Quick implementation of social networking

I need to implement a concept somewhat similar to social networking
sites like orkut/linkedIn etc etc... The story is below......
>
'A' can send request to any other member say 'B' to join his network.
The B joins his network. Now 'A' can approach the 'B' 's contacts by 2
ways directly or through 'B' and so on. Now suppose there is 'Z'
somewhere in the chain of 'B' , when I search for 'Z' it will display
me that you can reach 'Z' from 'B' using the chain or how can I view
the chain.
>
My question is "do i need to write/implement graph theory here or is
there any other rapid development approach available for this problem"

>
>The path thru B to Z is just "minimum cost spanning tree", which is one of
>the simplest graph theory algorithms available. Look that up, and you might
>even find an implementation in C++

On Fri, 15 Sep 2006 12:08:12 GMT in comp.lang.c++, "Phlip"
<phlipcpp@yahoo.comwrote,>
Sounds like it might be related to
http://www.boost.org/libs/graph/doc/..._contents.html

Can you explain with example and how do we extract data from the
database.

>
>
sgupta1221@gmail.com wrote:>
Buy the book! Boost::Graph is documented very well. It comes with CD.
>
ISBN: 0201729148
>
HTH,
Michiel Salters

Hi,
I dont think we exactly need Graph theory or "Minimum cost" Algos here
as in this case all the edges of the graph carry the same weightage.
We just need to detect the relation at minimum number of nodes/degrees.
If there is more than one path available to end contact with same
degree - we do not need to calulate anything based on "minumum cost" or
any other criteria as there is no different weightages of friendship.
>
As per my understanding, a sql query in the form of stored proc can be
enough which iteratively looks up friends at 1..n th level.
>
Your comments please...
>

aasingla@gmail.com wrote:[snip]
>
You are off-topic here where standard C++ is discussed. So find a forum
where you are topical or rephrase your question so that it is about
standard C++.
>
/Peter

Thanx Peter but i am not the questioner - I am just replying to the
question that is already posted in this group.