poj 1274 The Perfect Stall

poj 1274 The Perfect Stall

二分图的最大匹配，加一个源点，这个源点到左边所有顶点加一条边，权值为1. 。同理，加一个终点，所有右边的定点到这个终点有条边，权值为1. 这样二分图的最大匹配就转换成了最大流问题。

#include < iostream >
#include < queue >
#include < cstring >
using   namespace  std;
const   int  MAX = 405 ;
int  cap[MAX][MAX] = { 0 };
int  flow[MAX][MAX] = { 0 };
int  pre[MAX],m[MAX];
int  N, M,S,T;
const   int  INF = 10000000 ;

int  bfs( int  s)
{
    memset(m, 0 , sizeof  m);  
    memset(pre, 0 , sizeof  pre);
    queue < int >  q;
    q.push(s);
    m[s] = INF;
     while ( ! q.empty())
    {
             int  u = q.front(); q.pop();
             for ( int  v = 0 ; v <= N + M + 1 ; v ++ )
                     if ( ! m[v] && cap[u][v] > flow[u][v])
                    {
                         pre[v] = u;
                         m[v] =  m[u] > cap[u][v] - flow[u][v] ? cap[u][v] - flow[u][v]:m[u];
                         q.push(v);
                    }
    }
    
    if (m[T] == 0 ) return   0 ;
   
    for ( int  u = T; u != S ; u = pre[u])
   {
           flow[pre[u]][u] += m[T];
           flow[u][pre[u]] -= m[T];
   }
    return  m[T];
}

int  main()
{
    
     while (cin >> N >> M)
    {    memset(cap, 0 , sizeof  cap);
         memset(flow, 0 , sizeof  flow);
          for ( int  i = 1 ; i <= N; i ++ )
         {
             int  c,e;
            cin >> c;
             for ( int  j = 1 ; j <= c; j ++ )
                   {
                         cin >> e;
                         e = e + N;
                         cap[i][e] = 1 ;
                   } 
         } 
         S = 0 ; 
         T = N + M + 1 ;
          for ( int  i = 1 ; i <= N; i ++ )
            cap[ 0 ][i] = 1 ;
            
          for ( int  i = 1 ; i <= M; i ++ )
            cap[N + i][T] = 1 ;
    
          int  ans = 0 ;
          while ( 1 )
         {
             int  temp = bfs(S);
             if (temp == 0 ) break ;
             else  ans += temp;
         }
    
         cout << ans << endl;
    }   
    system( " pause " );
     return   0 ;
}