Largest possible cost of a shortest path in which every edge's length is either -1, 0, or 1 I'm doing an online course in which I'm struggling with the following (multiple-choice) question:

Suppose we run the FLoyd-Warshall algorithm on a directed graph $G =(V,E)$ in which every edge's length is either $-1,0,$ or $1$. Suppose further that $G$ is strongly connected, with at least one $u$-$v$ path for every pair $u,v$ of vertices. The graph $G$ may or may not have a negative-cost cycle. How large can the final entries $A[i,j,n]$ be, in absolute value? Choose the smallest number that is guaranteed to be a valid upper bound. (As usual, $n$ denotes $|V|$.) [WARNING: for this question, make sure you refer to the implementation of the Floyd-Warshall algorithm given in lecture, rather than to some alternative source.]

$+\infty$

$n^2$

$2^n$

$n-1$



I'm furthermore told that $n-1$ shouldn't be selected, with the hint &quot;Experiment with graphs that have negative-cost cycles&quot;.
Here is the Floyd-Warshall algorithm as described in the lectures:

Let $A$ be a 3-D array, indexed by $i,j,k$.
Define the base cases:
$$ A[i,j,0] = \begin{cases}0 &amp; i = j \\ c_{i,j} &amp; (i,j)\in E \\ +\infty &amp; \text{else}\end{cases}$$
For k = 1 to n
    For i = 1 to n
        For j = 1 to n
            A[i,j,k] = min(A[i,j,k-1], A[i,k,k-1] + A[k,j,k-1])

Correctness: From optimal substructure and induction, as usual.
Running time: $O(1)$ per sub-problem, $O(n^3)$ total.

In the course's nomenclature, $n$ denotes the total number of vertices in the graph, so $A(i, j, n)$ is the shortest path from node $i$ to $j$ which may use all $n$ nodes.
As I understand it, a shortest path between any two nodes has at most $n - 1$ edges, so if the maximum cost of an edge is 1, I would expect the maximum cost of a shortest path to be $n - 1$. I don't see how the presence of negative-cost cycles could change this?
a shortest path between any two nodes has at most n - 1 edges


This is only true for simple paths (paths with no repeat nodes). If there is a negative-cost cycle, you can take the cycle multiple times to get a shorter path.