A mother vertex in a directed graph G = (V,E) is a vertex v such that all other vertices G can be reached by a directed path from v Give an O(n+m) algorithm to test whether graph G contains a mother vertex.
(c) from Skiena manual
Found only O(n(n+m)) way
When googling I actually found the answer here . If this is homework, you should think twice before peeking :)
Algorithm::
a) Do DFS/BFS of the graph and keep track of the last finished vertex 'x' .
b) If there exist any mother vertex, then 'x' is one of them. Check if 'x' is a mother vertex by doing DFS/BFS from vertex 'x'.
Time Complexity O(n+m) + O(n+m) = O(n+m)
step1 . Do topological sorting of vertices of directed graph.
step2 . Now check whether we can reach all vertices from first vertex of topologically sorted vertices in step 1.
To perform a step 2, again initialize array discovered[i] to false and do dfs startin from first node of topologically sorted vertices.
If all vertices can be reached, then graph has mother vertex, and mother vertex will be the former of topologically sorted vertices.
time complexity: step1 takes O(n + m)
, step 2 takes O(n + m)
so total O(n+m) + O(n+m) = O(n+m)
I saw the solution. I dont think we need to find SCC. Just do a DFS from a random vertex and then do the DFS from the vertex with last finish time. If there is a mother vertex then it has to be this.
we can find the mother vertex in O(m+n) using KOSARAJU's algorithm
See this link for DFS using recursion and stack to track the visiting time of all vertices
Here is the algorithm fo finding the mother vertex in a Graph , G = (VE) :
Do DFS traversal of the given graph. While doing traversal keep track of last finished vertex 'v'. This step takes O(V+E) time.
If there exist mother vertex/vertices, then 'v' must be one (or one of them). Check if v is a mother vertex by doing DFS/BFS from v. This step also takes O(V+E) time.
geeksforgeeks.com的这篇文章是一个很好的起点/整体答案: 在图形中查找母顶点
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