you can use Johnson with binary min-heap implementation that is running in
O(V*E*log(V))

2010/8/16 Mikhail Dektyarev <[email protected]>

> No. O(V^3 + V^2*log(V)) = O(V^3).
> So Dijkstra always is not worse than Floyd (asymptotically). In practice,
>  one operation in Floyd's algorithm is much simpler than in Dijkstra, so if
> your graph is close to complete, Floyd's algorithm will be faster.
>
>
> On Mon, Aug 16, 2010 at 1:32 AM, Ricardo Hahn <[email protected]>wrote:
>
>> It depends on the graph.
>>
>> If the graph is complete, E = O(V²),
>> then the dijkstra would become O(V³ + V (V²) lg V ), that is worse than
>> Floyd Warshall
>>
>>
>> On Sun, Aug 15, 2010 at 6:26 PM, Mohamed Ghoneim 
>> <[email protected]>wrote:
>>
>>> Dear all,
>>>      I have a question concerning Dijkstra and Floyd Warshall, which way
>>> is better to get all pair shortest path, is it through using dijkstra and
>>> looping through all the verticies and using them to find the shortest path
>>> to any other point O(E * V + V^2 * log(V)) , or just to execute floyed
>>> warshall on the graph O(V^3)
>>>
>>> Thanks
>>> Mohamed Sayed Ghoneim
>>>
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>>
>>
>>
>> --
>> []'s,
>> Ricardo
>>
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>
>
>
> --
> Best regards, Дектярев Михаил
>
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