Finding the "fairest" meeting point in an undirected weighted graph

Question

Given an undirected weighted graph, a start node and an end node, is there a way to (without using brute force) find a meeting point in between the start and end nodes such that:

the difference between 1) the distance from start to meeting point and 2) the distance from end to meeting point

is minimized?

I cant seem to think of any way to do this without using brute force.

Thank you

Answer 1

Just launch Dijkstra's algorithm to find distances to all nodes from the start node, then from the end node. Then find a node for which a difference between calculated distances is minimized.

Answer 2

You can use Dijkstra's algorithm, starting from both end and start nodes at the same time -- so using one priority queue. Just add as extra information to each node in the queue, whether you reached it from the start or from the end node. When you reach a node that was already visited, check whether it was visited from the "other" side (using that extra information): if so, stop the algorithm, and add up the two distances; the one you had registered on the first visit, and the one of your current, second visit.