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Given an undirected（no lengths） graph G=(V,E) with |V|=n and |E|= m, and two vertices v,w, find the algorithm that outputs the number of shortest v-w-paths in G. The running time should be O(m+n)

I have been working with this problem but have difficulties to let the running time be O(m+n)

Since this graph is both undirected and unweighted, I have tried this way. Use BFS to determine the length of the shortest v-w-path. Then use DFS to find the number of the v-w-shortest paths such that two nodes are connected and the length of path equals to the output of BFS. But the running time of this plan is O(m+n)+O(m+n).

Also I've tried to modify the Dijkstra algorithm. Store the length of the shortest path and the number of the shortest path when there is a node adding to the set of visited nodes. And I 'm stuck with the computation of running time.

Bowen Sun
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2 Answers2

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This question may be looking for a modification of Dijsktra's algorithm. With Dijkstra's algorithm you maintain, for each node, the length of the shortest path to that node, and you update this at a node based on the shortest path to a neighbouring node and the length of the simple link from that neighbouring node to the node in question.

At each node you could keep, as well as the length of a shortest path to it, a table of the nodes that can precede it on a shortest path to it, and the number of shortest paths to that node, which should be the sum of the number of shortest paths to these neighbours. When you find a new shortest path to a node you either delete all of this information (if the new shortest path is shorter than before) or update the entry in that table for the second last node in the path, if the new path is the same length as the previous shortest path to that node.

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mcdowella
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  • Thank you so much. That's very helpful. Could you please give me some more information of the running time? I'm not good at determining the running time of a given algorithm and I really want to learn how to do it. Can you help me with it? – Bowen Sun Sep 13 '14 at 23:55
  • Going through the complexity info at Wikipedia, you can see that you need to visit each vertex only once, and that, for large graphs, you spend most of your time running priority queue calculations to work out which vertex to visit first. When you visit a vertex you need to add or modify a table entry and increment a sum of paths or perhaps clear a table entry. Clearing a table might mean deleting many entries but if you account for this cost as you create each entry this is just a small amount of cost per entry. So I doubt if all this extra work will affect the O(n) cost at all. – mcdowella Sep 14 '14 at 05:07
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You can find the number of different paths using modified BFS in O(N). Here is the solution.

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Zoli
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