**链接：****传送门 **
题意：给出 n 个点，求出这 n 个点中最远的两个点距离的平方
思路：最远点对一定会在凸包的顶点上，然后直接暴力找一下凸包顶点中距离最远的两个点
/*************************************************************************
> File Name: poj2187.cpp
> Author: WArobot
> Blog: http://www.cnblogs.com/WArobot/
> Created Time: 2017年05月08日 星期一 14时30分43秒
************************************************************************/
#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
using namespace std;
#define eps 1e-10
const int maxn = 50010;
struct point{ double x,y; };
double multi(point sp,point ep,point op){
return (sp.x-op.x)*(ep.y-op.y) >= (sp.y-op.y)*(ep.x-op.x);
}
bool operator < (const point &a,const point &b){
return a.y < b.y || ( a.y == b.y && a.x < b.x );
}
int Graham(point pnt[] , int n , point res[]){
int i , len , k = 0 , top = 1;
sort(pnt,pnt+n);
if( n == 0 ) return 0; res[0] = pnt[0];
if( n == 1 ) return 1; res[1] = pnt[1];
if( n == 2 ) return 2; res[2] = pnt[2];
for(int i = 2 ; i < n ; i++){
while( top && multi( pnt[i] , res[top] , res[top-1] ))
top--;
res[++top] = pnt[i];
}
len = top; res[++top] = pnt[n-2];
for(int i = n - 3 ; i >= 0 ; i--){
while( top!=len && multi( pnt[i] , res[top] , res[top-1] ))
top--;
res[++top] = pnt[i];
}
return top;
}
int Dist(point a,point b){
return (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
}
int main(){
int n;
while(~scanf("%d",&n)){
point pnt[maxn] , res[maxn];
for(int i = 0 ; i < n ; i++) scanf("%lf%lf",&pnt[i].x,&pnt[i].y);
int num = Graham( pnt , n , res );
int dis = 0;
for(int i = 0 ; i < num ; i++){
for(int j = i+1 ; j < num ; j++){
dis = max( dis , Dist( res[i] , res[j] ) );
}
}
printf("%d\n",dis);
}
return 0;
}