计算无向图中所有具有约束的顶点对

Question

I'm struggling the following an algorithmic puzzle:

given a graph with N vertex and N edges I've to count all pairs of vertex (A,B) with the following properties: A>B and exists at least one path made of all vertex labeled with numbers between A and B...

As example (1,5) is a valid pair of vertex if the following path exists: (1,2,3,4,5) but would be ok even if the following would exists (1,5,2,3,4) or (1,3,4,3,2,3,5)... I'm not interested in the length of the path or its order it just have to include all vertex which label is >= A and <= B. I've tried with modified bfs and dfs but had no success.

can some one help?

give some hint?

thanks

Answer 1

Delete all edges which are connecting two nodes with not consecutive numbers. (eg a edge connecting node 2 with node 6, but not a edge from node 2 to node 3). In the modified graph, each node has either degree 0, 1 or 2.

Now traverse the graph starting at node 1. Store all visited nodes in a list. If you reach a node with degree 1, start again on node 2. (and then on node 3 and so on).

Consider the following example output for a graph with 7 nodes and edges E={{1,2},{2,3},{3,4},{4,5},{6,7}}. Please note that 1..5 denotes a list with the following elements {1},{2},{3},{4},{5}.

node | list | valid pairs

1 | 1..5 | 4 {(1,5),(1,4),(1,3),(1,2)}

2 | 2..5 | 3 {(2,5),{2,4),(2,3)}

3 | 3..5 | 2

4 | 4..5 | 1

5 | 5..5 | 0

6 | 6..7 | 1

7 | 7..7 | 0

This should answer your question. Sorry for the bad formatting of the table. As you can see in the table, you can calculate the valid pairs for node 2,3,4 and 5 once you have all pairs for node 1. So there is some space to improve my algorithm.