p[i] maintains previous index from which b[i] has reached longest sequence
till i.
to get the actual list of non-decrease sequence, p has to be traversed
through back indices
for (u = b.size(), v = b.back(); u--; v = p[v])  b[u] = v;

surender
On Sat, Jul 16, 2011 at 9:06 AM, Neeraj Gupta <[email protected]>wrote:

>
>
> Hi
>
>> Can anyone help me in understanding the following code
>> http://www.algorithmist.com/index.php/Longest_Increasing_Subsequence.cpp
>>
>> I am not able to understand what is the exact purpose of vector p in the
>> above mentioned code.
>> A little detail explanation will be helpful.
>>
>> I have already read this the basic idea of the algorithm
>
> * Let Ai,j be the smallest possible tail out of all increasing
> subsequences of length j using elements a1, a2, ... , ai. Observe that,
> for any particular i, Ai,1, Ai,2, ... , Ai,j. This suggests that if we
> want the longest subsequence that ends with ai + 1, we only need to look for
> a j such that Ai,j < ai + 1 < = Ai,j + 1 and the length will be j + 1.
> Notice that in this case, Ai + 1,j + 1 will be equal to ai + 1, and all Ai +
> 1,k will be equal to Ai,k for k!=j+1. Furthermore, there is at most one
> difference between the set Ai and the set Ai + 1, which is caused by this
> search. Since A is always ordered in increasing order, and the operation
> does not change this ordering, we can do a binary search for every single a
> 1, a2, ... , an.
> *
>
>> ~Neeraj
>>
> Hi,
>
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