Some information about task DEPOT

The 25 test cases have the following characteristics.

Case   N  Shape              P  Note
----  --  -------       ------  ----
  1    3  3                  1   1, 2, 3, 4
  2    3  1 1 1              1   1, 2, 3, 4
  3    3  2 1                2   1, 2, 3, 4
  4    4  2 2                2   2, 3, 4
  5    5  2 1 1 1            4   2, 3, 4
  6    6  3 2 1             16   2, 3, 4 (largest output for N=6)
  7    7  3 2 1 1           35   2, 3, 4 (largest output for N=7)
  8    7  2 2 2 1           14   2, 3, 4
  9    8  4 3 1             70   2?, 3, 4
 10    8  4 2 1 1           90   2?, 3, 4 (largest output for N=8)
 11    8  3 2 2 1           70   2?, 3, 4
 12    9  3 2 2 2           84   3, 4
 13    9  4 3 2            168   3, 4
 14    9  4 2 2 1          216   3, 4 (largest output for N=9)
 15   10  4 2 2 2          300   3, 4
 16   10  4 3 2 1          768   3, 4 (largest output for N=10)
 17   10  5 2 2 1          525   3, 4
 18   11  6 5              132   3, 4
 19   11  5 3 2 1         2310   3, 4 (largest output for N=11)
 20   12  5 4 2 1         5775   3, 4
 21   12  5 3 2 1 1       7700   3, 4 (largest output for N=12)
 22   13  8 1 1 1 1 1      792   3, 4
 23   13  2 2 2 2 2 2 1    429   3, 4
 24   13  5 3 2 2 1      21450   3, 4 (largest output for N=13)
 25   13  5 4 2 1 1      21450   3, 4 (largest output for N=13)

where
  N     = number of containers in depot (from input)
  Shape = vector of subsequent row lengths (sum = N; input)
  P     = number of possible orders
  Note  = 1, solvable by pre-tabulating small cases
          2, solvable by enumerating permutations, building depots, and test
          3, solvable by straightforward top-down backtracking
          4, solvable by recursively breaking down the depot in reverse order

In general, the number of possibilities goes up with N as N!/sqrt(N).
There also exist fast heuristic approaches that will find some but not all
possibilities, and that may get a partial score.