bool dfs(int d)
{
    int i;
    for(i = 1; i <= count; i++)
       if(!check[i] && g[d][i])
       {
           check[i] = 1;
           if(!link[i] || dfs(link[i]))
           {
                link[i] = d;// xM[u] = v;
                return 1 ;
           }
        }
    return 0 ;
}
里面用这个就超时if(!check[i] && g[d][i])
而把他们交换一下就900ms  if( g[d][i]&&!check[i] )

• Re:怎么回事呀？？用dfs时要注意啦！！ (414) ldxiaoxi 2010-08-02 13:49:04