Faster second-best MST algorithm?

Question

I am struggling with this.

We can get MST using Kruskal\'s algorithm or Prim\'s algorithm for the MST.

And for \"second-best\" MST, I can:

1. first g

Answer 1

Let V be the vertex set and E be the edge set.

Let T be the MST obtained using any of the standard algorithms.

Let maxEdgeInPath(u,v) be the maximum edge on the unique path in T from vertex u to vertex v.

For each vertex u perform BFS on T. This gives maxEdgeInPath(u,x) for all x belonging to V-u.

Find an edge (x,y) which does not belong to T that minimizes w(x,y) - w(maxEdgeInPath(x,y))

Weight of 2ndMST is W(T) + w(x,y) - maxEdgeInPath(x,y)

This is based on the algorithm provided in this link. I'm not sure if this is correct and I hope someone would add a proof here.

Compexity: To compute BST for 1 vertex takes O(V+E) = O(V) as E = V-1 in T Hence overall time complexity is O(V^2)