This is the code that I have studied for finding SCC and I have several doubts.

vi dfs_num, dfs_low, S, visited; void tarjanSCC(int u) { dfs_low[u] = dfs_num[u] = dfsNumberCounter++; S.push_back(u); visited[u] = 1; for (int j = 0; j < (int)AdjList[u].size(); j++) { ii v = AdjList[u][j]; if (dfs_num[v.first] == UNVISITED) tarjanSCC(v.first); if (visited[v.first]) // condition for update
dfs_low[u] = min(dfs_low[u], dfs_low[v.first]); } if (dfs_low[u] == dfs_num[u]) { printf("SCC %d:", ++numSCC); while (1) { int v = S.back(); S.pop_back(); visited[v] = 0; // why visited[v]=0 is done? printf(" %d", v); if (u == v) break; } printf("\n"); } } // inside int main() dfs_num.assign(V, UNVISITED); dfs_low.assign(V, 0); visited.assign(V, 0); dfsNumberCounter = numSCC = 0; for (int i = 0; i < V; i++) if (dfs_num[i] == UNVISITED) tarjanSCC(i);

//code is from steven and felix book.

Can one node be contained in two SCCs?

5 6

1 4

1 3

2 4

3 4

4 5

5 1

consider this testcase — 5 is number of nodes and 6 is the number of directed edges. 1 4 means there is an directed edge from 1 to 4.

if I assume that one node is contained in only one SCC then 1->4->5 will be considered or 1->3->4->5 for 1, 4 and 5.

And how we update dfs_low[u] in case of visited node as dfs_low[u]=min(dfs_low[u], dfs_low[v.first]) or as dfs_low[u]=min(dfs_low[u], dfs_num[v.first]). Which one is correct and why ?

Any help would be appreciated.