You don't make any topological sort in the original algorithm. But in the case of an a-cyclic graph, then you can decrees the running time to O(V) (while the original running time is O(|V|*log(|V|)). The reason is that you sort in O(|V|) time, and then you can use that order, and don't need any heap (or priority queue).
So the over all time decreases to O(|V|).
Dijkstra's algorithm doesn't appear to require a topological sort. Perhaps doing so avoids a bug you have in your implementation.
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