$$$*2600$$$ but I personally think it's $$$*2300$$$ to $$$*2500$$$

Multisource BFS and simple BFS

Does placing a block in an empty cell matter?

Nope, it doesn't matter.

This task is simple, since blocking every cell is the optimal solution.

This is because it cannot escape anyway so just cover everything.

Max number of block placed $$$=$$$ Total number of '.' $$$-$$$ Min blocks needed to move to an exit.

This task is also relatively simple, since Max number of block placed $$$=$$$ Min blocks needed to move to an exit. We can simply find the shortest path from $$$V$$$ to the only exit and just cover the other unused blocks.

This can be done easily using a $$$BFS$$$ and check minimum distance to every reachable exit.

It is easy to observe that we don't need all exits, taking more than that is simply using up more spaces. We only need 2.

Maybe we consider moving to a cell $$$(i,j)$$$ first then we split into 2 paths going to 2 exits?

This is task is what makes the problem hard.

We notice that we can always consider moving to a cell $$$(i,j)$$$ first then then we split into 2 paths going to 2 exits.

So our path would be $$$V \rightarrow (i,j) \rightarrow A,B$$$ where $$$A$$$ and $$$B$$$ are the closest exits. So no of cells used $$$=$$$ $$$dis(V,(i,j)) + dis((i,j),A) + dis((i,j),B)$$$. In order to minimize this cells used, and the first $$$dis(V,(i,j))$$$ is constant, we should minimize the last $$$dis((i,j),A) + dis((i,j),B)$$$. This is sum of distances of the closest two exits.

So, we need to find two closest exits to a cell and that can be performed using a multisource $$$BFS$$$. Also note that we also have to make sure the same exit isn't recorded twice. That inspires us to let $$$who_{x,y,0/1} = $$$ id of exit. The rest is just implementation.

Note: make sure to set who is going to a cell before throwing it into the queue, or else you would get MLE like me.

Amazing problem, somehow looks familiar for me.

Anyways, I had the idea of converting the grid into a graph with $$$N*M$$$ nodes and adding an edge of where it came from in the $$$BFS$$$ from $$$V$$$. Then convert the graph to $$$DAG$$$ then we can perform $$$DP$$$ on the $$$DAG$$$. Then I noticed that it is quite hard to do so, while I'm currently still doubting it'll work or not.

It is rather implementation heavy for me so even though I had the idea from like the beginning, I needed many submissions to pass.