POJ 2187 （2013.9.15周赛E题凸包）

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)

思路：给定平面上的一些散点集，求最远两点距离的平方值。刚开始提交了多次，都是TLE，后面和队友们聊天，然后说到这题我才知道是凸包

那时竟然没有想到，唉……失误啊，学长们教了竟然不会用，好吧，看来我要重新研究以前学长们教的算法了，要做那些例题巩固一下算法才得了，罪过罪过呀……

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=50010;
struct point
{
    int x,y;
};
point list[MAXN];
int stack[MAXN],top;
int cross(point p0,point p1,point p2)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x);
}
int dis(point p1,point p2)
{
    return (p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y);
}
bool cmp(point p1,point p2)
{
    int tmp=cross(list[0],p1,p2);
    if(tmp>0) return true;
    else if(tmp==0&&dis(list[0],p1)<dis(list[0],p2)) return true;
    else return false;
}
void init(int n)
{
    int i,k;
    point p0;
    scanf("%d%d",&list[0].x,&list[0].y);
    p0.x=list[0].x;
    p0.y=list[0].y;
    k=0;
    for(i=1;i<n;i++)
    {
        scanf("%d%d",&list[i].x,&list[i].y);
        if( (p0.y>list[i].y) || ((p0.y==list[i].y)&&(p0.x>list[i].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);
}
void graham(int n)
{
    int i;
    if(n==1)
    {top=0;stack[0]=0;}
    if(n==2)
    {
        top=1;
        stack[0]=0;
        stack[1]=1;
    }
    if(n>2)
    {
        for(i=0;i<=1;i++) stack[i]=i;
        top=1;
        for(i=2;i<n;i++)
        {
            while(top>0&&cross(list[stack[top-1]],list[stack[top]],list[i])<=0)
            top--;
            top++;
            stack[top]=i;
        }
    }
}
int main()
{
    int n,i,j;
    while(scanf("%d",&n)!=EOF)
    {
        int max=0;
        init(n);
        graham(n);
        for(i=0;i<=top;i++)
            for(j=i+1;j<=top;j++)
                if(max<dis(list[stack[i]],list[stack[j]]))
                    max=dis(list[stack[i]],list[stack[j]]);
        printf("%d\n",max);
    }
    return 0;
}