I was trying to solve the following problem using Burnside's lemma, but when testing the solution for a $12\times 12$ matrix involving $20$ possibilities for each matrix element, it had an awful run time.
Does this problem have a better solution than the Burnside's lemma, so it can give the result with a better time complexity?
Suppose that a matrix is equivalent to another one if it can be obtained by swapping the rows and/or columns of the other one. The goal is to find the number of inequivalent $h\times w$ matrices with each of its elements having $s$ possible choices.
1 5 0 0
would be equivalent to itself and to:
0 0 1 5
and would also be equivalent to:
0 0 5 1
5 1 0 0