距离double distance(Point p){return hypot(x-p.x,y-p.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);}Point operator /(const double &k)const{return Point(x/k,y/k);}//`计算pa 和 pb 的夹角`//`就是求这个点看a,b 所成的夹角`//`测试 LightOJ1203`double rad(Point a,Point b){Point p = *this;return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));}//`化为长度为r的向量`Point trunc(double r){double l = len();if(!sgn(l))return *this;r /= l;return Point(x*r,y*r);}//`逆时针旋转90度`Point rotleft(){return Point(-y,x);}//`顺时针旋转90度`Point rotright(){return Point(y,-x);}//`绕着p点逆时针旋转angle`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);}
};
/** Stores two points* Line() - Empty constructor* Line(Point _s,Point _e) - Line through _s and _e* operator == - checks if two points are same* Line(Point p,double angle) - one end p , another end at angle degree* Line(double a,double b,double c) - Line of equation ax + by + c = 0* input() - inputs s and e* adjust() - orders in such a way that s < e* length() - distance of se* angle() - return 0 <= angle < pi* relation(Point p) - 3 if point is on line* 1 if point on the left of line* 2 if point on the right of line* pointonseg(double p) - return true if point on segment* parallel(Line v) - return true if they are parallel* segcrossseg(Line v) - returns 0 if does not intersect* returns 1 if non-standard intersection* returns 2 if intersects* linecrossseg(Line v) - line and seg* linecrossline(Line v) - 0 if parallel* 1 if coincides* 2 if intersects* crosspoint(Line v) - returns intersection point* dispointtoline(Point p) - distance from point p to the line* dispointtoseg(Point p) - distance from p to the segment* dissegtoseg(Line v) - distance of two segment* lineprog(Point p) - returns projected point p on se line* symmetrypoint(Point p) - returns reflection point of p over se**/
struct Line{Point s,e;Line(){}Line(Point _s,Point _e){s = _s;e = _e;}bool operator ==(Line v){return (s == v.s)&&(e == v.e);}//`根据一个点和倾斜角angle确定直线,0<=angle<pi`Line(Point p,double angle){s = p;if(sgn(angle-pi/2) == 0){e = (s + Point(0,1));}else{e = (s + Point(1,tan(angle)));}}//ax+by+c=0Line(double a,double b,double c){if(sgn(a) == 0){s = Point(0,-c/b);e = Point(1,-c/b);}else if(sgn(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);}//求线段长度double length(){return s.distance(e);}//`返回直线倾斜角 0<=angle<pi`double angle(){double k = atan2(e.y-s.y,e.x-s.x);if(sgn(k) < 0)k += pi;if(sgn(k-pi) == 0)k -= pi;return k;}//`点和直线关系`//`1 在左侧`//`2 在右侧`//`3 在直线上`int relation(Point p){int c = sgn((p-s)^(e-s));if(c < 0)return 1;else if(c > 0)return 2;else return 3;}// 点在线段上的判断bool pointonseg(Point p){return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;}//`两向量平行(对应直线平行或重合)`bool parallel(Line v){return sgn((e-s)^(v.e-v.s)) == 0;}//`两线段相交判断`//`2 规范相交`//`1 非规范相交`//`0 不相交`int segcrossseg(Line v){int d1 = sgn((e-s)^(v.s-s));int d2 = sgn((e-s)^(v.e-s));int d3 = sgn((v.e-v.s)^(s-v.s));int d4 = sgn((v.e-v.s)^(e-v.s));if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||(d4==0 && sgn((e-v.s)*(e-v.e))<=0);}//`直线和线段相交判断`//`-*this line -v seg`//`2 规范相交`//`1 非规范相交`//`0 不相交`int linecrossseg(Line v){int d1 = sgn((e-s)^(v.s-s));int d2 = sgn((e-s)^(v.e-s));if((d1^d2)==-2) return 2;return (d1==0||d2==0);}//`两直线关系`//`0 平行`//`1 重合`//`2 相交`int linecrossline(Line v){if((*this).parallel(v))return v.relation(s)==3;return 2;}//`求两直线的交点`//`要保证两直线不平行或重合`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));}//点到直线的距离double dispointtoline(Point p){return fabs((p-s)^(e-s))/length();}//点到线段的距离double dispointtoseg(Point p){if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)return min(p.distance(s),p.distance(e));return dispointtoline(p);}//`返回线段到线段的距离`//`前提是两线段不相交，相交距离就是0了`double dissegtoseg(Line v){return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));}//`返回点p在直线上的投影`Point lineprog(Point p){return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );}//`返回点p关于直线的对称点`Point symmetrypoint(Point p){Point q = lineprog(p);return Point(2*q.x-p.x,2*q.y-p.y);}
};
//圆
struct circle{Point p;//圆心double r;//半径circle(){}circle(Point _p,double _r){p = _p;r = _r;}circle(double x,double y,double _r){p = Point(x,y);r = _r;}//`三角形的外接圆`//`需要Point的+ / rotate() 以及Line的crosspoint()`//`利用两条边的中垂线得到圆心`//`测试：UVA12304`circle(Point a,Point b,Point c){Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));p = u.crosspoint(v);r = p.distance(a);}//`三角形的内切圆`//`参数bool t没有作用，只是为了和上面外接圆函数区别`//`测试：UVA12304`circle(Point a,Point b,Point c,bool t){Line u,v;double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.x);u.s = a;u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2));v.s = b;m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x);v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2));p = u.crosspoint(v);r = Line(a,b).dispointtoseg(p);}//输入void input(){p.input();scanf("%lf",&r);}//输出void output(){printf("%.2lf %.2lf %.2lf\n",p.x,p.y,r);}bool operator == (circle v){return (p==v.p) && sgn(r-v.r)==0;}bool operator < (circle v)const{return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0));}//面积double area(){return pi*r*r;}//周长double circumference(){return 2*pi*r;}//`点和圆的关系`//`0 圆外`//`1 圆上`//`2 圆内`int relation(Point b){double dst = b.distance(p);if(sgn(dst-r) < 0)return 2;else if(sgn(dst-r)==0)return 1;return 0;}//`线段和圆的关系`//`比较的是圆心到线段的距离和半径的关系`int relationseg(Line v){double dst = v.dispointtoseg(p);if(sgn(dst-r) < 0)return 2;else if(sgn(dst-r) == 0)return 1;return 0;}//`直线和圆的关系`//`比较的是圆心到直线的距离和半径的关系`int relationline(Line v){double dst = v.dispointtoline(p);if(sgn(dst-r) < 0)return 2;else if(sgn(dst-r) == 0)return 1;return 0;}//`两圆的关系`//`5 相离`//`4 外切`//`3 相交`//`2 内切`//`1 内含`//`需要Point的distance`//`测试：UVA12304`int relationcircle(circle v){double d = p.distance(v.p);if(sgn(d-r-v.r) > 0)return 5;if(sgn(d-r-v.r) == 0)return 4;double l = fabs(r-v.r);if(sgn(d-r-v.r)<0 && sgn(d-l)>0)return 3;if(sgn(d-l)==0)return 2;if(sgn(d-l)<0)return 1;return 0;}//`求两个圆的交点，返回0表示没有交点，返回1是一个交点，2是两个交点`//`需要relationcircle`//`测试：UVA12304`int pointcrosscircle(circle v,Point &p1,Point &p2){int rel = relationcircle(v);if(rel == 1 || rel == 5)return 0;double d = p.distance(v.p);double l = (d*d+r*r-v.r*v.r)/(2*d);double h = sqrt(r*r-l*l);Point tmp = p + (v.p-p).trunc(l);p1 = tmp + ((v.p-p).rotleft().trunc(h));p2 = tmp + ((v.p-p).rotright().trunc(h));if(rel == 2 || rel == 4)return 1;return 2;}//`求直线和圆的交点，返回交点个数`int pointcrossline(Line v,Point &p1,Point &p2){if(!(*this).relationline(v))return 0;Point a = v.lineprog(p);double d = v.dispointtoline(p);d = sqrt(r*r-d*d);if(sgn(d) == 0){p1 = a;p2 = a;return 1;}p1 = a + (v.e-v.s).trunc(d);p2 = a - (v.e-v.s).trunc(d);return 2;}//`得到过a,b两点，半径为r1的两个圆`int gercircle(Point a,Point b,double r1,circle &c1,circle &c2){circle x(a,r1),y(b,r1);int t = x.pointcrosscircle(y,c1.p,c2.p);if(!t)return 0;c1.r = c2.r = r1;return t;}//`得到与直线u相切，过点q,半径为r1的圆`//`测试：UVA12304`int getcircle(Line u,Point q,double r1,circle &c1,circle &c2){double dis = u.dispointtoline(q);if(sgn(dis-r1*2)>0)return 0;if(sgn(dis) == 0){c1.p = q + ((u.e-u.s).rotleft().trunc(r1));c2.p = q + ((u.e-u.s).rotright().trunc(r1));c1.r = c2.r = r1;return 2;}Line u1 = Line((u.s + (u.e-u.s).rotleft().trunc(r1)),(u.e + (u.e-u.s).rotleft().trunc(r1)));Line u2 = Line((u.s + (u.e-u.s).rotright().trunc(r1)),(u.e + (u.e-u.s).rotright().trunc(r1)));circle cc = circle(q,r1);Point p1,p2;if(!cc.pointcrossline(u1,p1,p2))cc.pointcrossline(u2,p1,p2);c1 = circle(p1,r1);if(p1 == p2){c2 = c1;return 1;}c2 = circle(p2,r1);return 2;}//`同时与直线u,v相切，半径为r1的圆`//`测试：UVA12304`int getcircle(Line u,Line v,double r1,circle &c1,circle &c2,circle &c3,circle &c4){if(u.parallel(v))return 0;//两直线平行Line u1 = Line(u.s + (u.e-u.s).rotleft().trunc(r1),u.e + (u.e-u.s).rotleft().trunc(r1));Line u2 = Line(u.s + (u.e-u.s).rotright().trunc(r1),u.e + (u.e-u.s).rotright().trunc(r1));Line v1 = Line(v.s + (v.e-v.s).rotleft().trunc(r1),v.e + (v.e-v.s).rotleft().trunc(r1));Line v2 = Line(v.s + (v.e-v.s).rotright().trunc(r1),v.e + (v.e-v.s).rotright().trunc(r1));c1.r = c2.r = c3.r = c4.r = r1;c1.p = u1.crosspoint(v1);c2.p = u1.crosspoint(v2);c3.p = u2.crosspoint(v1);c4.p = u2.crosspoint(v2);return 4;}//`同时与不相交圆cx,cy相切，半径为r1的圆`//`测试：UVA12304`int getcircle(circle cx,circle cy,double r1,circle &c1,circle &c2){circle x(cx.p,r1+cx.r),y(cy.p,r1+cy.r);int t = x.pointcrosscircle(y,c1.p,c2.p);if(!t)return 0;c1.r = c2.r = r1;return t;}//`过一点作圆的切线(先判断点和圆的关系)`//`测试：UVA12304`int tangentline(Point q,Line &u,Line &v){int x = relation(q);if(x == 2)return 0;if(x == 1){u = Line(q,q + (q-p).rotleft());v = u;return 1;}double d = p.distance(q);double l = r*r/d;double h = sqrt(r*r-l*l);u = Line(q,p + ((q-p).trunc(l) + (q-p).rotleft().trunc(h)));v = Line(q,p + ((q-p).trunc(l) + (q-p).rotright().trunc(h)));return 2;}//`求两圆相交的面积`double areacircle(circle v){int rel = relationcircle(v);if(rel >= 4)return 0.0;if(rel <= 2)return min(area(),v.area());double d = p.distance(v.p);double hf = (r+v.r+d)/2.0;double ss = 2*sqrt(hf*(hf-r)*(hf-v.r)*(hf-d));double a1 = acos((r*r+d*d-v.r*v.r)/(2.0*r*d));a1 = a1*r*r;double a2 = acos((v.r*v.r+d*d-r*r)/(2.0*v.r*d));a2 = a2*v.r*v.r;return a1+a2-ss;}//`求两圆相交的面积(精度更高)(需要long double)`double areacircle2(circle v){double a=hypot(p.x-v.p.x,p.y-v.p.y),b=r,c=v.r;double s1=pi*r*r,s2=pi*v.r*v.r;if(sgn(a-b-c)>=0)return 0;if(sgn(a+min(b,c)-max(b,c))<=0)return min(s1,s2);else{double cta1=2*acos((a*a+b*b-c*c)/(2*a*b));double cta2=2*acos((a*a+c*c-b*b)/(2*a*c));return cta1/(2*pi)*s1-0.5*sin(cta1)*b*b+cta2/(2*pi)*s2-0.5*sin(cta2)*c*c;}}//`求圆和三角形pab的相交面积`//`测试：POJ3675 HDU3982 HDU2892`double areatriangle(Point a,Point b){if(sgn((p-a)^(p-b)) == 0)return 0.0;Point q[5];int len = 0;q[len++] = a;Line l(a,b);Point p1,p2;if(pointcrossline(l,q[1],q[2])==2){if(sgn((a-q[1])*(b-q[1]))<0)q[len++] = q[1];if(sgn((a-q[2])*(b-q[2]))<0)q[len++] = q[2];}q[len++] = b;if(len == 4 && sgn((q[0]-q[1])*(q[2]-q[1]))>0)swap(q[1],q[2]);double res = 0;for(int i = 0;i < len-1;i++){if(relation(q[i])==0||relation(q[i+1])==0){double arg = p.rad(q[i],q[i+1]);res += r*r*arg/2.0;}else{res += fabs((q[i]-p)^(q[i+1]-p))/2.0;}}return res;}
};/** n,p Line l for each side* input(int _n) - inputs _n size polygon* add(Point q) - adds a point at end of the list* getline() - populates line array* cmp - comparision in convex_hull order* norm() - sorting in convex_hull order* getconvex(polygon &convex) - returns convex hull in convex* Graham(polygon &convex) - returns convex hull in convex* isconvex() - checks if convex* relationpoint(Point q) - returns 3 if q is a vertex* 2 if on a side* 1 if inside* 0 if outside* convexcut(Line u,polygon &po) - left side of u in po* gercircumference() - returns side length* getarea() - returns area* getdir() - returns 0 for cw, 1 for ccw* getbarycentre() - returns barycenter**/
struct polygon{int n;Point p[maxp];Line l[maxp];void input(int _n){n = _n;for(int i = 0;i < n;i++)p[i].input();}void add(Point q){p[n++] = q;}void getline(){for(int i = 0;i < n;i++){l[i] = Line(p[i],p[(i+1)%n]);}}struct cmp{Point p;cmp(const Point &p0){p = p0;}bool operator()(const Point &aa,const Point &bb){Point a = aa, b = bb;int d = sgn((a-p)^(b-p));if(d == 0){return sgn(a.distance(p)-b.distance(p)) < 0;}return d > 0;}};//`进行极角排序`//`首先需要找到最左下角的点`//`需要重载号好Point的 < 操作符(min函数要用) `void norm(){Point mi = p[0];for(int i = 1;i < n;i++)mi = min(mi,p[i]);sort(p,p+n,cmp(mi));}//`得到凸包`//`得到的凸包里面的点编号是0$\sim$n-1的`//`两种凸包的方法`//`注意如果有影响，要特判下所有点共点，或者共线的特殊情况`//`测试 LightOJ1203 LightOJ1239`void getconvex(polygon &convex){sort(p,p+n);convex.n = n;for(int i = 0;i < min(n,2);i++){convex.p[i] = p[i];}if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判if(n <= 2)return;int &top = convex.n;top = 1;for(int i = 2;i < n;i++){while(top && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0)top--;convex.p[++top] = p[i];}int temp = top;convex.p[++top] = p[n-2];for(int i = n-3;i >= 0;i--){while(top != temp && sgn((convex.p[top]-p[i])^(convex.p[top-1]-p[i])) <= 0)top--;convex.p[++top] = p[i];}if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判convex.norm();//`原来得到的是顺时针的点，排序后逆时针`}//`得到凸包的另外一种方法`//`测试 LightOJ1203 LightOJ1239`void Graham(polygon &convex){norm();int &top = convex.n;top = 0;if(n == 1){top = 1;convex.p[0] = p[0];return;}if(n == 2){top = 2;convex.p[0] = p[0];convex.p[1] = p[1];if(convex.p[0] == convex.p[1])top--;return;}convex.p[0] = p[0];convex.p[1] = p[1];top = 2;for(int i = 2;i < n;i++){while( top > 1 && sgn((convex.p[top-1]-convex.p[top-2])^(p[i]-convex.p[top-2])) <= 0 )top--;convex.p[top++] = p[i];}if(convex.n == 2 && (convex.p[0] == convex.p[1]))convex.n--;//特判}//`判断是不是凸的`bool isconvex(){bool s[3];memset(s,false,sizeof(s));for(int i = 0;i < n;i++){int j = (i+1)%n;int k = (j+1)%n;s[sgn((p[j]-p[i])^(p[k]-p[i]))+1] = true;if(s[0] && s[2])return false;}return true;}//`判断点和任意多边形的关系`//` 3 点上`//` 2 边上`//` 1 内部`//` 0 外部`int relationpoint(Point q){for(int i = 0;i < n;i++){if(p[i] == q)return 3;}getline();for(int i = 0;i < n;i++){if(l[i].pointonseg(q))return 2;}int cnt = 0;for(int i = 0;i < n;i++){int j = (i+1)%n;int k = sgn((q-p[j])^(p[i]-p[j]));int u = sgn(p[i].y-q.y);int v = sgn(p[j].y-q.y);if(k > 0 && u < 0 && v >= 0)cnt++;if(k < 0 && v < 0 && u >= 0)cnt--;}return cnt != 0;}//`直线u切割凸多边形左侧`//`注意直线方向`//`测试：HDU3982`void convexcut(Line u,polygon &po){int &top = po.n;//注意引用top = 0;for(int i = 0;i < n;i++){int d1 = sgn((u.e-u.s)^(p[i]-u.s));int d2 = sgn((u.e-u.s)^(p[(i+1)%n]-u.s));if(d1 >= 0)po.p[top++] = p[i];if(d1*d2 < 0)po.p[top++] = u.crosspoint(Line(p[i],p[(i+1)%n]));}}//`得到周长`//`测试 LightOJ1239`double getcircumference(){double sum = 0;for(int i = 0;i < n;i++){sum += p[i].distance(p[(i+1)%n]);}return sum;}//`得到面积`double getarea(){double sum = 0;for(int i = 0;i < n;i++){sum += (p[i]^p[(i+1)%n]);}return fabs(sum)/2;}//`得到方向`//` 1 表示逆时针，0表示顺时针`bool getdir(){double sum = 0;for(int i = 0;i < n;i++)sum += (p[i]^p[(i+1)%n]);if(sgn(sum) > 0)return 1;return 0;}//`得到重心`Point getbarycentre(){Point ret(0,0);double area = 0;for(int i = 1;i < n-1;i++){double tmp = (p[i]-p[0])^(p[i+1]-p[0]);if(sgn(tmp) == 0)continue;area += tmp;ret.x += (p[0].x+p[i].x+p[i+1].x)/3*tmp;ret.y += (p[0].y+p[i].y+p[i+1].y)/3*tmp;}if(sgn(area)) ret = ret/area;return ret;}//`多边形和圆交的面积`//`测试：POJ3675 HDU3982 HDU2892`double areacircle(circle c){double ans = 0;for(int i = 0;i < n;i++){int j = (i+1)%n;if(sgn( (p[j]-c.p)^(p[i]-c.p) ) >= 0)ans += c.areatriangle(p[i],p[j]);else ans -= c.areatriangle(p[i],p[j]);}return fabs(ans);}//`多边形和圆关系`//` 2 圆完全在多边形内`//` 1 圆在多边形里面，碰到了多边形边界`//` 0 其它`int relationcircle(circle c){getline();int x = 2;if(relationpoint(c.p) != 1)return 0;//圆心不在内部for(int i = 0;i < n;i++){if(c.relationseg(l[i])==2)return 0;if(c.relationseg(l[i])==1)x = 1;}return x;}
}t;
//`AB X AC`
double cross(Point A,Point B,Point C){return (B-A)^(C-A);
}
//`AB*AC`
double dot(Point A,Point B,Point C){return (B-A)*(C-A);
}
//`最小矩形面积覆盖`
//` A 必须是凸包(而且是逆时针顺序)`
//` 测试 UVA 10173`
double minRectangleCover(polygon A){//`要特判A.n < 3的情况`if(A.n < 3)return 0.0;A.p[A.n] = A.p[0];double ans = -1;int r = 1, p = 1, q;for(int i = 0;i < A.n;i++){//`卡出离边A.p[i] - A.p[i+1]最远的点`while( sgn( cross(A.p[i],A.p[i+1],A.p[r+1]) - cross(A.p[i],A.p[i+1],A.p[r]) ) >= 0 )r = (r+1)%A.n;//`卡出A.p[i] - A.p[i+1]方向上正向n最远的点`while(sgn( dot(A.p[i],A.p[i+1],A.p[p+1]) - dot(A.p[i],A.p[i+1],A.p[p]) ) >= 0 )p = (p+1)%A.n;if(i == 0)q = p;//`卡出A.p[i] - A.p[i+1]方向上负向最远的点`while(sgn(dot(A.p[i],A.p[i+1],A.p[q+1]) - dot(A.p[i],A.p[i+1],A.p[q])) <= 0)q = (q+1)%A.n;double d = (A.p[i] - A.p[i+1]).len2();double tmp = cross(A.p[i],A.p[i+1],A.p[r]) *(dot(A.p[i],A.p[i+1],A.p[p]) - dot(A.p[i],A.p[i+1],A.p[q]))/d;if(ans < 0 || ans > tmp)ans = tmp;}return ans;
}//`直线切凸多边形`
//`多边形是逆时针的，在q1q2的左侧`
//`测试:HDU3982`
vector<Point> convexCut(const vector<Point> &ps,Point q1,Point q2){vector<Point>qs;int n = ps.size();for(int i = 0;i < n;i++){Point p1 = ps[i], p2 = ps[(i+1)%n];int d1 = sgn((q2-q1)^(p1-q1)), d2 = sgn((q2-q1)^(p2-q1));if(d1 >= 0)qs.push_back(p1);if(d1 * d2 < 0)qs.push_back(Line(p1,p2).crosspoint(Line(q1,q2)));}return qs;
}
//`半平面交`
//`测试 POJ3335 POJ1474 POJ1279`
//***************************
struct halfplane:public Line{double angle;halfplane(){}//`表示向量s->e逆时针(左侧)的半平面`halfplane(Point _s,Point _e){s = _s;e = _e;}halfplane(Line v){s = v.s;e = v.e;}void calcangle(){angle = atan2(e.y-s.y,e.x-s.x);}bool operator <(const halfplane &b)const{return angle < b.angle;}
};
struct halfplanes{int n;halfplane hp[maxp];Point p[maxp];int que[maxp];int st,ed;void push(halfplane tmp){hp[n++] = tmp;}//去重void unique(){int m = 1;for(int i = 1;i < n;i++){if(sgn(hp[i].angle-hp[i-1].angle) != 0)hp[m++] = hp[i];else if(sgn( (hp[m-1].e-hp[m-1].s)^(hp[i].s-hp[m-1].s) ) > 0)hp[m-1] = hp[i];}n = m;}bool halfplaneinsert(){for(int i = 0;i < n;i++)hp[i].calcangle();sort(hp,hp+n);unique();que[st=0] = 0;que[ed=1] = 1;p[1] = hp[0].crosspoint(hp[1]);for(int i = 2;i < n;i++){while(st<ed && sgn((hp[i].e-hp[i].s)^(p[ed]-hp[i].s))<0)ed--;while(st<ed && sgn((hp[i].e-hp[i].s)^(p[st+1]-hp[i].s))<0)st++;que[++ed] = i;if(hp[i].parallel(hp[que[ed-1]]))return false;p[ed]=hp[i].crosspoint(hp[que[ed-1]]);}while(st<ed && sgn((hp[que[st]].e-hp[que[st]].s)^(p[ed]-hp[que[st]].s))<0)ed--;while(st<ed && sgn((hp[que[ed]].e-hp[que[ed]].s)^(p[st+1]-hp[que[ed]].s))<0)st++;if(st+1>=ed)return false;return true;}//`得到最后半平面交得到的凸多边形`//`需要先调用halfplaneinsert() 且返回true`void getconvex(polygon &con){p[st] = hp[que[st]].crosspoint(hp[que[ed]]);con.n = ed-st+1;for(int j = st,i = 0;j <= ed;i++,j++)con.p[i] = p[j];}
};
//***************************const int maxn = 1010;
struct circles{circle c[maxn];double ans[maxn];//`ans[i]表示被覆盖了i次的面积`double pre[maxn];int n;circles(){}void add(circle cc){c[n++] = cc;}//`x包含在y中`bool inner(circle x,circle y){if(x.relationcircle(y) != 1)return 0;return sgn(x.r-y.r)<=0?1:0;}//圆的面积并去掉内含的圆void init_or(){bool mark[maxn] = {0};int i,j,k=0;for(i = 0;i < n;i++){for(j = 0;j < n;j++)if(i != j && !mark[j]){if( (c[i]==c[j])||inner(c[i],c[j]) )break;}if(j < n)mark[i] = 1;}for(i = 0;i < n;i++)if(!mark[i])c[k++] = c[i];n = k;}//`圆的面积交去掉内含的圆`void init_add(){int i,j,k;bool mark[maxn] = {0};for(i = 0;i < n;i++){for(j = 0;j < n;j++)if(i != j && !mark[j]){if( (c[i]==c[j])||inner(c[j],c[i]) )break;}if(j < n)mark[i] = 1;}for(i = 0;i < n;i++)if(!mark[i])c[k++] = c[i];n = k;}//`半径为r的圆，弧度为th对应的弓形的面积`double areaarc(double th,double r){return 0.5*r*r*(th-sin(th));}//`测试SPOJVCIRCLES SPOJCIRUT`//`SPOJVCIRCLES求n个圆并的面积，需要加上init\_or()去掉重复圆（否则WA）`//`SPOJCIRUT 是求被覆盖k次的面积，不能加init\_or()`//`对于求覆盖多少次面积的问题，不能解决相同圆，而且不能init\_or()`//`求多圆面积并，需要init\_or,其中一个目的就是去掉相同圆`void getarea(){memset(ans,0,sizeof(ans));vector<pair<double,int> >v;for(int i = 0;i < n;i++){v.clear();v.push_back(make_pair(-pi,1));v.push_back(make_pair(pi,-1));for(int j = 0;j < n;j++)if(i != j){Point q = (c[j].p - c[i].p);double ab = q.len(),ac = c[i].r, bc = c[j].r;if(sgn(ab+ac-bc)<=0){v.push_back(make_pair(-pi,1));v.push_back(make_pair(pi,-1));continue;}if(sgn(ab+bc-ac)<=0)continue;if(sgn(ab-ac-bc)>0)continue;double th = atan2(q.y,q.x), fai = acos((ac*ac+ab*ab-bc*bc)/(2.0*ac*ab));double a0 = th-fai;if(sgn(a0+pi)<0)a0+=2*pi;double a1 = th+fai;if(sgn(a1-pi)>0)a1-=2*pi;if(sgn(a0-a1)>0){v.push_back(make_pair(a0,1));v.push_back(make_pair(pi,-1));v.push_back(make_pair(-pi,1));v.push_back(make_pair(a1,-1));}else{v.push_back(make_pair(a0,1));v.push_back(make_pair(a1,-1));}}sort(v.begin(),v.end());int cur = 0;for(int j = 0;j < v.size();j++){if(cur && sgn(v[j].first-pre[cur])){ans[cur] += areaarc(v[j].first-pre[cur],c[i].r);ans[cur] += 0.5*(Point(c[i].p.x+c[i].r*cos(pre[cur]),c[i].p.y+c[i].r*sin(pre[cur]))^Point(c[i].p.x+c[i].r*cos(v[j].first),c[i].p.y+c[i].r*sin(v[j].first)));}cur += v[j].second;pre[cur] = v[j].first;}}for(int i = 1;i < n;i++)ans[i] -= ans[i+1];}
};
int n,k,a,r;
int main()
{// ios::sync_with_stdio(false);
// cin.tie(0);cout.tie(0);cin>>a>>r;double r0=1.0*a*sqrt(6)/12.0;double r1=sqrt(r*r-r0*r0);double r2=1.0*a*sqrt(3)/6.0;double r3=1.0*a*sqrt(3)/3.0;if(r1>=r3){double ans=4.0*a*a*sqrt(3)/4.0;printf("%.5lf",ans);}else if(r1<=r2){double ans=4.0*pai*r1*r1;printf("%.5lf",ans);}else{t.n=3;t.p[0].x=0,t.p[0].y=r3;t.p[1].x=sqrt(r3*r3-r2*r2),t.p[1].y=-r2;t.p[2].x=-sqrt(r3*r3-r2*r2),t.p[2].y=-r2;Point aa(0,0);circle ar(aa,r1);double ans=4.0*t.areacircle(ar);printf("%.5lf",ans);}return 0;
}2021牛客寒假算法基础集训营1-E-三棱锥之刻-(计算几何)相关推荐
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