Aggressive cows （二分）

最新推荐文章于 2022-03-06 23:06:34 发布

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最新推荐文章于 2022-03-06 23:06:34 发布

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分类专栏：二分贪心文章标签：二分

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本文介绍了一种算法，用于解决给定数量的动物放置在不同位置时如何最大化它们之间的最小距离的问题。通过二分查找和贪心策略实现，适用于农场布局优化等场景。

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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

题目大概：

求m只动物间的最大距离。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
using namespace std;
int n,k;
int a[100005];

int go(int m)
{
    int h=a[1],sum=1;
    for(int i=2;i<=n;i++)
    {
        if(a[i]-h>=m){sum++; h=a[i];}

    }
    if(sum>=k)return 1;
    else return 0;
}

int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    sort(a+1,a+n+1);
    int l=0,r=(a[n]-a[1])/(k-1);
    int mid;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(go(mid))
        {
            l=mid+1;

        }
        else
        {
            r=mid-1;
        }

    }
    printf("%d\n",l-1);




    return 0;
}