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
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.