Dijkstra single-source shortest path $\Omega(n\log n)$? If I have a directed graph with $n$ weighted edges, is it possible to prove that Dijkstra's single-source shortest path algorithm takes $\Omega(n\log n)$ in the worst case?

I know heaps reduce Dijkstra's algorithm to $Ο(m \log n)$ and can be stored in an array.
Rooted, binary, as complete as possible tree with all nodes satisfy Heap property:
    node &lt;= all children of node.. Is this the right direction? 

**I have an implementation that runs $O( (n + m)\log n)$, could it work?
Let $m$ be the number of vertices and $n$ the number of edges. The best you can get is $O(n + m \log m)$ by using a Fibonacci heap.

A Fibonacci heap allows decreasing the key of an element in amortized constant time. For each edge in the graph, Dijkstra's algorithm executes at most one decrease key operation. All these operations take time $O(n)$ in total. On top of that we have $m$ minimum extractions, which take time $O(m \log m)$ in total. Therefore the final complexity is $O(n + m \log m)$.