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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