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  • POJ 3660 Cow Contest

    Description

    N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

    The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ AN; 1 ≤ BN; AB), then cow A will always beat cow B.

    Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

    Input

    * Line 1: Two space-separated integers: N and M
    * Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

    Output

    * Line 1: A single integer representing the number of cows whose ranks can be determined
     

    Sample Input

    5 5
    4 3
    4 2
    3 2
    1 2
    2 5
    

    Sample Output

    2
    

    Source

    思路:这道题用的是拓扑建图,用佛洛依德建立每个点与其他点的关系。如果能确定这个点与其他所有点的关系,那它就是可以明确排名的点,否则就无法建立关系——无法确定排名(自己画图)。
    #include <cstdio>
    #include <iostream>
    #include <string>
    #include <cstring>
    #include <cmath>
    #include <algorithm>
    #include <queue>
    #include <vector>
    #include <map>
    using namespace std;
    #define ll long long
    
    const int inf = 0x3f3f3f3f;
    int n, m, dis[100+8][100+8];
    
    int main()
    {
        int a, b;
        memset(dis, 0, sizeof(dis));
        scanf("%d%d", &n, &m);
        for(int i = 0; i<m; i++)
        {
            scanf("%d%d", &a, &b);
            dis[a][b] = 1;
            dis[b][a] = -1;
        }
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++)
                for(int k = 1; k <= n; k++)
                    if(dis[j][i] == dis[i][k] && dis[j][i])
                    {
                        dis[j][k] = dis[i][k];
                    }
        int ans = 0, sum;
        for(int i = 1; i <= n; i++)
        {
            sum = 0;
            for(int j = 1; j <= n; j++)
                if(dis[i][j] != 0)
                    sum++;
            if(sum == n-1)ans++;
        }
        printf("%d
    ", ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/RootVount/p/11209617.html
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