POJ 3281 Dining（最大流）

http://poj.org/problem?id=3281

题意：

有n头牛，F种食物和D种饮料，每头牛都有自己喜欢的食物和饮料，每种食物和饮料只能给一头牛，每头牛需要1食物和1饮料。问最多能满足几头牛的需求。

思路：

 #include<iostream>

 #include<algorithm>

 #include<cstring>

 #include<cstdio>

 #include<sstream>

 #include<vector>

 #include<stack>

 #include<queue>

 #include<cmath>

 #include<map>

 #include<set>

 using namespace std;

 typedef long long ll;

 typedef long long ull;

 typedef pair<int,int> pll;

 const int INF = 0x3f3f3f3f;

 const int maxn =  + ;

 struct Edge

 {

     int from,to,cap,flow;

     Edge(int u,int v,int w,int f):from(u),to(v),cap(w),flow(f){}

 };

 struct Dinic

 {

     int n,m,s,t;

     vector<Edge> edges;

     vector<int> G[maxn];

     bool vis[maxn];

     int cur[maxn];

     int d[maxn];

     void init(int n)

     {

         this->n=n;

         for(int i=;i<n;++i) G[i].clear();

         edges.clear();

     }

     void AddEdge(int from,int to,int cap)

     {

         edges.push_back( Edge(from,to,cap,) );

         edges.push_back( Edge(to,from,,) );

         m=edges.size();

         G[from].push_back(m-);

         G[to].push_back(m-);

     }

     bool BFS()

     {

         queue<int> Q;

         memset(vis,,sizeof(vis));

         vis[s]=true;

         d[s]=;

         Q.push(s);

         while(!Q.empty())

         {

             int x=Q.front(); Q.pop();

             for(int i=;i<G[x].size();++i)

             {

                 Edge& e=edges[G[x][i]];

                 if(!vis[e.to] && e.cap>e.flow)

                 {

                     vis[e.to]=true;

                     d[e.to]=d[x]+;

                     Q.push(e.to);

                 }

             }

         }

         return vis[t];

     }

     int DFS(int x,int a)

     {

         if(x==t || a==) return a;

         int flow=, f;

         for(int &i=cur[x];i<G[x].size();++i)

         {

             Edge &e=edges[G[x][i]];

             if(d[e.to]==d[x]+ && (f=DFS(e.to,min(a,e.cap-e.flow) ) )>)

             {

                 e.flow +=f;

                 edges[G[x][i]^].flow -=f;

                 flow +=f;

                 a -=f;

                 if(a==) break;

             }

         }

         return flow;

     }

     int Maxflow(int s,int t)

     {

         this->s=s; this->t=t;

         int flow=;

         while(BFS())

         {

             memset(cur,,sizeof(cur));

             flow +=DFS(s,INF);

         }

         return flow;

     }

 }DC;

 int n,F,D;

 int main()

 {

     //freopen("in.txt","r",stdin);

     while(~scanf("%d%d%d",&n,&F,&D))

     {

         int src=, dst=F+D+*n+;

         DC.init(dst+);

         for(int i=;i<=F;i++)  DC.AddEdge(src,i,);

         for(int i=;i<=D;i++)  DC.AddEdge(F+i,dst,);

         for(int i=;i<=n;i++)

         {

             DC.AddEdge(F+D+i,F+D+n+i,);

             int num1,num2;

             scanf("%d%d",&num1,&num2);

             while(num1--)

             {

                 int x;

                 scanf("%d",&x);

                 DC.AddEdge(x,F+D+i,);

             }

             while(num2--)

             {

                 int x;

                 scanf("%d",&x);

                 DC.AddEdge(F+D+n+i,F+x,);

             }

         }

         printf("%d\n",DC.Maxflow(src,dst));

     }

     return ;

 }