poj 2187 Beauty Contest

   http://poj.org/problem?id=2187  

平面上有N个牧场，牧场位置互不相同，计算距离最远的俩个牧场间的距离，输出最远的平方。

解：求n个点组成的凸包，凸包上找2个点一点是最远的。   

      Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 29979		Accepted: 9298

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

4
0 0
0 1
1 1
1 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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

struct node{
    double x,y;
    node(){};
    node(double _x, double _y)
     {
       x = _x;
        y = _y;
    }

    node operator - (const node & B) const
    {
        return node(x-B.x, y-B.y);
    }
}p[50000], ch[50000];
bool cmp(node a,node b)
{
    if(a.x==b.x)
           return a.y<b.y;
    return a.x<b.x;
}
int squarDist(node A, node B) /**距离的平方*/
{
    return (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);
}
double Cross(node a,node b)//叉积。
{
    return a.x*b.y-a.y*b.x;
}
int main()
{
    int n,i,j;
    scanf("%d",&n);
    for(i=0;i<n;i++)
     scanf("%lf%lf",&p[i].x,&p[i].y);
      /** 求凸包 */
      /**先按照 x 从小到大排序, 再按照 y 从小到大排序*/
    sort(p,p+n,cmp);
     int m=0;
    for(i=0;i<n;i++)
    {
        while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)/** 从前往后找"下凸包" */
            m--;
          ch[m++]=p[i];
    }
    int k=m;
    for(i=n-2;i>=0;i--)
    {
         while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)/**从后往前找"上凸包", 形成完整的封闭背包*/
             m--;
        ch[m++]=p[i];

    }
    if(n>1) /** 起点重复*/
       m--;
       int res=0;
  for(i=0;i<m;i++)
    for(j=i+1;j<m;j++)
     res=max(res, squarDist(ch[i], ch[j]));
    printf("%d\n",res);


}