题意：就是给N个点的坐标，然后求任意两个点距离的平方最大的值
枚举超时。
当明白了 最远距离的两个点一定在凸包上，一切就好办了。求出凸包，然后枚举
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const double eps = 1e-;
const int Max = + ;
struct Node
{
int x, y;
};
Node node[Max], ch[Max];
int cmp(Node tempx, Node tempy)
{
if(tempx.y == tempy.y)
return tempx.x < tempy.x;
return tempx.y < tempy.y;
}
int xmult(Node p1, Node p2, Node p3)
{
return (p2.x - p1.x) * (p3.y - p1.y) - (p3.x - p1.x) * (p2.y - p1.y);
}
int andrew(int n)
{
int len, top = ;
ch[] = node[];
ch[] = node[];
for(int i = ; i < n; i++)
{
while(top && xmult(ch[top - ], ch[top], node[i]) <= )
top--;
ch[ ++top ] = node[i];
}
len = top;
ch[ ++top ] = node[n - ];
for(int i = n - ; i >= ; i--)
{
while(len != top && xmult(ch[top - ], ch[top], node[i]) <= )
top--;
ch[ ++top ] = node[i];
}
return top;
}
int dist(Node p1, Node p2)
{
return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y) ;
}
int get_max(int x, int y)
{
if(x - y > )
return x;
return y;
}
int main()
{
int n;
while( scanf("%d", &n) != EOF) {
for(int i = ; i < n; i++)
scanf("%d%d", &node[i].x, &node[i].y);
sort(node, node + n, cmp);
int top = andrew(n); int ans = ;
if(top > )
{
for(int i = ; i < top; i++)
{
for(int j = ; j < top; j++)
{
ans = get_max(ans, dist(ch[i], ch[j]));
}
}
}
else
{
ans = dist(ch[], ch[]);
}
printf("%d\n", ans);
}
return ;
}