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Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 38702		Accepted: 11983

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

思路：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int N = 50005;
struct Point{
    int x,y;
}p[N],Stack[N];
int n;

int mult(Point a,Point b,Point c){
    return (a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y);
}
int dis(Point a,Point b){
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int cmp(Point a,Point b){
    if(mult(a,b,p[0])>0) return 1;
    if(mult(a,b,p[0])==0&&dis(b,p[0])>dis(a,p[0])) return 1;
    return 0;
}
int Graham(){
    sort(p+1,p+n,cmp);
    int top = 2;
    Stack[0]=p[0];
    Stack[1]=p[1];
    Stack[2]=p[2];
    for(int i=3;i<n;i++){
        while(top>=1&&mult(p[i],Stack[top],Stack[top-1])>=0){
            top--;
        }
        Stack[++top]=p[i];
    }
    return top;
}
int rotating_calipers(int top)//旋轉卡殼
{
    int q=1,ans=0,p;
    Stack[++top]=Stack[0]; ///將最後一個點和第一點的邊也要算進去,於是多加一個點表示第一個點
    for(p=0; p<top; p++)
    {
 ///找到以Stack[p] 和 Stack[p+1]這兩個點爲底邊的最大三角形 (叉積之積爲三角形有向面積)(圖1)
        while(mult(Stack[p],Stack[p+1],Stack[q+1])>mult(Stack[p],Stack[p+1],Stack[q]))
            q=(q+1)%top;///在底邊的兩個點中選擇與頂點距離大的
        ans=max(ans,max(dis(Stack[p],Stack[q]),dis(Stack[p+1],Stack[q+1])));
    }
    return ans;
}
int main()
{
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        int k = 0;
        for(int i=1;i<n;i++){
            if(p[k].y>p[i].y||(p[k].y==p[i].y)&&(p[k].x>p[i].x)){
                k=i;
            }
        }
        swap(p[0],p[k]);
        //printf("%d %d\n",p[0].x,p[0].y);
        int top = Graham();
        int ans =rotating_calipers(top);
        printf("%d\n",ans);
    }
    return 0;
}