Analysing Dijkstra Algorithm by using different varieties of Data Structure

Question

I want to analyse Dijkstra Algorithm by using different varieties of Data Structure.

My solution

1. Adjacency matrix to Store the Graph and Binary heap for Priority Queue.

$O(V\,\,log\,V+V^{2}\,\,log \,\,V)=O(V^{2}\,\,log \,\,V)$.

Explanation-:

In Binary Heap we have $O(log\,\,V)$ for both Extract min and decrease key.

So Extract min for $V$ vertex =$O(V\,\,log \,\,V)$ and $E$ times decrease key,but here graph is dense (Adjacency matrix) so $E=V^{2}$ so total decrease key=$O(V^{2}\,\,log \,\,V)$

1. Adjacency matrix to Store the Graph and fibonacci heap for Priority Queue.

$O(V\,\,log\,V+V^{2})=O(V^{2})$.

Explanation-:

In Fibonacci Heap we have $O(log\,\,V)$ for Extract min and $O(1)$ for decrease key.

So Extract min for $V$ vertex =$O(V\,\,log \,\,V)$ and $E$ times decrease key,but here graph is dense (Adjacency matrix) so $E=V^{2}$ so total decrease key=$O(V^{2})$

1. Adjacency matrix to Store the Graph and Sorted Linked List for Priority Queue.

$O(V\,\,*V^{}+V^{2}\,\,V)=O(V^{3})$.

Explanation-:

In Sorted Linked list we have $O(V)$ for both Extract min and $O(V)$ for decrease key.(NOT SURE)

So Extract min for $V$ vertex =$O(V^{2})$ and $E$ times decrease key,but here graph is dense (Adjacency matrix) so $E=V^{2}$ so total decrease key=$O(V^{2}*V)$

I am uncertain about point 1 and 2 and confused regarding sorted linked list's priority queue.Is there some efficent way through which i can optimize my algorithm.Can anyone please help me out to clear my confusion? Thanks in advance !

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