• poj 2345 Aggressive cows


    Aggressive cows
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 4737   Accepted: 2316

    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.
    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    int pos[200000],n,m;
    using namespace std;
    int judge(int w)
    {
         int i,cnt=pos[0],sum=1;
         for(i=1;i<n;i++)//贪心
         {
              if(pos[i]-cnt>=w)
              {
                    sum++;
                if(sum>=m)
                return 1;
                cnt=pos[i];
              }
    
         }
         return 0;
    }
    int main()
    {
         int i;
         while(scanf("%d %d",&n,&m)!=EOF)
         {
              memset(pos,0,sizeof(pos));
    
              for(i=0;i<n;i++)
              scanf("%d",&pos[i]);
              sort(pos,pos+n);
              int l,r,mid;
              l=0;
              r=pos[n-1]-pos[0];
              while(l<=r)//二分答案
              {
                   mid=(l+r)/2;
                   if(judge(mid))
                   l=mid+1;
                   else
                   r=mid-1;
    
              }
              printf("%d
    ",l-1);
         }
    
         return 0;
    }
  • 相关阅读:
    第四周学习总结
    第三周学习总结
    第二周学习总结
    第一周学习总结
    解决pycharm中导入Wordcloud库的时候失败的问题
    开发报告5
    TypeError: _cache_value_encoder: not supported type: <class 'numpy.ndarray'>
    pandas中的rename_axis用法
    金叉死叉量化交易--matplotlib绘图案例
    一、EDA分析
  • 原文地址:https://www.cnblogs.com/llei1573/p/3228925.html
Copyright © 2020-2023  润新知