zoukankan      html  css  js  c++  java
  • POJ

    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.
     1 #include <iostream>
     2 using namespace std;
     3 #include<string.h>
     4 #include<set>
     5 #include<stdio.h>
     6 #include<math.h>
     7 #include<queue>
     8 #include<map>
     9 #include<algorithm>
    10 #include<cstdio>
    11 #include<cmath>
    12 #include<cstring>
    13 #include <cstdio>
    14 #include <cstdlib>
    15 #include<stack>
    16 #include<vector>
    17 long long n,k;
    18 long long a[110000];
    19 long long panduan(long long zhi)
    20 {
    21     int add=1;
    22     int kaishi=1,jieshu=1;
    23     while(1)
    24     {
    25         if(a[jieshu]-a[kaishi]<zhi&&jieshu<=n)
    26         {
    27             jieshu++;
    28         }
    29         else if(a[jieshu]-a[kaishi]>=zhi)
    30         {
    31             add++;
    32             kaishi=jieshu;
    33         }
    34         else
    35             break;
    36     }
    37     //cout<<zhi<<"_"<<add<<endl;
    38     return add;
    39 }
    40 int  main()
    41 {
    42     scanf("%lld %lld",&n,&k);
    43     for(long long  i=1;i<=n;i++)
    44         scanf("%lld",&a[i]);
    45     sort(a+1,a+1+n);
    46     long long kaishi=0,jieshu=1e18;
    47     long long mid;
    48     long long t;
    49     while(kaishi<jieshu-1)
    50     {
    51         //cout<<kaishi<<"_"<<jieshu<<endl;
    52         mid=(kaishi+jieshu)>>1;
    53         if(panduan(mid)>=k)
    54             kaishi=mid;
    55         else
    56             jieshu=mid;
    57     }
    58     cout<<kaishi<<endl;
    59     return 0;
    60 }
    View Code
     
  • 相关阅读:
    【洛谷4657】[CEOI2017] Chase(一个玄学的树形DP)
    Tarjan在图论中的应用(二)——用Tarjan来求割点与割边
    Tarjan在图论中的应用(一)——用Tarjan来实现强连通分量缩点
    jquery放大镜
    自定义上传按钮样式
    一些设计理论资料
    jquery滚动条
    全栈工程师到底有什么用(转)
    巧用CSS文件愚人节恶搞(转)
    仿双色球-随机产生7个数字
  • 原文地址:https://www.cnblogs.com/dulute/p/7966712.html
Copyright © 2011-2022 走看看