poj2187 Beauty Contest 凸包

最新推荐文章于 2020-11-11 23:32:12 发布

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Bessie，FarmerJohn的荣誉奶牛，赢得了世界奶牛之美大赛的冠军，因此她将环游全球50,000多个农场进行交流。为了确保旅行期间有足够的食物，她需要计算从一个农场到另一个农场的最大直线距离，从而确定行李箱的大小。本文通过求解凸包问题来帮助Bessie找到最远距离的两座农场。

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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

凸包求直径，凸包上距离最远的两个点的距离的平方。。。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
using namespace std;
#define MAX 50005
#define INF 1000000000
int n,L;
struct point  //构造点
{
    double x,y;
}p[MAX];
point save[50005];  //存点
int xmult(point p1,point p2,point p0)
{
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
bool cmp(const point& a,const point &b)
{
    if(a.y == b.y)return a.x < b.x;
    return a.y < b.y;
}
int Graham(point *p,int n)
{
    int i;
    sort(p,p + n,cmp);
    save[0] = p[0];
    save[1] = p[1];
    int top = 1;
    for(i = 2;i < n; i++)
    {
        while(top && xmult(save[top],p[i],save[top-1])>=0) top--;//去掉=就是包含共线。。
        save[++top] = p[i];
    }
    int mid = top;
    for(i = n - 2; i >= 0; i--)
    {
        while(top>mid&&xmult(save[top],p[i],save[top-1])>=0) top--;//去掉=就是包含共线。。
        save[++top]=p[i];
    }
    return top;
}
double dis(point a,point b)
{
    double x=a.x-b.x;
    double y=a.y-b.y;
    return x*x+y*y;
}
int main()
{
    scanf("%d",&n);
    int i,j;
    for(i=0;i<n;i++) scanf("%lf %lf",&p[i].x,&p[i].y);
    int tot=Graham(p,n);
    double ans=0;
    for(i=0;i<tot;i++)
    {
        for(j=i;j<tot;j++)
        {
            ans=max(ans,dis(save[i],save[j]));
        }
    }
    printf("%.0f\n",ans);
    return 0;
}