poj 2553 The Bottom of a Graph

poj 2553 The Bottom of a Graph

Accepted

1004K

297MS

C++

1585B

求出度为0的强连通分量，可能有多个，按节点序号从小到大输出就行了。

还是用两次DFS 加 名字不好记 加 还不知道怎么发音的的那个算法，第一次原图DFS确定访问顺序，第二次对翻转过边的方向得到的新图DFS

确定强连通分量。

还不太熟，写了一些时间。

#include < iostream >
#include < vector >
using   namespace  std;
vector < int >  adj[ 5001 ];
vector < int >  adj_op[ 5001 ];
bool  visit[ 5001 ];
int  father[ 5001 ],outagree[ 5001 ];
vector < int >  finish;
int  n,m;
void  dfs1( int  v)
{
    visit[v] = true ;
      for ( int  i = 0 ; i < adj[v].size(); i ++ )
          if ( ! visit[adj[v][i]])
             dfs1(adj[v][i]);
    finish.push_back(v);
}

void  dfs2( int  v,  int  fa)
{
    visit[v] = true ;
     for ( int  i = 0 ; i < adj_op[v].size(); i ++ )
         if ( ! visit[adj_op[v][i]])
        {
             father[adj_op[v][i]] = fa;
             dfs2(adj_op[v][i],fa);
        }

}
int  main()
{
     while (cin >> n)
    {
         if (n == 0 ) break ;
        cin >> m;
         int  i,j,s,t;
         for (i = 1 ; i <= n; i ++ )
        {
            adj[i].clear();
            adj_op[i].clear();
        }

         for (i = 1 ; i <= m; i ++ )
        {
            cin >> s >> t;
            adj[s].push_back(t);
            adj_op[t].push_back(s);
        }
        
        finish.clear();
        memset(visit,  0 ,  sizeof  visit);
         for (i = 1 ; i <= n; i ++ )
             if ( ! visit[i])
                dfs1(i);

        memset(visit,  0 ,  sizeof  visit);
        memset(father, 0 , sizeof  father);
         int  cnt = 0 ;
         for (i = finish.size() - 1 ; i >= 0 ; i -- )
             if ( ! visit[finish[i]])
            {
                cnt ++ ;
                father[finish[i]] = cnt;
                dfs2(finish[i],cnt);
            }
          
          // for(i=1; i<=n; i++)
         //      cout<<i<<' '<<father[i]<<endl;
         memset(outagree, 0 , sizeof  outagree);
        
          for (i = 1 ; i <= n; i ++ )
              for (j = 0 ; j < adj[i].size(); j ++ )
             {
                  if (father[i] != father[ adj[i][j] ])
                     outagree[ father[i] ] ++ ;
             }
        
              int  tt = 0 ;
              for (i = 1 ; i <= n; i ++ )
                  if (outagree[father[i]] == 0 )
                 {
                      if (tt != 0 )cout << '   ' ;
                     cout << i;
                     tt ++ ;
                 }
             cout << endl;
    }

     return   0 ;
}