Updating a Minimum spanning tree when a new edge is inserted

Answer 1

I believe your step Find in T the shortest path from a to b is an order E operation.

Answer 2

I think a BFS would suffice. And it's complexity is O(|V| + |E|) but in a tree |E| is less than |V| so it's O(2|V|) that is O(|V|)

Answer 3

You've got the right idea, though you can do better than BFS for the shortest-path search if you store the tree the right way.

Say one node r in T is the root (you can pick any node and BFS from there to generate this structure if you have marked the tree edges in a matrix or adjacency-list graph structure), and each other node has a parent pointer and a depth count. To find the shortest path between a and b in T:

1. Let a be the 'deeper' node; swap if needed.
2. Traverse the parent links from a until either b or r is reached, storing the path traversed, marking the nodes visited.
3. If you reach b, the shortest path is as traversed.
4. If you reach r, then also traverse from b to the root; if you reach node reached in the traversal from a to r, stop. Join the two where they meet and you have the shortest path in T.

Proof of the validity of this algorithm is left as an exercise to the reader. This is O(|V|) like BFS, but will also generally be faster. Only a few MST configurations would actually require O(|V|) time in practice. Of course, generating the parent-link tree takes O(|V|) time to begin with, so this only help in some circumstances, such as if you use an MST-building algorithm that naturally creates this structure in the process of determining the MST.

As another commenter said, note that if there is a MST for a graph it is connected, so |V| <= |E| and thus O(|V|) < O(|E|).

Also, to fix the tree in O(|V|) time, if needed, simply find the longest edge on the cycle and remove it, replacing it with the new edge. Doing this efficiently with a parent-link MST is also an exercise for the reader.