求凸包上最大距离 poj2187

最新推荐文章于 2017-10-24 23:55:20 发布

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最新推荐文章于 2017-10-24 23:55:20 发布

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分类专栏：几何

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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之间的最大距离。通过构建凸包并运用几何方法，有效地解决了这一问题。

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Beauty Contest

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 38389		Accepted: 11866

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

Hint

Farm 1 (0, 0) and farm 3 (1, 1) have the longest distance (square root of 2)

Source

USACO 2003 Fall

就是求最大距离。。先求出凸包，这里的排序方法和我写的上一篇不太一样，这里是根据最左下角已确定的点与之叉乘是否大于零若共线则优先排距离小的，~~留个坑，我的再想想为啥~~，

之后求每一对对踵点间最大距离，（其实是每个点都找一遍，利用三角形面积，等底，求最大高）稍优化的暴力。。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<iostream>
#define N 50005
using namespace std;
struct node
{
	int x,y;
}a[N],point;
int cheng(node a,node b,node c)
{
	return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}
int dis(node a,node b)
{
	return (b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y);
}
int cmp(node b,node c)
{
	int t=cheng(a[1],b,c);
	if(t==0)
	   return dis(a[1],b)<dis(a[1],c);
	return t>0;
}
int n,vis[N],head=-1,zhan[N],ans;
int main()
{
	scanf("%d",&n);
	point.x=10005,point.y=10005;int j;
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&a[i].x,&a[i].y);
	    if(a[i].y<point.y)
	         point=a[i],j=i;
	    else
	       if(a[i].y==point.y&&a[i].x<point.x)
	            point=a[i],j=i;
	}
	if(n==2)
	{
		printf("%d\n",dis(a[1],a[2]));
		return 0;
	}
	a[j]=a[1];a[1]=point;
	sort(a+2,a+n+1,cmp);
	zhan[++head]=1;zhan[++head]=2;
	for(int i=3;i<=n;i++)
	{
	    while(head>0&&cheng(a[zhan[head-1]],a[zhan[head]],a[i])<=0)head--;
	    zhan[++head]=i;	
	}
	j=1;zhan[++head]=1;
	for(int i=0;i<head;i++)
	{
		while(cheng(a[zhan[i]],a[zhan[i+1]],a[zhan[j+1]])>cheng(a[zhan[i]],a[zhan[i+1]],a[zhan[j]]))
		   j=(j+1)%head;
		ans=max(ans,dis(a[zhan[i]],a[zhan[j]]));
	}
	printf("%d\n",ans);
}