This is part of a larger project that I am working on. It is not a contest problem.

Consider a complete graph $$$G = (V, E)$$$, where $$$V = {0, 1, 2, \ldots, n}$$$. The graph $$$G$$$ is a complete graph, so we can go from Vertex $$$i$$$ to Vertex $$$j$$$ for all $$$i, j \in V$$$. At each vertex $$$v \in V$$$, there is a task that we must complete. The task at vertex $$$v$$$ takes exactly $$$q_v > 0$$$ minutes, and we must complete all tasks. We start at Vertex $$$0$$$ and we have $$$q_0 = 0$$$.

Of course, it will take us exactly $$$\sum_{i = 1}^{n} q_i$$$ total time for us to complete all of the tasks on our own (it takes $$$0$$$ time to traverse edges). But now let's suppose that we have one helper. We can "dispatch" this helper at any one node (say $$$v$$$), and we can leave the node, go work on other tasks, and pick up the helper after $$$q_v$$$ minutes to drop it off at another node. The helper cannot move on its own; it must be picked up and dropped off by you. Essentially, we can "parallelize" this process to some extent.

Is there some sort of algorithm that allows one to find the best strategy that arises when we are allowed one helper?

I've already thought about greedy strategies (i.e, always drop the helper at the vertex that has the largest $$$q_i$$$, and go work on the quickest tasks available before coming back to pick up the helper), but there's no reason to believe that this is optimal. I thought that a DP on trees approach may help here, but I wanted to know what others thought.

Thank you.