Is this variation of the traveling salesman problem NP-hard Consider the following setting. You have $n$ cities, and there is a cost to travel from a city $i$ to a city $j$ given by $c_{ij}&gt;0$ where $c_{ij}\neq c_{ji}$. Moreover, if you are traveling to city $i$, you are allowed to stay exactly $d_i&gt;0$ days. The problem is:
Find a travel schedule, this is: a number $m$ and a sequence of different cities $i_1,\dots,i_m\in\{1,\dots,n\}$, such that the cost $\sum_{k=1}^{m-1} c_{i_k,i_{k+1}}$ is minimal and that the whole trip lasts at least $L$, this is $\sum_{k=1}^m d_{i_k}\geq L$.
There are many similarities between this problem and the traveling salesman one. The key differences are the non symmetric costs $c_{ij}\neq c_{ji}$ and that instead of having to travel to all cities once, one just needs to travel to enough cities to make the trip last enough.
I understand that to show that this problem is NP-hard, I need to show that another NP-hard problem can be reduced to this one in polynomial time. I suspect that the original traveling salesman problem may work to do so through some clever trick I haven't found yet. Moreover, since there is plenty of literature on the traveling salesman problem, I suspect a similar problem to this may already be shown to be NP-hard.
My question is: Do you see how to show this problem to be NP-hard? Or, do you recognize this problem to be something standard in some piece of literature you can point me out?
I think if you choose all $d$'s to be exactly $1$ and choose $L$ to be exactly $n$ (number of cities), any solution must visit all cities, so the reduction will work out.