Travelling salesman problem with small edge weights Are there any advantages in finding the shortest tour for the problem if edge weights are much smaller than the number of vertices?

Let's say the maximum edge weight is $n$, and the number of vertices is $n!$, and the adjacency matrix is dense and all distances $d_{ij}$ satisfy $0 \le d_{ij} \le n$. Are there any known methods which work well in similar cases?
Travelling salesman remains NP-complete even if you restrict to two different edge weights $a&lt;b$.  Reduce from Hamiltonian cycle by giving weight&nbsp;$a$ to every edge and weight&nbsp;$b$ to every non-edge.  Then the resulting complete graph has a TSP tour of length&nbsp;$na$ (where $n$&nbsp;is the number of vertices) if, and only if, the original graph has a Hamiltonian cycle.