Shortest path hitting a given vertex I believe this problem to be NP-Complete, but I'm unable to find any references on possible reductions. Given a weighted graph (either undirected or directed, I cannot find results for either but am interested in both), and vertices s, t, u, find the shortest path from s to t which includes u. To clarify, the path should NOT repeat edges. Hence simply combining the shortest s-u and u-t paths is not valid.
It is NP-complete for directed graphs. Even deciding if there exist a path from $s$ to $t$ using vertex $u$ is NP-complete.

The 2-node disjoint path problem is NP-hard for directed graphs$^1$. 


  Given a directed graph $G$ and vertex pairs $(s_1,t_1)$ and
  $(s_2,t_2)$, does there exist two node disjoint paths that goes from
  $s_1$ to $t_1$ and $s_2$ to $t_2$?


We can reduce this problem to your problem:
Split each vertex of $G$, so each vertex $v$ in $G$ become $v^-$ and $v^+$, and edge $uv$ become $u^+v^-$, also there are edges $v^-v^+$. Add one new vertex $u$, and edges $t_1^+u$ and $us_2^-$. Let this graph be $G'$.

There exist a path from $s_1^-$ to $t_2^+$ in $G'$ that uses $u$ if and only if there exist a node disjoint path between $(s_1,t_1)$ and $(s_2,t_2)$ in $G$.


The directed subgraph homeomorphism problem, Steven Fortune, John Hopcroft, James Wyllie.