Shortest path in matrix with obstacles with cheat paths

Question

First of all this is an assingment and I am not looking for direct answers but instead the complexity of the best solution as you might be thinking it .

This is the know

Answer 1

You can just build a new graph with nodes labeled by (N, w) where N is a node in the original graph (so a position in your matrix), and w=0 or 1 is whether you're carrying a weight. It's then quite easy to add all possible edges in this graph

This new graph is of size 2*V, not V^2 (and the number of edges is around 4*V+number(B)).

Then you can use a shortest path algorithm, for instance Dijkstra's algorithm: complexity O(E + V log(V)) which is O(V log(V)) in your case.

Answer 2

Your matrix is a representation of a graph. Without the cheat paths it is quite easy to implement a nice BFS. Implementing the cheat paths is not a big deal. Just add the same matrix as another 'layer' on top of the first one. bottom layer is 'carry', top layer is 'no carry'. You can move to the other layer only at B-points for the given cost. This is the same BFS with a third dimension.

You have n^2 nodes and (n-1)^2 edges per layer and additionally a maximum of n^2 eges connecting the layers. That's O(n^2).