POJ 2186 Popular Cows

POJ 2186 Popular Cows

先求出有向图强连通分量，缩点之后重新构图，新图为一个 有向无环图，如果在这个DAG是只有一个出度为0的点，那么这个点所表示的强连通分量中的所有点都是符合要求的。
一开始使用list来构建邻接表，结果TLE。后来改成用数组模拟，效率较高。
以下是我的代码：

#include < stack >
#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxn( 10007 );

int  n,m,n2,id[kMaxn];
int  first[kMaxn],edge[ 5 * kMaxn],next[ 5 * kMaxn];

int  dfscnt,dfsn[kMaxn],low[kMaxn];
stack < int >  s;

void  dfs( int  u)
{
    dfsn[u] = low[u] =++ dfscnt;
    s.push(u);
     for ( int  i = first[u];i !=- 1 ;i = next[i])
         if ( ! dfsn[edge[i]])
        {
            dfs(edge[i]);
            low[u] = min(low[u],low[edge[i]]);
        }
         else
            low[u] = min(low[u],dfsn[edge[i]]);
     if (dfsn[u] == low[u])
    {
        n2 ++ ;
         int  v;
         do
        {
            v = s.top();s.pop();
            id[v] = n2;
        } while (v != u);
    }
}

bool  Tarjan( int   & num)
{
    n2 = dfscnt = 0 ;
    memset(dfsn, 0 ,kMaxn * sizeof ( int ));
     for ( int  i = 1 ;i <= n;i ++ )
         if ( ! dfsn[i])
            dfs(i);

     int  counter( 0 ),degree[kMaxn];
     for ( int  i = 1 ;i <= n2;i ++ )
        degree[i] = 0 ;
     for ( int  u = 1 ;u <= n;u ++ )
         for ( int  i = first[u];i !=- 1 ;i = next[i])
             if (id[u] != id[edge[i]])
                degree[id[u]] ++ ;
     for ( int  i = 1 ;i <= n2;i ++ )
         if ( ! degree[i])
        {
            counter ++ ;
            num = i;
        }
     return  (counter == 1 );
}

int  main()
{
     while (scanf( " %d%d " , & n, & m) == 2 )
    {
        memset(first, - 1 ,kMaxn * sizeof ( int ));
         for ( int  i = 1 ;i <= m;i ++ )
        {
             int  u,v;
            scanf( " %d%d " , & u, & v);
            edge[i] = v;
            next[i] = first[u];
            first[u] = i;
        }

         int  num;
         if (Tarjan(num))
        {
             int  ans( 0 );
             for ( int  i = 1 ;i <= n;i ++ )
                 if (id[i] == num)
                    ans ++ ;
            printf( " %d\n " ,ans);
        }
         else
            printf( " %d\n " , 0 );
    }

     return   0 ;
}