poj3281-Dining ，最大流量，内置图

id=3281">点击打开链接
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;

}