POJ 2187 Beauty Contest(凸包暴力法)

Beauty Contest Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 24124   Accepted: 7364 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)  Source USACO 2003 Fall

求出凸包上的點，然後暴力枚舉

開始過不了，後來把如果點的個數在100以內，就直接暴力A了！

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <cmath>using namespace std;#define eps 1e-8#define PI 3.14159265struct point{int x;int y;}po[55500],temp;int n,pos;bool zero(double a){return fabs(a) < eps;}int dis(point &a,point &b)//返回兩點之間距離的平方{return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);}int across(point &a,point &b,point &c)//求a b and a c 的X積{return (b.x-a.x)*(c.y-a.y) - (b.y-a.y)*(c.x-a.x);}int cmp(const void *a,const void *b){return across(po[0],*(point*)a,*(point*)b) > 0 ? -1 : 1;}int select(){int i,j,k=1;for(i=2;i<n;i++){if(across(po[0],po[k],po[i])==0){if(dis(po[0],po[k]) < dis(po[0],po[i]))po[k]=po[i];}elsepo[++k]=po[i];}return k+1;}int graham(int num){int i,j,k=2;//////////////////////////////////////////po[num]=po[0];//fangbian num++;int ans=0;int t;if(num<20){for(i=0;i<num;i++)for(j=0;j<num;j++){t=dis(po[i],po[j]);if(t > ans)ans=t;}printf("%d\n",ans);return 0;}for(i=3;i<num;i++){while(across(po[k-1],po[k],po[i]) < -eps){k--;}po[++k]=po[i];//就這個迴圈結束，不需要了！}/*for(i=0;i<k;i++)printf("%lf %lf\n",po[i].x,po[i].y);*/for(i=0;i<k;i++)for(j=0;j<k;j++){t=dis(po[i],po[j]);if(t > ans)ans=t;}printf("%d\n",ans);return 0;}int main(){int i,j,k;point my_temp;int ans;int t;while(scanf("%d",&n)!=EOF){ans=0;scanf("%d%d",&po[0].x,&po[0].y);temp=po[0];pos=0;for(i=1;i<n;i++){scanf("%d%d",&po[i].x,&po[i].y);if(po[i].y < temp.y)temp=po[i],pos=i;}if(n<100){for(i=0;i<n;i++)for(j=0;j<n;j++){t=dis(po[i],po[j]);if(t > ans)ans=t;}printf("%d\n",ans);continue;}my_temp=po[0];po[0]=po[pos];po[pos]=my_temp;qsort(po+1,n-1,sizeof(po[0]),cmp);graham(select());}return 0;}