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;

}

//