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A longest path maximizes path length, either the number of edges or the sum of specified edge weights, under the stated endpoints and simple-path convention. In a general graph, imposing vertex simplicity makes the optimization computationally difficult; a Directed acyclic graph admits a direct dynamic programming solution. Allowing arbitrary walks instead requires care with reachable positive cycles. The convention permitting a zero-edge path gives a possible value zero even when all nonempty paths have negative weight.

Find a topological sorting of the Directed acyclic graph. Initialize $L[s]=0$ and $L[v]=-\infty$ for $v\ne s$. Process vertices in that order. At a reachable vertex $u$, update each outgoing edge by $L[v]\leftarrow \max \{L[v],L[u]+w(u,v)\}$, and retain a predecessor whenever an improvement is made.

Every path to a vertex ends at an earlier vertex in the topological sorting. Thus, when $v$ is processed, all possible final-edge candidates for $v$ have already been considered. Induction along the order proves that $L[v]$ is the longest path length from $s$, with $-\infty$ exactly for unreachable vertices. Signed weights cause no problem because the graph has no directed cycle. The running time is $O(|V|+|E|)$, and predecessor pointers recover a maximizing path.