@nima
could not understand your soln..

try wid an example..


On Thu, Aug 27, 2009 at 1:40 PM, Nima Aghdaii <[email protected]>wrote:

> you can use dynamic programming.
> we wanna solve: f(n1, n2)
> f(n1, n2) :
> res=0
> for each u such that we have n1->u
>     res += f(u, n2) // which is calculated before
>
> (actually you need a one dimensional array for the source)
> O(V+E)
>
> Good luck,
> Nima
>
>
>
> On Wed, Aug 26, 2009 at 1:33 AM, ankur aggarwal 
> <[email protected]>wrote:
>
>> given a directed graph and node n1 and n2
>> find all possible path between them
>>
>> i think graph is acyclic ..
>> otherwise there are infinte path we can write
>>
>> eg
>>
>> for  node "a" and "d" are there
>> we have a cycle  node "b"  to "c" and "c" to "b"
>>
>> a-> b->c->b->d
>> a-> b->c->b->c->b->d
>> etc
>>
>>
>>
>
>
> --
> Nima Aghdaii
>
> "The ideal situation occurs when the things that we regard as beautiful are
> also regarded by other people as useful."
> -Donald Knuth-
>
>
>
> >
>

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