A topological ordering depends on the direction of edges, not on shortest-path distances.
For every edge, $u \rightarrow v$ a topological ordering requires $u$ before $v$
But it is possible for $v$ to have a shorter distance from the source than $u$ because $v$ may have another shorter path.
Consider this DAG:
$s \rightarrow v$ with weight $1$
$s \rightarrow u$ with weight $10$
$u \rightarrow v$ with weight $1$
The shortest-path distances are:
$d(s)=0$
$d(v)=1$
$d(u)=10$
Therefore, ordering vertices by increasing distance gives:
$s,v,u$
But the graph contains:
$u \rightarrow v$
So every valid topological ordering must place:
$u$ before $v$
The distance ordering puts $v$ before $u$.
Therefore it is not a valid topological ordering.
Answer: B
