It's even more epic because Petr predicted it before the rating update. (He asked admins about this possible problem)

Something like this? Or even shorter?

if (w > h) swap(w,h);
if (w <= 0) return 0;
if (h <= 1) return 1;
int n=0;
while ((1ll<<n) < h) n++;
return max(n, min(n+1, calc(w, h-(1ll<<(n-1)))+1));

Let's try:

ll sum = (w + h) - 1;
int n = 0;
while (sum >> n) n++;
return n;

Let's draw a directed graph with n vertices and n edges: from x to f(x). Let's look at the required function g(x) as a mapping on this graph: we map each vertex to some other vertex. The condition f(g(x))=g(f(x)) means that mapping should be such that whenever there's an edge from a to b, there should be an edge from g(a) to g(b), so we're kind of embedding the graph onto itself.

The graph has exactly one outgoing edge for each vertex, so its structure is some cycles with some directed trees growing into the cycles. Consider a vertex X on the cycle of length A. Where can it be mapped? Because of the "edges are mapped to edges" condition, it must be mapped to another vertex on a cycle, and the length of that cycle should divide A. Let's just iterate over all vertices Y that X can be mapped to.

For each particular Y, the mapping of all vertices on X's cycle is uniquely determined. As for the trees growing into this cycle, we can compute the number of ways to embed them using the following dynamic programming: we just count the number of ways to embed a subtree going to each vertex P in such a way that P is mapped to Q for each vertex Q. When we want to embed a subtree in such a way that P maps to Q, that means that all predecessors of P must be mapped to predecessors of Q, so we can use our dynamic programming results to compute the number of ways to map each branch of the tree, and multiply them.

The working time of the solution is O(N^2).