POJ 3660 Cow Contest

对每个结点 BFS 一遍，找出能被它打败和可以打败它的结点的总和 s ，如果 s == n-1 ，它的排名可以确定；

C++ vector 邻接表；

# include <cstdio>
# include <iostream>
# include <cstring>
# include <vector>

# define N 105

using namespace std;

int n, m;
char vis[N];
vector <int> pre[N];
vector <int> aft[N];

int bfs(int u, int d)
{
    int i, ret, cx, nx, front, rear, size;
    int Q[1000*N];

    Q[1] = u;
    memset(vis, 0, sizeof(vis));
    vis[u] = 1;

    front = 1;
    rear  = 2;

    ret = 0;
    while (front < rear)
    {
        cx = Q[front++];
        size = d>0 ? aft[cx].size():pre[cx].size();
        for (i = 0; i < size; ++i)
        {
            nx = d>0 ? aft[cx][i]:pre[cx][i];
            if (!vis[nx])
            {
                Q[rear++] = nx;
                vis[nx] = 1;
                ++ret;
            }
        }
    }

    return ret;
}

int main()
{
    int i, cnt, u, v;

    while (~scanf("%d%d", &n, &m))
    {
        for (i = 1; i <= n; ++i)
        {
            pre[i].clear();
            aft[i].clear();
        }
        for (i = 0; i < m; ++i)
        {
            scanf("%d%d", &u, &v);
            pre[v].push_back(u);
            aft[u].push_back(v);
        }
        cnt = 0;
        for (i = 1; i <= n; ++i)
        {
            if (bfs(i, 1)+bfs(i, -1) == n-1) ++cnt;
        }
        printf("%d\n", cnt);
    }

    return 0;
}

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