poj 2187Beauty Contest(凸包)

最新推荐文章于 2020-12-24 16:43:14 发布

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本文介绍了一种算法，用于计算二维平面上多个农场间的最大距离，以帮助一头名为Bessie的牛确定旅行中携带的食物数量。通过极角排序和凸包算法优化计算过程。

Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 35138		Accepted: 10894

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

Sample Output

2
 /*
9
1 2
2 2
0 1
1 1
2 1
3 1
0 0
1 0
2 0

10
*/
#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
int n,top;
struct point
{
    int x,y;
}pt[51000],Stack[51000],pointa;
int cross(point a,point b,point c)
{
    return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
int cmp(point a,point b)
{
    if(a.y==b.y)
        return a.x<b.x;
      return a.y<b.y;
}

int cmp1(point a,point b)
{
    double k = cross(pointa,a,b);
    if(k>0) return 1;
    if(k<0) return 0;
    return pow(a.x-pointa.x,2)+pow(a.y-pointa.y,2)>pow(b.x-pointa.x,2) + pow(b.y-pointa.y,2);//当共线时候，距离按照从大到小排时
                                                                                             //下文cross()<=0;
                                                                                             //当从小到大排时cross（）<0;
}                                                                                            //注意   从大到小排序  共线时只记录远的不记录近的，从小到大排都记录
void gram()
{
    sort(pt,pt+n,cmp);
    pointa = pt[0];
    pt[n] = pt[0];
    sort(pt+1,pt+n,cmp1);
    int i;
    top=2;
    Stack[0]=pt[0];
    Stack[1]=pt[1];
    Stack[2]=pt[2];
    for(i=3;i<=n;i++)
    {
        while(top>=2&&cross(Stack[top-1],Stack[top],pt[i])<=0)//如果是<=0那么极角排序时当共线时距离从大到小排
            top--;                                            //如果<0那么极角排序时当共线时距离从小到大排
        Stack[++top] = pt[i];
    }
}
int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
            scanf("%d%d",&pt[i].x,&pt[i].y);
        gram();
        int max = 0;

             for(i=0;i<=top;i++)
            {
                for(j=0;j<=top;j++)
                {
                    if(max < pow( Stack[i].x-Stack[j].x,2 ) + pow( Stack[i].y-Stack[j].y,2 ) )
                        max = pow(Stack[i].x-Stack[j].x,2)+pow(Stack[i].y-Stack[j].y,2);
                }
            }
            printf("%d\n",max);
    }
    return 0;
}