poj2456Aggressive cows(二分+贪心)

最新推荐文章于 2019-10-24 17:16:40 发布

Chibi_Maruko

最新推荐文章于 2019-10-24 17:16:40 发布

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本文链接：https://blog.youkuaiyun.com/j2013210855/article/details/38375103

这篇博客详细介绍了如何使用二分查找和贪心策略来解决POJ2456题目的过程。通过分析输入输出样例，博主展示了算法的具体实现，并提供了关键的思路提示。

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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

* 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

Hint

OUTPUT DETAILS:

FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3.

Huge input data,scanf is recommended.

题意大致是：农夫约翰搭了一间有N间牛舍的小屋。牛舍被排在一条线上，第i 号牛舍在xi的位置。但是他的M头牛对小屋很不满意，因此经产互相攻击。约翰为了防止牛之间互相伤害

，因此决定把每头牛都放在离其他牛尽量远的牛舍。也就是要最大化最近的两头牛之间的距离。

关于最大化最小值或者最小化最大值的问题，通常用二分搜索法。

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 100100
int arr[N],n,m;
int cheak(int len)
{
    int num=1,x,i;
    x=arr[1];
    for(i=2;i<=n;i++)
    {
        if(arr[i]-x>=len)
        {
            num++;
            x=arr[i];
        }
    }
    if(num>=m)
        return 1;
    else
        return 0;
}
int main()
{
    while(cin>>n>>m)
    {
        int i;
        for(i=1;i<=n;i++)
            scanf("%d",&arr[i]);
        sort(arr+1,arr+n+1);
        int be,en,ans,mid;
        be=0;en=arr[n]-arr[1];
        while(be<=en)
        {
            mid=(en+be)/2;
            if(cheak(mid)==1)
            {
                ans=mid;
                be=mid+1;
            }
            else
               en=mid-1;
        }
        printf("%d\n",ans);
    }
    return 0;
}