m also new to algorithm...
anyway...
what algorithm r u using???

if u use DFS then there u have to mark the edges u have already visited.
hope u have the concepts clear for DFS . if so den u might hv got what i
mean....
if not den i cant make u understand.... :-(
On 20 Apr 2012 18:55, "vivek dhiman" <[email protected]> wrote:

> Registered order can you explain a bit..
>
>
> An other geeks and people.... what happened ??
> reply
>
> 2012/4/20 Registered user <[email protected]>
>
>> i think if u mark the path which u want to block as visited beforehand
>> then u may get it, m  not sure but i think it should work.
>>
>> or if possilbe give the sample input / output cases , den i may solve
>> it....
>>
>> On 4/20/12, Andrey Ponomarev <[email protected]> wrote:
>> > Hypothesis 1: after eliminating one edge e' the shortest path from a
>> > to b must contain vertex v, such that
>> > neighter sp(a,v), nor sp(v,b) contain e' in the original graph, if
>> > there is such. (Here sp stands for 'shortest path')
>> > Hypothesis 2: if v is some vertex of the modified graph, such that
>> > neighter sp(a,v), nor sp(v,b) contain e' in
>> > the original graph, the shortest path from a to b in the modified
>> > graph is len(sp(a,v)) + len(sp(v,b)).
>> >
>> > If both hypotheses are true then you may try following. In the
>> > original graph build 2 shortest path trees:
>> > one from a and one from b with edges reversed. The first tree let you
>> > at O(1) find len(sp(a,v)), аnd
>> > the second let you  at O(1) find len(sp(v,b)). When you have an edge
>> > (v1, v2) removed you should walk
>> > tree1 from v2 and tree2 from v1 and mark all vertices you pass by.
>> > Then you should find one unmarked
>> > vertex with the minimum sum of len(sp(a,v)) + len(sp(v,b)) - it will
>> > be the shortest path and it took O(n).
>> >
>> > 20 апреля 2012 г. 10:48 пользователь vivek dhiman
>> > <[email protected]> написал:
>> >> ok..
>> >> So if an edge is removed from existing path.Do i have to again find the
>> >> shortest path ?
>> >>
>> >>
>> >> On Fri, Apr 20, 2012 at 11:08 AM, Luke Pebody <[email protected]>
>> >> wrote:
>> >>>
>> >>> Here's a start for you. Find the shortest route. This works as an
>> answer
>> >>> if you remove any edge not in that route. Now you just have to solve
>> it
>> >>> for
>> >>> any removed edges that are in that route.
>> >>>
>> >>>
>> >>>
>> >>> On 20 Apr 2012, at 05:19, vivek dhiman <[email protected]>
>> wrote:
>> >>>
>> >>> Hi all geeks
>> >>>
>> >>> So I have a simple question.
>> >>>
>> >>> Suppose you have a graph  G(V,E).
>> >>> You are supposed to find the shortest path from a vertex 's' to vertex
>> >>> 'e'
>> >>> for 'n' different cases.
>> >>>
>> >>> In each case one of the edges  'Ei' (any one edge) of the graph will
>> be
>> >>> blocked/deleted only for that case and we have to find the shortest
>> path
>> >>> in
>> >>> the graph with that edge removed.
>> >>>
>> >>> Guys finding the shortest path is easy. But how can I make the algo so
>> >>> fast that even if I remove one of the edges my algo should still be
>> very
>> >>> fast. O(n log n) or  faster.
>> >>> Remember we are not deleting the edges permanently. We are just
>> temporary
>> >>> removing one edge per case.
>> >>> In each case only one edge is removed.
>> >>> Suppose we blocked one edge E in one case. We have to find the
>> shortest
>> >>> path for the graph.
>> >>> In next case, we will reconnect the last edge and we will
>> block/remove a
>> >>> new edge. And again for this new case we have to find the shortest
>> path.
>> >>>
>> >>> Another way of understanding the problem is suppose there are cities
>> >>> connected to each other.
>> >>> And every day one of the roads gets blocked because of heavy rain.
>> what
>> >>> is
>> >>> the shortest path every day from city s to e.
>> >>> Also one more important thing to note that each road can be used only
>> >>> once.
>> >>> But there could be more than 1 direct road from city a to city b.
>> >>>
>> >>> FInd the shortest path distance from city s to e on a day when all
>> direct
>> >>> roads from city f to city h are blocked. If there is no connecting
>> path
>> >>> return -1
>> >>>
>> >>> Thanks and Regards
>> >>> Vivek Dhiman
>> >>>
>> >>> --
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>> >>
>> >>
>> >>
>> >> --
>> >> Regards
>> >> Vivek Dhiman
>> >>
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>
>
> --
> Regards
> Vivek Dhiman
>
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