题意:给你n个坐标,求最远的两点距离 思路:用凸包算法求处,各个定点,再用旋转凸包卡壳 #include <iostream> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; #define N 50010 struct node{ int x,y,d; }p[N]; int dist(node a,node b){ return (a.x-b.x)*(…

链接:http://poj.org/problem?id=2187 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the wor…

题面 Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill between farm…

Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of N (2 <= N <= 50,000) farms around the world in order to spread goodwill bet…

旋转卡壳求凸包的直径的平方 板子题 #include<cstdio> #include<vector> #include<cmath> #include<algorithm> using namespace std; struct Point { int x, y; Point(int x=0, int y=0):x(x),y(y) { } }; typedef Point Vector; Vector operator - (const Point&…

题目:http://poj.org/problem?id=2187 学习材料:https://blog.csdn.net/wang_heng199/article/details/74477738 https://www.jianshu.com/p/74c25c0772d6 可以再倒着枚举一遍那样求凸包. 用叉积算面积来旋转卡壳. 注意在面积等于的时候就不要往后走了,不然只有两个点的数据就会死循环. #include<cstdio> #include<cstring> #inclu…

Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 32708   Accepted: 10156 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a…

题目:http://poj.org/problem?id=2187 学习资料:https://blog.csdn.net/wang_heng199/article/details/74477738 https://www.jianshu.com/p/74c25c0772d6 注意求凸包时先下后上,保持逆时针: 别忘了给点排序囧. 代码如下: #include<cstdio> #include<cstring> #include<algorithm> #include&l…

题目: http://poj.org/problem?id=1113 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/F Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26219   Accepted: 8738 Description Once upon a time there was a greedy King who…

链接:传送门 题意:给出 n 个点,求出这 n 个点中最远的两个点距离的平方 思路:最远点对一定会在凸包的顶点上,然后直接暴力找一下凸包顶点中距离最远的两个点 /************************************************************************* > File Name: poj2187.cpp > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ > Cre…

/* poj 2187 Beauty Contest 凸包:寻找每两点之间距离的最大值 这个最大值一定是在凸包的边缘上的! 求凸包的算法: Andrew算法! */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; struct Point{ Point(){} Point(int x, int y){ this->…

题目链接:http://poj.org/problem?id=2187 Time Limit: 3000MS Memory Limit: 65536K Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bessie will make a tour of…

链接: http://poj.org/problem?id=2187 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22013#problem/E Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24254   Accepted: 7403 Description Bessie, Farmer John's prize cow, h…

D - Beauty Contest Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a r…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24283   Accepted: 7420 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 36113   Accepted: 11204 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bes…

题目传送门 题意:求两点的距离平方的最大值 分析:凸包模板题 /************************************************ * Author :Running_Time * Created Time :2015/10/25 9:31:11 * File Name :A.cpp ************************************************/ #include <cstdio> #include <algorithm&…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27276   Accepted: 8432 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 26180   Accepted: 8081 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

题目链接 题意:给你n个点的坐标,n<=50000,求最远点对 #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include <queue> #include <map> #include <algorithm> #i…

题链: http://poj.org/problem?id=2187 题解: 计算几何,凸包,旋转卡壳 一个求凸包直径的裸题,旋转卡壳入门用的. 代码: #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #define MAXN 50050 using namespace std; const double eps=1…

Beauty Contest 题意:给你一个数据范围在2~5e4范围内的横纵坐标在-1e4~1e4的点,问你任意两点之间的距离的最大值的平方等于多少? 一道卡壳凸包的模板题,也是第一次写计算几何的题,就看了些模板,关于预备知识;我是直接找到左下角的点,排好序之后,就直接形成凸包,之后调用rotating_calipers()求解:里面注意在凸包构造好之后,因为是++top的,所以在后面卡壳里面%top会出现问题,所以循环之后再一次++top;开始看graham()里面的while循环看了很久,其…

http://poj.org/problem?id=2187 题意 给n个坐标,求最远点对的距离平方值. 分析 模板题,旋转卡壳求求两点间距离平方的最大值. #include<iostream> #include<cmath> #include<cstring> #include<queue> #include<vector> #include<cstdio> #include<algorithm> #include<…

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24431   Accepted: 7459 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a result, Bess…

貌似直接求出凸包之后$O(n^2)$暴力就可以了? #include <cstdio> #include <cstring> #include <string> #include <cstdlib> #include <algorithm> #include <iostream> #include <stack> #include <cmath> using namespace std; struct P{ i…

/** 求出凸包枚举每个点的矩距离即可 因为凸包上的点可定不多.. 学习: 刚开始WA 了一次,,因为用int 存的, 一看discuss 里提供的数据,想起来,,应该是越界了.. 后来用longlong 就过了.. **/ #include <iostream> #include <algorithm> #include <cmath> using namespace std; struct point { long long x,y; //point (){} po…

http://poj.org/problem?id=2187 显然直径在凸包上(黑书上有证明).(然后这题让我发现我之前好几次凸包的排序都错了QAQ只排序了x轴.....没有排序y轴.. 然后本题数据水,暴力也能过... (之前一直以为距离是单增的,其实并不是,应该是三角形面积单增...) 考虑旋转卡壳 一篇好的文章:http://www.cnblogs.com/Booble/archive/2011/04/03/2004865.html 首先对踵点就是两条平行线夹紧凸包的两个点(或者3个点或4…

id=2187">Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27218   Accepted: 8410 Description Bessie, Farmer John's prize cow, has just won first place in a bovine beauty contest, earning the title 'Miss Cow World'. As a…

Poj 2187 旋转卡壳求解 传送门 旋转卡壳,是利用凸包性质来求解凸包最长点对的线性算法,我们逐渐改变每一次方向,然后枚举出这个方向上的踵点对(最远点对),类似于用游标卡尺卡着凸包旋转一周,答案就在这其中的某个方向上. 直接暴力和旋转卡壳速度对比(仅此题) #include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #inc…

Poj 2187 凸包模板求解 传送门 由于整个点数是50000,而求凸包后的点也不会很多,因此直接套凸包之后两重循环即可求解 #include <queue> #include <cmath> #include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #define ll long lo…