The idea is to build a graph with only the edges that are candidates for the shortest path from $v$ to $u$ assuming that $f$ is the required solution.
If during the process we find an edge that lead to a shorter path then we return false.
Otherwise, at the end, we check that all the nodes are still connected (with a DFS) hence that there exists a path with the required weight.
if $f(v)\neq 0$ then return false
$\forall e=(v_1,v_2)\in E $ do
if $f(v_2) > f(v_1)+w(e)$ then return false
if $f(v_2) < f(v_1)+w(e)$ then remove $e$ from $E$
done
build $cV$ the set of reachable vertices in the new graph (with a DFS)
return $V=cV$
Correction:
- if there exists $u\in V$ such that $f(u)>\delta(v,u)$ then there exists $e=(v_1,v_2)\in E$ such that $f(v_2) > f(v_1)+w(e)$. Let $e_1,...,e_n$ be the shortest path from $v$ to $u$ if $f(v_{i+1}) \leq f(v_i)+w(e_i)$ then $f(u)\leq \delta(v,u)$ contradiction.
Hence the algorithm return false.
otherwise if there exists $u\in V$ such that $f(u)<\delta(v,u)$ then $u$ is disconnected from $v$ in the resulting graph.
Hence the algorithm return false. $u$ is connected, hence there exist a path $e_1,...,e_n$ from $v$ to $u$ such that $f(v_{i+1}) = f(v_i)+w(e_i)$ hence $f(u)\geq \delta(v,u)$ contradiction.
otherwise if for all $u\in V$ $f(u)=\delta(v,u)$ then all the edges belonging to the shortest path verify $f(v_{i+1}) = f(v_i)+w(e_i)$ hence are still in the graph at the end hence all the vertices are still connected. Also there are no edges such that $f(v_2) > f(v_1)+w(e)$ otherwise $f(v_2)>\delta(v,v_2)$.
Complexity:
$O(|E|)$ to build the new graph plus $O(|E|)$ for the DFS plus $O(|V|)$ for set equality hence $O(|V|+|E|)$.
