[Solved]: Find the lightest weight of a path between $u$ to $v$ that contains no more then $k$ edges (If there's one)

Problem Detail: 

Given a directed and strongly connected graph $G=(V,E)$, weight function $w: E \to \mathbb{R}$ and two distinct vertices $u,v \in V$. We know that there aren't negative cycles. I need to find algorithm, efficient as possible,such that for every value of $k$, $2 \leq k \leq |V|-1$, it will find the lightest weight of a path between $u$ to $v$ that contains no more then $k$ edges (If there's one).

I don't know what to do. I want to use Bellman Ford somehow, I just don't sure how.

Thanks a lot.

Asked By : Jozef

Answered By : Raphael

Consider the dynamic programming formulation of Bellman-Ford:

$\qquad \displaystyle \mathrm{bf}(i,j) = \begin{cases} 0 &, i = 0 \land j = s \\ \infty &, i = 0 \land j \neq s \\ \min\limits_{k \in V}\, \big(\, \mathrm{bf}(i-1,k) + c(k,j)\,\big) &, \text{else} \end{cases}$

What meaning does $\mathrm{bf}(i,j)$ have, that is what does DP-matrix entry $\mathrm{bf}[i,j]$ contain?

If you unfold the definition (i.e. proof by induction) you see that $\mathrm{bf}(i,j)$ is the weight of the shortest path from starting node $s$ to $j$ with at most $i$ edges -- exactly what you are looking for.

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Question Source : http://cs.stackexchange.com/questions/3352

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