# POJ 2187：Beauty Contest （凸包+旋转卡壳）

Beauty Contest

 Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 35396 Accepted: 10961

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

AC代码：

#include<iostream>
#include<algorithm>
#include<cmath>
#include<stdio.h>
using namespace std;
#define max(a,b) (a>b?a:b)
#define PI acos(-1)
struct point
{
double x;
double y;
double distance(const point &b)const    //计算距离
{
return (x-b.x)*(x-b.x)+(y-b.y)*(y-b.y);
}
point operator-(point s)
{
return {x-s.x,y-s.y};
}
} ;
point a[55000],data[55000];

int crossLeft(point p0,point p1,point p2)   //判读左转还是右转
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}
bool cmp(point p1,point p2)         //极角序
{
int tmp=crossLeft(a[0],p1,p2);
if(tmp>0) return true;
else if(tmp==0&&a[0].distance(p1)<a[0].distance(p2))return true;
else return false;
}
bool cmp_x(point p1,point p2)       //坐标序
{
if(p1.x!=p2.x)return p1.x<p2.x;
return p1.y<p2.y;
}
void init(int n)
{
int i,k=0;
point p;
scanf("%lf%lf",&a[0].x,&a[0].y);
p=a[0];
for(i=1; i<n; i++)
{
scanf("%lf%lf",&a[i].x,&a[i].y);
if( (p.y>a[i].y) || ((p.y==a[i].y)&&(p.x>a[i].x)) )
{
p=a[i];
k=i;
}
}
a[k]=a[0];
a[0]=p;
sort(a+1,a+n,cmp);
}

int main()
{
int N;
while(~scanf("%d",&N))
{
init(N);
for(int i=0; i<2; i++)
data[i]=a[i];         //前两个点入栈
int top=1;
for(int i=2; i<N; i++)
{
data[++top]=a[i];     //新点进栈
while(top>1&&crossLeft(data[top-2],data[top-1],data[top])<=0)  //向后遍历判断是否存在凹的区域
data[top-1]=data[top],top--;                               //存在，出栈一个点
}
++top;
/* 旋转卡壳部分 */
if(top==2)printf("%.0f\n",data[0].distance(data[1]));               //凸包退化的情况
else
{
double maxx=0.0;
int i=0,j=0;                        //某方向上的对踵点对
for(int k=0; k<top; k++)
{
if(!cmp_x(data[i],data[k]))i=k;
if(cmp_x(data[j],data[k]))j=k;
}
int si=i,sj=j;
while(i!=sj||j!=si)                 //旋转180°
{
maxx=max(maxx,data[i].distance(data[j]));
if(crossLeft({0,0},data[(i+1)%top]-data[i],data[(j+1)%top]-data[j])<0)
i=(i+1)%top;
else j=(j+1)%top;
}
printf("%.0f\n",maxx);
}
/* END */
}
}