POJ2187——Beauty Contest

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

求平面点集最远点对，方法是凸包+旋转卡壳。

旋转卡壳求最远点对是通过：先确定一条边，然后去找第三个点，使得三角形的面积最大。这时距离最远。

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<math.h>

using namespace std;

const int maxn=50010;

struct point 
{
 int x,y;
}list[maxn],stack[maxn];

int dist(point a,point b)
{
 return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}

int cross(point p0,point p1,point p2)
{
 return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

int cmp(point a,point b)
{
 int cur=cross(list[0],a,b);
 if(cur > 0)
 return 1;
 else if(cur==0 && dist(a,list[0]) < dist(b,list[0]))
 return 1;
 else
 return 0;
}

void init(int n)
{
 int k=0;
 point p0;
 scanf("%d%d",&list[0].x ,&list[0].y);
 p0.x=list[0].x;
 p0.y=list[0].y;
 for(int i=1;i<n;i++)
 {
 scanf("%d%d",&list[i].x,&list[i].y);
 if(list[i].y < p0.y || (p0.y == list[i].y && list[i].x < p0.x))
 {
 p0.x=list[i].x;
 p0.y=list[i].y;
 k=i;
 }
 }
 list[k]=list[0];
 list[0]=p0;
 sort(list+1,list+n,cmp);
}

int graham(int n)
{
 int top;
 if(n==2)
 {
 stack[0]=list[0];
 stack[1]=list[1];
 top=1;
 }
 else
 {
 stack[0]=list[0];
 stack[1]=list[1];
 top=1;
 for(int i=2;i<n;i++)
 {
 while(top && cross(stack[top-1],stack[top],list[i]) <=0)
 top--;
 top++;
 stack[top]=list[i];
 }
 }
 return top;
}

int xuanzhuankake(int top)
{
 int q=1;
 int ans=0;
 stack[top+1]=stack[0];
 for(int p=0;p<=top;p++)
 {
 while(cross(stack[p],stack[p+1],stack[q]) < cross(stack[p],stack[p+1],stack[q+1]))
 q=(q+1)%(top+1);//找到使得面积比当前大的点，当q+1==top+1时，q实际是0，所以要取模
 ans=max(ans,max(dist(stack[p],stack[q]),dist(stack[p+1],stack[q+1])));
 }
 return ans;
}

int main()
{
 int n;
 while(~scanf("%d",&n))
 {
 init(n);
 int top=graham(n);
 printf("%d\n",xuanzhuankake(top));
 }
 return 0;
}