Not able to devise a concrete solution for the given graph query
Your problem is very similar to interval scheduling, and it seems it can be solved similarly using a greedy approach. Note first that you can solve each row separately. For each row, you scan the row from left to right. Assign each policeman to the leftmost thief that he can capture and isn't already taken (if any). Intuitively, this should work, and you can try proving that it does using the standard proof method of comparing the greedy solution to an optimal one.