poj 2187 凸包or旋转qia壳法

题意：

给n（50000）个点，求这些点与点之间距离最大的距离。

解析：

先求凸包然后暴力。

或者旋转卡壳大法。

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long longusing namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double ee = exp(1.0);/
struct Point
{double x, y;Point(double x = 0, double y = 0) : x(x), y(y) {}
};bool cmp(Point A, Point B)
{if (A.x == B.x)return A.y < B.y;return A.x < B.x;
}typedef Point Vector;Vector operator + (Vector A, Vector B)
{return Vector(A.x + B.x, A.y + B.y);
}Vector operator - (Point A, Point B)
{return Vector(A.x - B.x, A.y - B.y);
}Vector operator * (Vector A, double p)
{return Vector(A.x * p, A.y * p);
}Vector operator / (Vector A, double p)
{return Vector(A.x / p, A.y / p);
}//便于点排序
bool operator < (const Point& a, const Point& b)
{return a.x < b.x || (a.x == b.x && a.y < b.y);
}//用于判断相等
int dcmp(double x)
{if (fabs(x) < eps){return 0;}else{return x < 0 ? -1 : 1;}
}bool operator == (const Point& a, const Point& b)
{return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}//点积
double Dot(Vector A, Vector B)
{return A.x * B.x + A.y * B.y;
}//向量长度
double Length(Vector A)
{return sqrt(Dot(A, A));
}//向量夹角
double Angle(Vector A, Vector B)
{return acos(Dot(A, B) / Length(A) / Length(B));
}//叉集
double Cross(Vector A, Vector B)
{return A.x * B.y - A.y * B.x;
}//求两个向量相夹的面积
double Area2(Point A, Point B, Point C)
{return Cross(B - A, C - A);
}//旋转rad弧度
Vector Rotate(Vector A, double rad)
{return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}//向量的单位法线即向量左转90°
Vector Normal(Vector A)
{double L = Length(A);return Vector(-A.y / L, A.x / L);
}//求交点坐标
Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)
{Vector u = P - Q;double t = Cross(w, u) / Cross(v, w);return P + v * t;
}//点P到直线AB的距离
double DistanceToLine(Point P, Point A, Point B)
{Vector v1 = B - A;Vector v2 = P - A;return fabs(Cross(v1, v2)) / Length(v1);
}//点到直线的距离
double DistanceToSegment(Point P, Point A, Point B)
{if (A == B){return Length(P - A);}Vector v1 = B - A;Vector v2 = P - A;Vector v3 = P - B;if (dcmp(Dot(v1, v2)) < 0)return Length(v2);else if (dcmp(Dot(v1, v3)) > 0)return Length(v3);elsereturn fabs(Cross(v1, v2)) / Length(v1);
}//判断线段是否相交
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)
{double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1);double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}//线段是否在端点相交 + 线段相交的判定
bool OnSegment(Point p, Point a1, Point a2)
{return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}//传入顶点集 计算多边形的面积
double ConvexPolygonArea(Point* p, int n)
{double area = 0;for (int i = 1; i < n - 1; i++){area += Cross(p[i] - p[0], p[i + 1] - p[0]);}return area / 2.0;
}//° -> 弧度
double torad(double deg)
{return deg / 180 * pi;
}//求凸包 返回点个数 ch为凸包的点
int ConvexHull(Point* p, int n, Point* ch)
{sort(p, p + n);int m = 0;for (int i = 0; i < n; i++){while (m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0)m--;ch[m++] = p[i];}int k = m;for (int i = n - 2; i >= 0; i--){while (m > k && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0)m--;ch[m++] = p[i];}if (n > 1)m--;return m;
}Point readPoint()
{double x, y;scanf("%lf %lf", &x, &y);return Point(x, y);
}//
const int maxn = 50000 + 10;int n;
Point p[maxn << 1];
Point ch[maxn << 1];double dist(Point a, Point b)
{return Dot(a - b, a - b);
}void bruteForce()
{double ans = 0;int m = ConvexHull(p, n, ch);for (int i = 0; i < m; i++){for (int j = i + 1; j < m; j++){ans = max(ans, dist(ch[i], ch[j]));}}printf("%.0lf\n", ans);
}void rotatingCalipers()
{double ans = 0;int m = ConvexHull(p, n, ch);if (m == 2){printf("%.0lf\n", dist(ch[0], ch[1]));return;}int i = 0, j = 0;for (int k = 0; k < m; k++){if (ch[k] < ch[i])i = k;if (ch[j] < ch[k])j = k;}int si = i, sj = j;while (i != sj || j != si){ans = max(ans, dist(ch[i], ch[j]));if (Cross(ch[(i + 1) % m] - ch[i], ch[(j + 1) % m] - ch[j]) < 0){i = (i + 1) % m;}else{j = (j + 1) % m;}}printf("%.0lf\n", ans);
}int main()
{
#ifdef LOCALfreopen("in.txt", "r", stdin);
#endif // LOCALwhile (~scanf("%d", &n)){for (int i = 0; i < n; i++){p[i] = readPoint();}rotatingCalipers();}return 0;
}

poj 2187 凸包or旋转qia壳法相关推荐

POJ 2187 凸包+旋转卡壳
思路: 求个凸包旋转卡壳一下就求出来最远点对了注意共线情况也就是说凸包如果有一堆点共线保留端点即可 //By SiriusRen #include <cmath> #incl ...
凸包初学 Andrew 和 Melkman （模板）旋转卡(qia)壳(ke)
凸包初学定义: 给你n个散落的点,让你求出最小的凸多边形将所有的点包括起来,或者点在边上. 稳定凸包,不能再扩充,每条边都用3个点必备知识: 会用叉积判断点与直线的关系(这里指点在线的那一边, ...
POJ - 2187 Beauty Contest （求距离最远点对-凸包+旋转卡壳/枚举（旋转卡壳学习））
链接:https://vjudge.net/problem/POJ-2187 题意:求求距离最远点对. 思路:肯定为凸包上的点,可枚举,也可根据凸包性质旋转卡壳求对踵点. 参考博客: https:// ...
[Poj 2187] 计算几何之凸包(二) {更高效的算法}
{ 承上一节继续介绍点集的凸包 (下文中所有凸包若不做特殊说明均指点集的凸包) 这一节介绍相比更高效的算法 } ========================================= ...
poj 2187（旋转卡壳）
传送门模板题,求凸包,用旋转卡壳求出最远点对. 因为把int弄成double,WA了好几次,差点对idy大神给的模板失去信心...不过事实证明idy的写法没问题,orz! #include<c ...
poj 2187 Beauty Contest(凸包求解多节点的之间的最大距离)
1 /* poj 2187 Beauty Contest 2 凸包:寻找每两点之间距离的最大值 3 这个最大值一定是在凸包的边缘上的! 4 5 求凸包的算法: Andrew算法! 6 */ 7 #in ...
凸包练习: POJ 2187(JAVA)
分治化求凸包,请参看:http://128kj.iteye.com/admin/blogs/1748622 POJ 2187题意: 给出一个点集,求两点之间最长的距离的平方,最长距离的两个点一定在凸包 ...
【POJ 2187】如何修改Graham算法，才能使之顺利解决三点一条直线的问题。
暑期集训中做了 POJ 2187 题,噩梦一场,但是这次仔细把一些 Graham 算法在处理存在三点在一条直线上问题的改进方法找到了. 题目意思很明确:给你平面上的一坨点,求最远两点距离的平方. 思路 ...
漫步微积分三十四——体积计算：圆柱壳法
还有一种去体积的方法,往往它比上篇文章的方法更加方便. 为了理解这种方法,考虑图1左边所示的区域,也就是,第一象限数轴和所示示曲线y=f(x)y=f(x)围成的区域.如果这个区域绕xx轴旋转,那么图中 ...

poj 2187 凸包or旋转qia壳法

poj 2187 凸包or旋转qia壳法相关推荐

最新文章

热门文章