时间复杂度：O(n+m).

void tarjan(int u)
{
    dfn[u] = low[u] = ++ ixp ;
    stp[++tot] = u;
    st[u] = true;
    for (int i = h[u]; ~i; i = ne[i])
    {
        int y = e[i];
        if(!dfn[y])
        {
            tarjan(y);
            low[u] = min(low[y],low[u]);
        }
        else if(st[y]) low[u] = min(low[u],dfn[y]);
    }
    if(dfn[u] == low[u])
    {
        scv++;
        int y;
        do
        {
            y = stp[tot--];
            st[y] = false;
            id[y] = scv;
        }while(y != u);
    }
}

int main()
{
    for(int i = 1; i <= n; i ++ )
       if(!dfn[i])
          tarjan(i);
}