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  • Aggressive cows 二分不仅仅是查找

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 7083   Accepted: 3522

    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.

    Source

     
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 #include <algorithm>
     6 #include <string>
     7 #include <vector>
     8 #include <set>
     9 #include <map>
    10 #include <queue>
    11 #include <stack>
    12 #include <sstream>
    13 #include <iomanip>
    14 using namespace std;
    15 const int INF=0x4fffffff;
    16 const int EXP=1e-6;
    17 const int MS=100005;
    18 
    19 int pos[MS];
    20 int N,M;
    21 
    22 bool judge(int d)
    23 {
    24       int last=0;
    25       int now;
    26       for(int i=1;i<M;i++)     //贪心法给第i头牛找宿舍
    27       {
    28             now=last+1;
    29             while(now<N&&pos[now]-pos[last]<d)
    30                   now++;
    31             if(now>=N)
    32                   return false;
    33             last=now;
    34       }
    35       return true;
    36 }
    37 
    38 void solve()
    39 {
    40       sort(pos,pos+N);
    41       int l=0,r=INF;
    42       while(r-l>1)         //  注意是区间,很多时候开区间更方便一些。
    43       {
    44             int mid=(l+r)/2;
    45             if(judge(mid))
    46                   l=mid;
    47             else
    48                   r=mid;
    49       }
    50       printf("%d
    ",l);
    51 }
    52 
    53 int main()
    54 {
    55       while(scanf("%d%d",&N,&M)!=EOF)
    56       {
    57             for(int i=0;i<N;i++)
    58                   scanf("%d",&pos[i]);
    59             solve();
    60       }
    61       return 0;
    62 }
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  • 原文地址:https://www.cnblogs.com/767355675hutaishi/p/4375584.html
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