right!!

On Wed, Nov 25, 2009 at 4:38 PM, Aditya Shankar <
iitm.adityashan...@gmail.com> wrote:

>
>
> On Wed, Nov 25, 2009 at 2:13 PM, Ankur Parashar <ankurlnm...@gmail.com>wrote:
>
>> Use Flyod Warshal.....All pair shortest path algo.
>> initialise d[0] = 1 (i,j) is and edge
>> now apply flyod warhsal
>
> That is O(n^3) right?
>
>>
>>
>> On Wed, Nov 25, 2009 at 8:59 AM, Aditya Shankar <
>> iitm.adityashan...@gmail.com> wrote:
>>
>>>
>>> On Tue, Nov 24, 2009 at 10:46 PM, Rohit Saraf <
>>> rohit.kumar.sa...@gmail.com> wrote:
>>>
>>>> i had already understood what you claimed
>>>>
>>>> But MY CLAIM: every pair of nodes in the connected component is
>>>> connected to every other node in the same connected component. U CANNOT
>>>> UPDATE ALL THE PAIRS AS EASILY AS U ARE SAYING. For any two nodes in DFS of
>>>> Graph corresponding entry is 1. Got it?
>>>>
>>> If the problem is to _fill_ up matrix entries, unless you can get around
>>> using adjacency lists/matrices, your complexity will be O(n^2). However, we
>>> can use an implicit data structure, since all nodes are connected to each
>>> other. Just use a vector of sets, each set containing the vertices of a
>>> connected component.
>>>
>>> Regards
>>> Aditya Shankar
>>>
>>>
>>>
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