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    <p>You can use adjacency lists to speed your solution up (but not for dense graphs), but even then, you are not going to get O(V log V) without a Fibonacci heap..</p> <p>Maybe the <a href="http://en.wikipedia.org/wiki/Kruskal%27s_algorithm" rel="nofollow noreferrer">Kruskal algorithm</a> would be simpler for you to understand. It features no priority queue, you only have to sort an array of edges once. It goes like this basically:</p> <ul> <li>Insert all edges into an array and sort them by weight</li> <li>Iterate over the sorted edges, and for each edge connecting nodes i and j, check if i and j are connected. If they are, skip the edge, else add the edge into the MST.</li> </ul> <p>The only catch is to be quickly able to say if two nodes are connected. For this you use the Union-Find data structure, which goes like this:</p> <pre><code>int T[MAX_#_OF_NODES] int getParent(int a) { if (T[a]==-1)return a; return T[a]=getParent(T[a]); } void Unite(int a,int b) { if (rand()&amp;1) T[a]=b; else T[b]=a; } </code></pre> <p>In the beginning, just initialize T to all -1, and then every time you want to find out if nodes A and B are connected, just compare their parents - if they are the same, they are connected (like this <code>getParent(A)==getParent(B)</code>). When you are inserting the edge to MST, make sure to update the Union-Find with <code>Unite(getParent(A),getParent(B))</code>.</p> <p>The analysis is simple, you sort the edges and iterate over using the UF that takes O(1). So it is O(E logE + E ) which equals O(E log E).</p> <p>That is it ;-)</p>
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    1. COI know Kruskals algorithm, but I wanted to understand this one also :-). If I use adjacency lists for Prim then I get: The while + for loop loops over all edges and inserts them into a heap. This should be then E*log(E). Is this the best complexity you can get with this approach(using a heap, not Fibbonaci heap)?
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    2. COYeah, you check each edge at most twice (from both nodes) and the queue has E edges at maximum, which results in O(E log E). But we do not write it like that, because constant factors are irrelevant in O() notation. So O(E log E) = O(E log (V^2)) = O(E * 2 log V) = O(E log V).
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    3. COThis(the comment above) is the answer that I wanted, thank you I understand now :-)
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