Long time did not write a blog ( blame too much work, I'm definitely not playing too Hey a )
So today is a tall write on something: strongly connected
First, some definitions related strongly connected // degrees from Mother
1. FIG strongly connected (Strongly Connected Graph) is meant a directed graph G, if for each pair vi, vj, vi ≠ vj, from vi to vj vi to vj and from there the path G is called strong FIG communication.
2. strongly connected subgraph to FIG great called strongly connected components (strongly connected components).
Of course, look at the definition is certainly not read, so I give chestnuts explain
We have the following diagram, for example, this is a particularly strong graph with the classic, three framed up three places on are strongly connected component
We look at DFS, a departure from, we traverse from right to left, so the path is 1--> 3 -> 5 -> 6 to 6, we found no way out, went back to 5, and 6 can not reach any point, it is a strongly connected component alone. Similarly, 5 is a strongly connected components. And 1--> 3 -> 4 -> 1--> 2, we can reach each other, so this is a strongly connected component.
Tarjan algorithm
Next, it is a strong communication in a common algorithm.
Tarjan algorithm is based on depth-first search algorithm of FIG., Each strongly connected component of a subtree of the search tree. When searching, the current node in the search tree unprocessed added a stack, the stack can be determined whether a stack node is strongly connected components when backtracking.
Defined DFN (u) is the order of the node number (time stamp) u search, Low (u) for the u or u sub-tree can be traced back to the sequence number of the earliest stack node.
When DFN (u) = when the Low (u), u is a root in search of all the sub-tree node is a strongly connected component.
Then demonstrate what algorithm:
从1开始DFS，把遍历到的节点加入栈中。搜索到节点u=6时，DFN[6]=LOW[6]，找到了一个强连通分量。退栈到u=v为止，{6}为一个强连通分量。
返回到5，发现DFN[5]=LOW[5]，退栈后{5}为一个强连通分量。
继续回到1，最后访问2。访问边(2,4)，4还在栈中，所以LOW[2]=DFN[4]=5。返回1后，发现DFN[1]=LOW[1]，把栈中节点全部取出，组成一个连通分量{1,3,4,2}。
所以，三个强连通分量全部都找出来了。
模板如下：
1 void Tarjan(int u){ 2 dfn[u]=low[u]=++num; 3 st[++top]=u; 4 for (int i=fir[u]; i; i=nex[i]){ 5 int v=to[i]; 6 if (!dfn[v]){ 7 Tarjan(v); 8 low[u]=min(low[u],low[v]); 9 } 10 else if (!co[v]) 11 low[u]=min(low[u],dfn[v]); 12 } 13 if (low[u] == dfn[u]){ 14 co[u]=++col; 15 while (st[top]!=u){ 16 co[st[top]]=col; 17 --top; 18 } 19 --top; 20 } 21 }