Exhaustively enumerating all possible ways of visiting the tools and chefs will time out. With clever pruning, it might be possible to get such an approach accepted, but there is a simpler way to guarantee acceptance. It is an easy modification of the O(n^2 * 2^n) dynamic programming approach to the Hamiltonian cycle problem. Let S be a subset of the tools and cooks. Let d[S,k] denote the minimum distance required to visit all locations not in S starting from location k while observing the “carry at most 2 tools at a time” rule. Of course, S should be such that if a cook c is in S, then their requested tool t is also in S (call this property *).
Given such a pair (S,k), we can easily determine how many tools the Chef is holding at location k after visiting locations in S. To compute d[S,k], we recursively try to compute d[S+z,z] for each location z not in S as long as S+z still has property * and doesn’t leave the Chef holding more than 2 tools.
Then, d[S,k] is simply the minimum over d[S+z,z] + dist(z,k) over all valid z in S. The number of states is at most n * 3^n since each cook-tool pair has 3 states (neither visited, tool visited, or tool and cook visited) and for a fixed S and a fixed cook-tool pair, at most one of the cook or tool may appear as the location k in a valid pair (S,k).