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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