Given a vertex, find a path between it and each other vertex that minimizes the maximum weight in the path

Question

I'm revising for an upcoming exam and was wondering if someone could help me with a practice problem:

We model a set of cities and highways as an undirected weighted graph $G = (V,E,l)$, where the vertices $V$ represent cities, edges $E$ represent highways connecting cities, and for every undirected edge $e = \{v, w\}\in E$ the number $\ell[e]$ denotes the number of litres of fuel that your motorcycle needs in order to ride the distance between cities $v$ and $w$. There are fuel stations in every city but none on the highways between the cities. Therefore, if the capacity of your motorcycle's fuel tank is $L$ litres, then you can only follow a path (a sequence of adjacent highways), if for every highway $e$ on this path, we have $L\geq \ell[e]$, because you can only refuel in the cities.

Design an algorithm with a running of time of $O(m\log n)$ that, given an undirected weighted graph $G=(V,E,\ell)$ modelling a setting of cities and highways, and a city $s \in V$ as inputs, computes for every other city $v \in V$, the smallest fuel tank capacity $M[v]$ that your motorcycle needs in order to be able to reach city $v$ from city $s$.

I'm not sure how to solve this problem, but I do have a few ideas. I thought of possibly running Kruskal's algorithm since that would give me the MST of the graph. I'm not sure how I would then efficiently get the answer I need to produce.

Alternatively, I was thinking of using dynamic programming in some way since that's one of the things that we use a lot in the course, but I'm not really sure how to go about it.

Answer 1

Let $T = \{s\}$, and let $Q$ be a new empty priority queue. Insert each edge incident on $s$ in $Q$. While there are still nodes with $d$ from $s$ yet to be computed, extract and remove the minimum from $Q$.

If such edge connects a node $u$ in $T$ with a node $v$ not in $T$, add each edge incident on $v$ that still isn't in $Q$ to $Q$, then set $M[v]$ to $\max\{M[u], w(u, v)\}$.

Otherwise, disregard the extracted edge and extract more edges until you find a suitable one.

The overall complexity is $O(|E| \log |E|) = O(|E| \log |V|^2) = O(|E| \log |V|)$.

To informally prove correctness, we can observe that whenever we set $M[v]$ to a certain value $x$, we can certainly reach $v$ from $s$ with a path in which each edge weighs at most $x$. On the other hand, no better path exists, because we already gave each edge lighter than $x$ a chance to connect to $s$.

Answer 2

It's modification of Dijkstra algorithm. In Dijkstra you take minimum of distance from beginning to visited plus distance to not visited adjacent, here you take minimum of max { visited, not visited adjacent }.