DSAA之图论双连通及割点问题（十一）

1. 定义

A connected undirected graph is biconnected if there are no vertices whose removal disconnects the rest of the graph.

If a graph is not biconnected, the vertices whose removal would disconnect the graph are known as articulation points.

　　在理解下面的算法前，再次记忆双连通是指不存在割点的无向连通图。

2. 解决算法

Depth-first search provides a linear-time algorithm to find all articulation points in a connected graph.

First, starting at any vertex, we perform a depth first search and number the nodes as they are visited. For each vertex v, we call this preorder number num(v).这也是下面的算法assign_num所做的事情，非常简单就是图的DFS遍历及给每个节点赋值

Then, for every vertex v in the depth-first search spanning tree, we compute the lowest-numbered vertex, which we call low(v).之前说过不太理解DFS生成树的coding办法，其实DFS生成树是DFS搜索过程的反映，只要在DFS搜索过程中做相应的处理就相当于后序或者中序遍历DFS生成树

We can efficiently compute low by performing a postorder traversal of the depth-first spanning tree.

By the definition of low, low(v) is the minimum of以下三者中的最小值，所以每条都要计算

num(v)

The first condition is the option of taking no edges

the lowest num(w) among all back edges (v, w)

the second way is to choose no tree edges and a back edge就是所有背向边中最小的num(v)

the lowest low(w) among all tree edges (v, w)

the third way is to choose some tree edges and
possibly a back edge.

Any other vertex v is an articulation point if and only if v has some child w such that low(w) $\geq$ num(v).这里是最关键的割点的判断条件

Notice that this condition is always satisfied at the root; hence the need for a special test.

　　下面的伪代码由DSAA给出，将所有的步骤整合到一起，在DFS框架中给节点编号，并在调用递归之后判断每个节点的low()，同时判断是否有割点出现。这种编程方式将前序遍历和后续遍历整合到一起，值得学习。

3. 伪代码

void find_art( vertex v ){
  vertex w;
  visited[v] = TRUE;
  /* Rule 1 同时也给每个节点一个前序编号*/
  low[v] = num[v] = counter++;
  for each w adjacent to v{
    {/* forward edge 没有访问的节点，一定是当前节点的子代*/
    if( !visited[w] ){
      //标记当前节点w的父代为v
      parent[w] = v;
      find_art( w );
      //通过上面的find_art(w)已经获得low(w)值，判断是否w-v满足割点性质，此时是后序遍历。使用nums[v] ！=1 排除root节点
      if( nums[v] != 1 &&low[w] >= num[v] )
        printf ( "%v is an articulation point\n", v );
      //Rule 3 更新当前节点的low值
      low[v] = min( low[v], low[w] ); 
    }
    /* back edge 下面解释*/
    else if( parent[v] != w )
      //Rule 2 更新当前节点的low值
      low[v] = min( low[v], num[w] ); 
    }
}

　　上面伪代码中最后的else if( parent[v] != w )需要重点考虑下，因为无向图边都是双向的，所以在w已经访问过的时候，一种可能是在之前w-v已经考虑过该边了，而此时再将v-w当成背向边将导致错误，所以这里需要判断下。借助下图可以很容易理解：（比如：CD边与DC边）
这里写图片描述

时间复杂度

　　由于上面的整体框架还是DFS，遍历每一个边和节点一次，所以整体上的时间复杂度为 $O(|E|+|V|)$ 。

4. 最后

　　留下了个问题，为什么low(w) $\geq$ num(v)时，当前节点为割点？这个问题，暂时笔者当算法结论记住。