@ all.
what if we use greedy approch..like if we slice when in remaining fruits on 
screen any one fruit is going to disappear?will it work?



On Friday, February 1, 2013 9:07:14 AM UTC+5:30, emmy wrote:
>
> Ohk.. I finally got it.
> Thanks guys!!
> This was great help.
>
>
> On Mon, Jan 28, 2013 at 4:11 PM, Nikhil Karnwal 
> <[email protected]<javascript:>
> > wrote:
>
>> 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]<javascript:>
>> > 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]<javascript:>
>>> > 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]<javascript:>
>>>> > 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]<javascript:>
>>>>> > 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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>>>>>  
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