Beauty Contest(graham求凸包算法)

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 25256		Accepted: 7756

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

题意：给n个点的坐标，计算这些点两两距离最大的那个距离；

思路：如果直接枚举，肯定超。所以可以先形成凸包，距离最大的那两个端点一定是凸包中的点。所以形成凸包后再枚举就可以了。
　　　这里用了graham算法，

 #include<stdio.h>
 #include<string.h>
 #include<math.h>
 #include<algorithm>
 using namespace std;
 const int maxn = ;
 int top,stack[maxn];
 int n;
 struct Point
 {
     double x;
     double y;
 } point[maxn];

 double cross(const Point &a, const Point &b, const Point &c)//三个点的叉积，结果大于0说明bc的极角大于ac的极角，等于0说明共线；
 {
     return (a.x-c.x)*(b.y-c.y) - (a.y-c.y)*(b.x-c.x);
 }

 double dis(const Point &a, const Point &b)
 {
     return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
 }

 int cmp(const Point &a, const Point &b)//对点排序找出最左最下的那个点作为point[0];
 {
     if(a.y == b.y)
         return a.x < b.x;
     return a.y < b.y;
 }

 void Graham()
 {
     sort(point,point+n,cmp);
     for(int i = ; i <= ; i++)
         stack[i] = i;
     top = ;
     for(int i = ; i < n; i++)
     {
         while(top && cross(point[i],point[stack[top]],point[stack[top-]]) >= )
             top--;
         stack[++top] = i;
     }
     int count = top;
     stack[++top] = n-;
     for(int i = n-; i >= ; i--)
     {
         while(top != count && cross(point[i],point[stack[top]],point[stack[top-]])>=)
             top--;
         stack[++top] = i;
     }
 }

 int main()
 {
     while(~scanf("%d",&n))
     {
         for(int i = ; i < n; i++)
             scanf("%lf %lf",&point[i].x,&point[i].y);

         Graham();

         double ans = ,distance;
         for(int i = ; i < top; i++)
         {
             for(int j = ; j < i; j++)
             {
                 distance = dis(point[stack[i]],point[stack[j]]);
                 if(ans < distance)
                     ans = distance;
             }
         }
         printf("%.0lf\n",ans);
     }
     return ;
 }

这是经典的计算几何学问题，判断向量p1=(x1,y1)到p2=(x2,y2)是否做左转，只需要判断x1*y2-x2*y1的正负，如果结果为正，则从p1到p2做左转。也就是向量的叉积。

Graham算法是这样的

1.将各点排序（），为保证形成圈，把P0在次放在点表的尾部；

2.准备堆栈：建立堆栈S，栈指针设为t，将0、1、2三个点压入堆栈S；

3.对于下一个点i

　　　只要S[t-1]、S[t]、i不做左转

　　　　　就反复退栈；

　　　将i压入堆栈S

4.堆栈中的点即为所求凸包；