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  • Aggressive Cows 二分

    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<iostream>
    #include<algorithm>
    #include<cstdio>
    using namespace std;
    #define MAXN 100001
    #define INF 1000000001
    /*
        将奶牛分配到到多个点,使得他们之间的最小距离最大
        check(x)函数  最小距离为x的时候能否放下n头奶牛
    */
    int n, c, a[MAXN];
    bool check(int x)
    {
        int tmp = a[0], cnt = 1;
        for(int i=1;i<n;i++)
        { 
            if (a[i] - tmp >= x)
            {
                cnt++;
                tmp = a[i];
            }
        }
        return (cnt >= c);
    }
    int main()
    {
        while (scanf("%d%d", &n, &c) != EOF)
        {
            int  Max = -1;
            for (int i = 0; i < n; i++)
            {
                scanf("%d", &a[i]);
                Max = max(Max, a[i]);
            }
            int beg = 1, end = Max,ans=0;
            sort(a, a + n);
            while (beg <= end)
            {
                int mid = (beg + end) / 2;
                if (check(mid))
                {
                    ans = mid;
                    beg = mid + 1;
                }
                else
                    end = mid - 1;
            }
            printf("%d
    ", ans);
        }
    }
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  • 原文地址:https://www.cnblogs.com/joeylee97/p/6702323.html
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