学步园

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分析：

求最大流

建图：

拆点 牛拆成左边与食物相连的左牛 和 右边与饮料相连的右牛 

1、s->食物 连边

2、食物->左牛

3、左牛->右牛

4、右牛->饮料

5、饮料->t

边的方向为 s->食物->左牛->右牛->饮料->t

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

const int maxn = 500 + 5;
const int INF = 100000000;
struct Edge{
    int from, to, cap, flow;
};

struct Dinic{
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool vis[maxn];
    int d[maxn];
    int cur[maxn];

    void init(int n)
    {
        this->n = n;
        for(int i=0; i<=n; ++i) G[i].clear();
        edges.clear();
    }

    void AddEdge(int from, int to, int cap)
    {
        edges.push_back((Edge){from, to, cap, 0});
        edges.push_back((Edge){to, from, 0, 0});
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BFS()
    {
        memset(vis, 0, sizeof vis );
        queue<int> Q;
        Q.push(s);
        vis[s] = 1;
        d[s] = 0;
        while(!Q.empty())
        {
            int x = Q.front(); Q.pop();
            for(int i=0; i<G[x].size(); ++i)
            {
                Edge& e = edges[G[x][i]];
                if(!vis[e.to] && e.cap > e.flow)
                {
                    vis[e.to] = 1;
                    d[e.to] = d[x] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x, int a)
    {
        if(x == t || a == 0) return a;
        int flow = 0, f;
        for(int& i = cur[x]; i<G[x].size(); ++i)
        {
            Edge& e = edges[G[x][i]];
            if(d[x] + 1 == d[e.to] && (f=DFS(e.to, min(a,e.cap-e.flow)))>0)
            {
                e.flow += f;
                edges[G[x][i]^1].flow -= f;
                flow += f;
                a -= f;
                if(a==0) break;
            }
        }
        return flow;
    }

    int Maxflow(int s, int t)
    {
        this->s = s; this->t =t;
        int flow = 0;
        while(BFS()){
            memset(cur, 0, sizeof cur );
            flow += DFS(s, INF);
        }
        return flow;
    }
};

Dinic solver;
int main()
{
    int N, F, D;
    int i, j;
    scanf("%d%d%d", &N, &F, &D);
    int s = 0, t = 2*N + F + D;
    solver.init(t+1);
    for(i=1; i<=N; ++i)
        solver.AddEdge(i, N+i, 1);
    for(i=1; i<=N; ++i)
    {
        int f, d, x;
        scanf("%d%d", &f, &d);
        for(j=1; j<=f; ++j)
        {
            scanf("%d", &x);
            solver.AddEdge(2*N+x, i, 1);
        }
        for(j=1; j<=d; ++j)
        {
            scanf("%d", &x);
            solver.AddEdge(N+i, 2*N+F+x, 1);
        }
    }
    for(i=1; i<=F; ++i)
        solver.AddEdge(s, 2*N + i, 1);
    for(i=1; i<=D; ++i)
        solver.AddEdge(2*N + F + i, t, 1);
    int ans = solver.Maxflow(s, t);
    printf("%d\n", ans);
    return 0;
}