This is not really an answer but using a naive search you can get
. 1 . 2 3 3
4 1 5 2 3 3
. 1 5 2 . 6
7 1 5 2 8 6
. 1 . 2 8 6
Basically you start from the top left corner and use it as the top left corner of your next rectangle, then you check how far you can extend it to the right and down, then find the topmost and leftmost cell of the remaining bits and so on.
This is probably very ineffective in some cases but it's quick as you don't have to recalculate anything..
