Aggressive cows（最大化最小值问题）

最新推荐文章于 2023-06-04 20:59:44 发布

free-dancer

最新推荐文章于 2023-06-04 20:59:44 发布

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分类专栏：二分

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解决农场牛棚布局问题，通过二分查找法找到使牛棚间最小距离最大化的最优解。给定牛棚位置，算法首先对位置进行排序，接着在最小与最大距离间进行二分搜索，判断是否能满足特定数量的牛分配到不同牛棚的要求。

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Problem Description

Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,...,xN (0 <= xi <= 1,000,000,000).

His C (2 <= C <= N) cows don't like this barn layout and become aggressive towards each other once put into a stall. To prevent the cows from hurting each other, FJ want to assign the cows to the stalls, such that the minimum distance between any two of them is as large as possible. What is the largest minimum distance?

Input

* Line 1: Two space-separated integers: N and C< br>< br>* Lines 2..N+1: Line i+1 contains an integer stall location, xi

Output

* Line 1: One integer: the largest minimum distance

Sample Input

Sample Output

题目大意：

//有n个点，给出每个点的位置，然后没c个点一组，这一组里的距离的最小值为minn,在若干个这样的组合中求出各个minn值中的最大值

思路：

将位置排排序，在最小距离和最大的距离中间，进行二分，看看能放的牛的牛棚的个数，和牛的数量比较，然后改变二分的区间。


#if 0
#include<iostream>
#include<cmath>  
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int a[100005];
int n,c;
int fun(int mid)
{
	int cnt=1,sum=0;
	for(int i=1; i<n; i++)
	{
		if(sum+a[i]-a[i-1]<mid)
		{
			sum+=a[i]-a[i-1];
		}
		else
		{
			sum=0;
			cnt++;	
		}	
	}
	
	if(cnt>=c)
	{
		return 1;	
	}
	else
	{
		return 0;	
	}	
}

int main()
{
	
	while(scanf("%d%d",&n,&c)==2)
	{
		memset(a,0,sizeof(a));
		for(int i=0; i<n; i++)
		{
			scanf("%d",&a[i]);
		}
		
		sort(a,a+n);
		int r=a[n-1]-a[0],l=a[n-1]-a[0];
		
		for(int i=1; i<n; i++)
		{
			if(a[i]-a[i-1]<l)
				l=a[i]-a[i-1];
		}
		
		while(r>=l) 
		{
			int mid=(l+r)/2;     
			if(fun(mid))            //多了 
			{
				l=mid+1;
			}
			else
			{
				r=mid-1;
			}
		}
		
		printf("%d\n",r);
	
	}
}
#endif