POJ 3660

题目大意:有N个人，进行M场比赛，M场比赛只给两个参赛选手，第一个就是赢了的参赛选手，求排名固定的参赛选手

结题思路:如果一个人被赢了m次，赢了别人n次，如果m+n==N-1，那么满足条件,Floyd

#include <iostream>
#include <cstring>
#include <algorithm>
#define MAXVERTEXNUM 110
#define INF 1000000
using namespace std;

int Nv, Ne;
int D[MAXVERTEXNUM][MAXVERTEXNUM];

void Floyd()
{
    for (int k = 1; k <= Nv; ++k)
        for (int i = 1; i <= Nv; ++i)
            for (int j = 1; j <= Nv; ++j)
                if (D[i][j] > D[i][k] + D[k][j])
                    D[i][j] = D[i][k] + D[k][j];
}

int main()
{
    cin >> Nv >> Ne;
    for (int i = 1; i <= Nv; ++i)
        for (int j = 1; j <= Nv; ++j)
        {
            if (i == j)
                D[i][j] = 0;
            else
                D[i][j] = INF;
        }

    for (int i = 1; i <= Ne; ++i)
    {
        int V1, V2;
        cin >> V1 >> V2;
        D[V1][V2] = 1;
    }

    Floyd();

    int getRes[MAXVERTEXNUM];
    memset(getRes, 0, sizeof(getRes));
    for (int i = 1; i <= Nv; ++i)
    {
        for (int j = 1; j <= Nv; ++j)
        {
            if (i != j && D[i][j] != INF)
            {
                getRes[j]++;
                getRes[i]++;
            }
        }
    }

    int res = 0;
    for (int i = 1; i <= Nv; ++i)
        if (getRes[i] == Nv - 1)
            res++;

    cout << res << endl;

    return 0;
}