zoukankan      html  css  js  c++  java
  • 最大化最小值poj2456Aggressive cows

    Aggressive cows
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 15528   Accepted: 7440

    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

    5 3
    1
    2
    8
    4
    9

    Sample Output

    3

    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.


    check(d)=可以安排所有的牛的位置使得最近的两头牛的距离都不小于d

    求满足check(d)的最大的d的值。


    #include <iostream>
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    const int maxn=1e5+10;
    int house[maxn];
    int n,m;
    const int inf=99999999;
    bool check(int d)
    {
        int last=0;
        for(int i=1;i<m;i++)
        {
            int crt=last+1;
            while(crt<n&&house[crt]-house[last]<d)
            {
                crt++;
            }
            if(crt==n)return false;
            last=crt;
        }
        return true;
    }
    void solve()
    {
        sort(house,house+n);
        int lb=0,ub=inf;
        while(ub-lb>1)
        {
            int mid=(ub+lb)/2;
            if(check(mid))
                lb=mid;
            else
                ub=mid;
        }
        printf("%d
    ",lb);
    }
    
    int main()
    {
        cin>>n>>m;
        for(int j=0;j<n;j++)
            scanf("%d",&house[j]);
        solve();
        return 0;
    }
    

  • 相关阅读:
    应用连接Redis报错处理
    粗谈RESTFul API接口-认识粗谈RESTFul API接口-认识
    策略模式有效解决过多的if-else
    slice的共享内存时需要注意的问题
    swoole中使用wss
    swoole版本和PHP版本的对应关系
    php中关于新引入object的坑
    Java Object Serialization Specification.
    php将索引数组转换成关联数组
    Gson解析json字符串的坑
  • 原文地址:https://www.cnblogs.com/bryce1010/p/9387410.html
Copyright © 2011-2022 走看看