That's true if you implement Dijkstra using Fibonacci heap. But if you
use simple heap (as most people do during programming contests) then
you'll get O((E*V+V^2)*log V). For complete graph it will be O(V^3 * log
V).
But I absolutely agree with conclusions. Floyd is preferable for dense
graphs while Dijkstra will be choice for sparse ones.
On 15.08.2010 17:36, Mikhail Dektyarev wrote:
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]
<mailto:[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] <mailto:[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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