POJ 3281 网络流 拆点 Dining

题意：

有F种食物和D种饮料，每头牛有各自喜欢的食物和饮料，而且每种食物或者饮料只能给一头牛。

求最多能有多少头牛能同时得到它喜欢的食物或者饮料。

分析：

把每个牛拆点，中间连一条容量为1的边，保证一头牛不会被多个食物或者饮料分配。

然后把饮料和牛连边，食物和另外一边的牛连边，最后增加一个源点和汇点跑最大流。

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

 const int maxn =  + ;
 const int maxnode =  + ;
 const int INF = 0x3f3f3f3f;

 int N, F, D;

 int n, s, t;

 struct Edge
 {
     int from, to, cap, flow;
     Edge(int u, int v, int c, int f):from(u), to(v), cap(c), flow(f) {}
 };

 vector<Edge> edges;
 vector<int> G[maxnode];

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

 void AddEdge(int u, int v, int c)
 {
     edges.push_back(Edge(u, v, c, ));
     edges.push_back(Edge(v, u, , ));
     int m = edges.size();
     G[u].push_back(m - );
     G[v].push_back(m - );
 }

 bool vis[maxnode];
 int d[maxnode];
 int cur[maxnode];

 bool BFS()
 {
     d[s] = ;
     queue<int> Q;
     Q.push(s);
     memset(vis, false, sizeof(vis));
     vis[s] = true;

     while(!Q.empty())
     {
         int u = Q.front(); Q.pop();
         for(int i = ; i < G[u].size(); i++)
         {
             Edge& e = edges[G[u][i]];
             int v = e.to;
             if(!vis[v] && e.cap > e.flow)
             {
                 vis[v] = true;
                 d[v] = d[u] + ;
                 Q.push(v);
             }
         }
     }

     return vis[t];
 }

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

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

 int main()
 {
     while(scanf("%d%d%d", &N, &F, &D) == )
     {
         n = N *  + D + F + ;
         s = , t = n - ;
         init();

         //build graph
         for(int i = ; i <= F; i++) AddEdge(s, N*+i, );
         for(int i = ; i <= D; i++) AddEdge(N*+F+i, t, );
         for(int i = ; i <= N; i++) AddEdge(i, i + N, );

         for(int i = ; i <= N; i++)
         {
             int f, d, x; scanf("%d%d", &f, &d);
             while(f--)
             {
                 scanf("%d", &x);
                 AddEdge(N*+x, i, );
             }
             while(d--)
             {
                 scanf("%d", &x);
                 AddEdge(N + i, N*+F+x, );
             }
         }

         printf("%d\n", Maxflow());
     }

     return ;
 }

代码君