@dave : correct..
i guess this will work :-
sort in decreasing area.
then run LIS such that for i < j , length( i ) > length( j ) && width(
i ) > width( j )
On 8/26/12, Dave <dave_and_da...@juno.com> wrote:
> @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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>> > --
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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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