Let $T=(\mathcal{V},\mathcal{E})$ be an udirected acyclic graph and $|\mathcal{V}|=n$. Let $\mathcal{V'}$ be $\mathcal{V'}\subset \mathcal{V}$ where $|\mathcal{V'}|=2m\leq n$. There are $2m \choose 2$ pairs of nodes in the set $\mathcal{V'}$. For each pair of nodes $(w,w')$ in the set $\mathcal{V'}$, there is a unique path form $w$ to $w'$ lets call it $\text{path}(w,w')$ and it contains all the edges in the path of $w,w'$. I would like to find a collection of pairs of nodes of $\mathcal{V'}$, where each node of $\mathcal{V'}$ appears in one pair exactly, thus we have $m$ pairs in total in the final desired collection. Lets call that collection of pairs $F(\mathcal{V'})$, also, $F(\mathcal{V'})$ should be found such that the total number of edges in the paths $\text{path}(w,w')$ is minimum, where $(w,w')\in F(\mathcal{V'}):$$$\min_{F(\mathcal{V'})} \sum_{(w,w')\in F(\mathcal{V'})} |\text{path}(w,w')|. $$ Notice that a necessary condition for the choice of pairs $F(\mathcal{V'})$ is that the paths should be disjointed (have no common edges), otherwise we have a total number of edges greater than the minimum possible.
I would like to know what the algorithm is that finds $F(\mathcal{V'})$ in polynomial time. Is this related to the Hamiltonian cycle problem? My graph is acyclic which should help to get a polynomial time algorithm. *By disjointed paths I mean paths with no common edges, they may have common vertices. **By minimum path cover I mean the total number of edges in the collection of paths is minimum since the number of paths is always chosen to be $m$.
