• nyoj 211&&poj 3660


    Cow Contest

    时间限制:1000 ms  |  内存限制:65535 KB
    难度:4
    描述

    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 ≤ A ≤ N; 1 ≤ B ≤ NA ≠ B), 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.

    输入
    * 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

    There are multi test cases.The input is terminated by two zeros.The number of test cases is no more than 20.
    输出
    For every case:
    * Line 1: A single integer representing the number of cows whose ranks can be determined
    样例输入
    5 5
    4 3
    4 2
    3 2
    1 2
    2 5
    0 0
    样例输出
    2
    ////能打败的个数加上被打败的个数恰好等于n-1,则能确定,
    ////否则无法确定,抽象为简单的floyd传递闭包算法,
    ////在加上每个顶点的出度与入度 (出度+入度=顶点数-1,则能够确定其编号)。
    #include<queue>
    #include<stack>
    #include<vector>
    #include<math.h>
    #include<stdio.h>
    #include<numeric>//STL数值算法头文件
    #include<stdlib.h>
    #include<string.h>
    #include<iostream>
    #include<algorithm>
    #include<functional>//模板类头文件
    using namespace std;
    
    const int INF=1e9+7;
    const int maxn=102;
    
    int n,m;
    int tu[maxn][maxn];
    int main()
    {
        int i,j,k,x,y,ans;
        while(~scanf("%d %d",&n,&m),n+m)
        {
            memset(tu,0,sizeof(tu));
            while(m--)
            {
                scanf("%d %d",&x,&y);
                tu[x][y]=1;
            }
            for(k=1; k<=n; k++)
                for(i=1; i<=n; i++)
                    for(j=1; j<=n; j++)
                        if(tu[i][k]&&tu[k][j])
                            tu[i][j]=1;
            int ans=0;
            for(i=1; i<=n; i++)
            {
                int cont=n-1;
                for(j=1; j<=n; j++)
                    if(tu[i][j]||tu[j][i])
                        cont--;
                if(!cont)
                    ans++;
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
    
    
    









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  • 原文地址:https://www.cnblogs.com/nyist-xsk/p/7264868.html
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