in the sorted array, if suppose fr[i].start==fr[i+1].start then u only consider a cut at f[i+1] (that means ith fruit is also cut with the (i+1)th fruit. Thus u keep on incrementing the sorted array until u find sm j s.t. f[j].start!=f[j+1].start
Then u check for the condition : dp[i]=max(dp[i],dp[j-1]+match); On Thu, Feb 28, 2013 at 8:20 PM, marti <[email protected]> wrote: > Please reply ...what changes? > > > On Thursday, January 24, 2013 1:35:38 PM UTC+5:30, emmy wrote: > >> please help me with the following problem: >> http://www.spoj.com/problems/**JUICE/<http://www.spoj.com/problems/JUICE/> >> >> bit mask will require very large memory. >> If I sort the intervals in decreasing order of their start time.. I still >> cant make it to a dp >> If I sort the intervals in decreasing order of their finish times I am >> still not getting a recurrence for dp that is valid >> If I convert this to an interval graph, I have to find the maximum number >> of vertices that can be included in some clique of size >=3. But this also >> seems like a brute force approach. >> >> Which approach to try?? please help!! >> >> -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > For more options, visit https://groups.google.com/groups/opt_out. > > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/groups/opt_out.
