@Atul007: No, because a (1,4) tile will not fit on a (2,3) tile.
Dave
On Sunday, August 26, 2012 7:45:27 AM UTC-5, atul007 wrote:
> @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 <kbkailas...@gmail.com <javascript:>> 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.8...@gmail.com<javascript:>>
>
> > wrote:
> >
> >> its a LIS problem.
> >>
> >> need to think for n-dimension...
> >>
> >> On 8/26/12, Ravi Ranjan <ravi.c...@gmail.com <javascript:>> 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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