I am given a directed acyclic graph $G = (V,E)$, which can be assumed to be topologically ordered (if needed). Each edge $e$ in G has two types of costs - a nominal cost $w(e)$ and a spiked cost $p(e)$. I am also given two nodes in $G$, node $s$ and node $t$.

The goal is to find a path from $s$ to $t$ that minimizes the following cost: $$\sum_e w(e) + \max_e \{p(e)\},$$ where the sum and maximum are taken over all edges of the path.

Standard dynamic programming methods show that this problem is solvable in $O(E^2)$ time. Is there a more efficient way to solve it? Ideally, an $O(E\cdot \operatorname{polylog}(E,V))$ algorithm would be nice.

This is the $O(E^2)$ solution I found using dynamic programming, if it'll help.

First, order all costs $p(e)$ in an ascending order. This takes $O(E\log(E))$ time.

Second, define the state space consisting of states $(x,i)$ where $x$ is a node in the graph and $i\in \{1,2,...,|E|\}$. It represents "We are in node $x$, and the highest edge weight $p(e)$ we have seen so far is the $i$-th largest".

Let $V(x,i)$ be the length of the shortest path (in the classical sense) from $s$ to $x$, where the highest $p(e)$ encountered was the $i$-th largest. It's easy to compute $V(x,i)$ given $V(y,j)$ for any predecessor $y$ of $x$ and any $j \in \{1,...,|E|\}$ (there are two cases to consider - the edge $y\to x$ is has the $j$-th largest weight, or it does not).

At every state $(x,i)$, this computation finds the minimum of about $\deg(x)$ values. Thus the complexity is $$O(E) \cdot \sum_{x\in V} \deg(x) = O(E^2),$$ as each node is associated to $|E|$ different states.

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    $\begingroup$ @BernardoSubercaseaux, you can add a constant to all $p(e)$ to achieve this. $\endgroup$ – Dmitry Aug 2 at 23:02
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    $\begingroup$ I spent way too much time on this, you can get $O(|E||V|W)$ for integer weights where the maximum weight is $W$, and you can do a bit better if you allow approximate algorithms. Both are using decremental/incremental single-source shortest path algorithms, so you can't really expect more fruitful stuff to come from that direction. $\endgroup$ – orlp Aug 3 at 3:34
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    $\begingroup$ This problem can't be solved in $O(|E|^{2- \varepsilon})$ under a widely used conjecture (k-Cycle Hypothesis). The $k$-Cycle Hypothesis can be stated in the following way. For a fixed positive integer $k$ consider the problem $k$-Cycle: detect whether there is a cycle with length exactly $k$ in the given directed unweighted graph. Then, the $k$-Cycle Hypothesis states that, for any $\varepsilon > 0$, there exists a fixed integer $k$, such that there is no algorithm that solves $k$-Cycle in $O(|E|^{2 - \varepsilon})$ time. $\endgroup$ – Kaban-5 Aug 3 at 15:33
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    $\begingroup$ Now, let's reduce $k$-Cycle instance $(V, E, k)$ to our problem. We will construct a DAG with $(k + 1) |V| + 2$ vertices and $k |E| + 2|V|$ edges. The vertex set will consist of a source vertex $t$, a sink vertex $t$ and vertices $(v, \ell)$ for each $v \in V$, $\ell \in [1, k + 1]$ ($k + 1$ layers, each being a copy of $V$). There will be edges $(v, \ell) \to (u, \ell + 1)$ for all $(v, u) \in E$, $\ell \in [1, k]$ with $p = w = 0$, edges $(s, v)$ for every $v \in V$ with $w = v, e = 3|V| - 2v$ and edges $(v, t)$ with $w = v, e = 3|V| - 2v$. $\endgroup$ – Kaban-5 Aug 3 at 15:40
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    $\begingroup$ @Dmitry: I meant that, for each edge in the original graph, we create $k$ copies of said edge with their $w$ and $p$ parameters being equal to $0$ (so a path of length $k$ in the original graph corresponds to a path from the first layer to the last layer). Moreover, we add $|V|$ edges from $s$ to the first layer and $|V|$ edges from the last layer to $t$, $s \to t$ paths in the new graph correspond to paths of length $k$ in the original graph. Finally, we need to choose parameters $w$ and $p$ for those "$s \to$ first layer", "last layer $\to t$" edges. $\endgroup$ – Kaban-5 Aug 3 at 17:47

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