tarjan算法

void tarjan(int u)
{
    int v;
    dfn[u] = low[u] = ++cnt;//开始时dfn[u] == low[u]
    S[top++] = u;//进栈
    vis[u] = true;
    for (int i=head[u]; i!=-1; i=Edge[i].next)
    {
        v = Edge[i].to;
        if (dfn[v] == 0)//如果v点还未遍历
        {
            tarjan(v);//向下遍历
            low[u] = low[u] < low[v] ? low[u] : low[v];//确保low[u]最小
        }
        else if (vis[v] && low[u] > dfn[v])//v在栈中，修改low[u]
            low[u] = dfn[v];
    }
    if (dfn[u] == low[u])//u为该强连通分量中遍历所成树的根
    {
        ++scc;
        do
        {
            v = S[--top];//栈中所有到u的点都属于该强连通分量，退栈
            vis[v] = false;
            belong[v] = scc;
        } while (u != v);
    }

}

可通过如下题目巩固：

poj2186

poj2553

poj1236

hdu1269

hdu4280

poj1273

poj2112

poj3469