The opaque rooms divide the board into vertical and horizontal segments. Inside of those segments, you can place only one fighter, and each cell of the board will belong to exactly one horizontal segment and one vertical segment. So you can build a bipartite graph where the left part are horizontal segments, the right part are vertical segments and if a cell belongs to horizontal segment h and vertical segment v, there's an edge [u, v] with capacity 1. The answer will be the max flow of that graph.

You must be careful with the implementation though, because it might give you TLE.