Poj 2112 二分 多重匹配

/*
    1y，有点意外
    对应匈牙利算法看，不同的是右边的点可匹配多个
*/
#include < cstdio >
#include < cstring >

const   int  MAXN = 1005 ;

struct  Node{
     int  v,next;
}nodes[MAXN * MAXN];

int  n,m;
int  G[MAXN];
int  alloc,N;
bool  vi[MAXN];
int  link[MAXN][MAXN];

void  add( int  a, int  b){
    alloc ++ ;
    nodes[alloc].v = b,nodes[alloc].next = G[a];
    G[a] = alloc;
}
bool  find( int  u){
     for ( int  son = G[u];son !=- 1 ;son = nodes[son].next){
         int  v = nodes[son].v;
         // v对应多个匹配
         if (vi[v]) continue ;
        vi[v] = 1 ;
         if (link[v][ 0 ] < N){
            link[v][ ++ link[v][ 0 ]] = u;
             return   true ;
        }
         for ( int  i = 1 ;i <= link[v][ 0 ];i ++ )
             if (find(link[v][i])){
                link[v][i] = u;
                 return   true ;
            }
    }
     return   false ;
}
bool  chk(){
     int  ans = 0 ;
     for ( int  i = 0 ;i < m;i ++ )link[i][ 0 ] = 0 ;
     for ( int  i = 0 ;i < n;i ++ ){
        memset(vi, 0 , sizeof (vi));
         if ( ! find(i)) return   false ;
    }
     return   true ;
}
int  main(){
     char  str[ 20 ],ch;
     while (scanf( " %d%d " , & n, & m),n){
        memset(G, - 1 , sizeof (G));
        alloc = 0 ;
         for ( int  i = 0 ;i < n;i ++ ){
            scanf( " %s " ,str);
             while ( 1 ){
                 while (ch = getchar(),ch == '   ' );                
                 int  tmp = ch - ' 0 ' ;
                 while (ch = getchar(),ch <= ' 9 ' && ch >= ' 0 ' )tmp = 10 * tmp + ch - ' 0 ' ;
                add(i,tmp);
                 if (ch == ' \n ' || ch == EOF) break ;
            }
        }
         int  low = 0 ,high = n + 1 ;
         while (high - low > 1 ){
             int  mid = (high + low) >> 1 ;
            N = mid;
             if (chk())high = mid;
             else  low = mid;
        }
        printf( " %d\n " ,high);
    }
     return   0 ;
}