POJ2187--凸包--Beauty Contest

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 between farmers and their cows. For simplicity, the world will be represented as a two-dimensional plane, where each farm is located at a pair of integer coordinates (x,y), each having a value in the range -10,000 ... 10,000. No two farms share the same pair of coordinates. 

Even though Bessie travels directly in a straight line between pairs of farms, the distance between some farms can be quite large, so she wants to bring a suitcase full of hay with her so she has enough food to eat on each leg of her journey. Since Bessie refills her suitcase at every farm she visits, she wants to determine the maximum possible distance she might need to travel so she knows the size of suitcase she must bring.Help Bessie by computing the maximum distance among all pairs of farms. 

Input

* Line 1: A single integer, N 

* Lines 2..N+1: Two space-separated integers x and y specifying coordinate of each farm 

Output

* Line 1: A single integer that is the squared distance between the pair of farms that are farthest apart from each other. 

Sample Input

40 00 11 11 0

Sample Output

2

Hint

Farm 1 (0, 0) and farm 3 (1, 1) have the longest distance (square root of 2) 

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <iomanip>using namespace std;struct Point{int x;int y;}point[50008],res[50008];bool cmp(Point a,Point b){if(a.x>b.x)return 0;else if(a.x==b.x)return a.y>b.y?0:1;else return 1;}bool ral(Point p1,Point p2,Point p3){return ((p3.y-p1.y)*(p2.x-p1.x)>(p2.y-p1.y)*(p3.x-p1.x));}int distant(Point a,Point b){return (b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y);}double max(double a,double b){return a>b?a:b;}int main(){int n;while(scanf("%d",&n)!=EOF){for(int i=0;i<n;i++){scanf("%d%d",&point[i].x,&point[i].y);}sort(point,point+n,cmp);int m=-1;for(int i=0;i<n;i++){while(m>=1&&(!ral(res[m-1],res[m],point[i]))){m--;}res[++m]=point[i];}int k=m;for(int i=n-2;i>=0;i--){while(m>k&&(!ral(res[m-1],res[m],point[i]))){m--;}res[++m]=point[i];}//凸包建立完毕//现在凸包上有0到m这m+1个点。接下来就是枚举这m+1个点两两之间的距离求最大值int maxlen=0;for(int i=0;i<=m;i++){for(int j=i+1;j<=m;j++){maxlen=max(maxlen,distant(res[i],res[j]));//这里我distant原先是distance。CE了N久}}cout<<maxlen<<endl;}return 0;}