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  • POJ 2456 Aggressive cows

    Aggressive cows
    Time Limit: 1000MS Memory Limit: 65536K
    Total Submissions: 24838 Accepted: 11537
    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

    USACO 2005 February Gold

    二分,看代码能懂,没什么坑~

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    const int INF = 1e9 + 5;
    const int maxn = 1e5 + 5;
    int n, k, x[maxn];
    int check(int d)  
    {
        int last = 0;
        for(int i = 1; i < k; i++) 
        {
            int cur = last + 1;  
            while(cur < n && x[cur] - x[last] < d)
                cur++;
            if(cur == n) return 0;
            last = cur;
        }
        return 1;
    }
    int main()
    {
        while(~scanf("%d%d", &n, &k))
        {
            for(int i = 0; i < n; i++)
                scanf("%d", &x[i]);
            sort(x, x+n);
            int l = 0, r = INF;
            while(r - l > 1)
            {
                int mid = (l + r) / 2;
                if(check(mid)) l = mid;
                else r = mid;
            }
            printf("%d
    ", l);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798945.html
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