@kailash : you can simply find area of each slab area=x*y ,,,store it;
then just run LIS on this area.
On 8/26/12, Kailash Bagaria <kbkailashbaga...@gmail.com> wrote:
> Yeah Atul is right.....
> Here is my solution:--
> 1) first rearrange the dimensions of slabs i.e. put bigger dimension in y
> and smaller dimenson in x (rotate the slab)
> 2) then arrange all slabs in increasing order of x dimension
> 3) and then find the longest increasing sub-sequence based on y
> dimension......
>
> Ex:- (2,5),(5,1),(1,3),(1,2),(6,1)
>
> Step-1=> (2,5),(1,5),(1,3),(1,2),(1,6)
>
> Step-2=> (1,2),(1,3),(1,5),(1,6),(2,5)
>
> Step-3=> LIS=4 { (1,2),(1,3),(1,5),(1,6) OR (1,2),(1,3),(1,5),(2,5)
> }
>
> Correct me if i wrong...
>
> On Sun, Aug 26, 2012 at 3:54 PM, atul anand <atul.87fri...@gmail.com>
> wrote:
>
>> its a LIS problem.
>>
>> need to think for n-dimension...
>>
>> On 8/26/12, Ravi Ranjan <ravi.cool2...@gmail.com> wrote:
>> > You are given many slabs each with a length and a breadth. A slab i can
>> be
>> > put on slab j if both dimensions of i are less than that of j. In this
>> > similar manner, you can keep on putting slabs on each other. Find the
>> > maximum stack possible which you can create out of the given slabs
>> >
>> > and for general n-dimesions
>> >
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>
>
> --
>
> --
>
> ‘Kailash Bagaria’
> B-tech 4th year
> Computer Science & Engineering
> Indian Institute of Technology, Roorkee
> Roorkee, India (247667)
>
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