no ...when u figure out those m matches just sort them ..now let k=[0,m]
if currently u are assuming that kth match is already sliced then all
before that k matches are already sliced.and this u can do by moving
incrementally from 0 to mth matches.

On Mon, Jan 28, 2013 at 1:43 PM, foram lakhani <[email protected]>wrote:

> by match u mean that number of fruits that overlap with i th fruit but are
> not sliced before time i.
> which means we have to consider 2^m cases where m is the maximum number of
> fruits that overlap with ith fruit.
> And this we have to do for each fruit.
>
> On Mon, Jan 28, 2013 at 12:54 AM, Nikhil Karnwal <[email protected]>wrote:
>
>> Actually dp[i] represent the state in which u make a slice at appearing
>> time of i th fruits.and match represent the overlapping fruits
>> with i.
>> so
>> for each i dp[i]=max(dp[i],dp[j]+match);
>> for all j=[0,i] and match =overlapped fruits with i which are not sliced
>> until the cut of j.
>> Hope this will help.
>> Thanks
>>
>> On Thu, Jan 24, 2013 at 10:18 PM, foram lakhani 
>> <[email protected]>wrote:
>>
>>> Thanx for the reply but I didnt get you. What does dp[i] represent ?
>>>
>>>
>>>  On Thu, Jan 24, 2013 at 9:50 PM, sunny <[email protected]> wrote:
>>>
>>>> take a structure or pair of start time and finish time and then sort
>>>> the the structure you know the comparator function  the for each for each
>>>> dp[i] select j belongs to  (0,i) and then count the overlap value
>>>>
>>>>
>>>>                                 if(j==0)
>>>> dp[i]=max(dp[i],match);
>>>> else
>>>>  dp[i]=max(dp[i],dp[j-1]+match);
>>>>
>>>>
>>>> with this recursive relation my code got accepted in .24 sec others
>>>> approach you mentioned may lead to the TLE
>>>>
>>>> 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!!
>>>>>
>>>>>  --
>>>>
>>>>
>>>>
>>>
>>>  --
>>>
>>>
>>>
>>
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