@jalaj..he has asked for a linear algo

On Sat, Jul 17, 2010 at 12:47 AM, jalaj jaiswal
<[email protected]>wrote:

> can be done in O(n^2) also
> below is a rough pseudocode
> initialize visited[v]=0
> for every vertex v{
>      call dfs(v)
>      check if al visited are 1 or not
>       if all visited break;
>       else do visited[v]=0 again.
> }
>
> correcf me if i'm missing anything
>
>
> On Fri, Jul 16, 2010 at 5:38 AM, Gene <[email protected]> wrote:
>
>> Construct the transitive closure of the graph.  You can use Warshall's
>> algorithm, which is O(v^3) in run time.  If any row of the adjacency
>> matrix is all 1's, the corresponding vertex can reach all others.  You
>> can ignore the diagonal element if you don't care whether vertices are
>> reachable from themselves (i.e. whether they are contained in a
>> cycle).
>>
>> On Jul 12, 1:06 pm, Love-143 <[email protected]> wrote:
>> > 1.Give an efficient algorithm which takes as input a directed graph G =
>> > (V,E), and determines whether or not there is a vertex s is in V from
>> which
>> > all other vertices are reachable.?
>>
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>
>
> --
> With Regards,
> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
>
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