Weighted undirected graph

Question

Consider a Weighted undirected graph connected with 'V'  vertices and 'E' edges.What is the worst case time complexity to check if 2 particular vertices 'x' and 'y' are present in the graph, if present then calculate the minimum distance between them?

Comments

O(v) search in adjacency list and O((E+v)logV) for dijkstra's ?

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it is distance only, isn't it? not weighted shortest path. then we can use BFS na?

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Worst case O(ElogV) ??

Search and then finding shortest distance

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@Ashwin Kulkarni @junk_mayav

I was also thinking Finding 2 vertices-->O(V)
Then Dijkstra-->O(ElogV)

But, then I thought that we need only the distance between x and y then why are we computing shortest path to all vertices.

But, Is there any algo to find shortest path between just 2 vertices which can do better than dijkstra's ?

Then , here I got stuck :P.

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Haha I also got stuck exactly at this point :p

No need to find shortest for all vertices. But can't suggest any other that this :(

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dijkstra's only give one shortest path, isn't it? we need distance here. distance is the least number of edges among all the shortest paths exists between the two vertices. is there anything in dijkstra's that we get the shortest path with minimum edges?

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dijkstra's can be modified to output shortest path with minimum edges. 

https://stackoverflow.com/questions/20041931/find-the-shortest-path-with-the-least-number-of-edges

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@junk_mayavi

https://www.fi.muni.cz/~hlineny/Vyuka/GT/Grafy-lect-en-3.pdf

Distance in weighted graph is considering the weights of the edges.

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@VS thank you... I was wrong about weighted distance concept :)