My code, accepted in practice room gives 2, but the answer is clearly 1, but
I believe the input is invalid

This list will include every directory already on your computer other than
> the root directory. (The root directory is on every computer, so there is no
> need to list it explicitly.)


I that means /home *must* be listed, so the case is
1
2 1
/home/gcj
/home
/home/x

So even if someones code doesn't check for the case you posted should give
correct answer to a valid input.

Carlos Guía


On Sat, May 22, 2010 at 6:06 PM, Mikhail Dektyarev <
[email protected]> wrote:

> Correct output is
> Case #1: 1
> because we need to create catalog "/home/x"
>
>
> On Sat, May 22, 2010 at 11:55 PM, Paulo Cezar Pereira Costa <
> [email protected]> wrote:
>
>> I was taking a look at Gluk's code for problem A,
>> seems like his code generate a wrong output, or I didn't understood the
>> problem statement..
>>
>> The case:
>> 1
>> 1 1
>> /home/gcj
>> /home/x
>>
>> What's the correct output ?
>>
>> On Sat, May 22, 2010 at 3:41 PM, Bharath Balakrishnan <
>> [email protected]> wrote:
>>
>>> I am not sure about a tree. But a greedy approach that creates the
>>> required directories in increasing order of the number of  separators '/' in
>>> the name, works.
>>>
>>>
>>> On Sun, May 23, 2010 at 12:03 AM, Abdelrhman Abotaleb <
>>> [email protected]> wrote:
>>>
>>>> Is the answer of the first problem needs a general tree !?
>>>> could any one provide me with a suitable splitting c++ code with a
>>>> dilemeter
>>>> thanks a lot
>>>>
>>>>
>>>> --
>>>> Regards,
>>>> Abdelrhman.M. Abotaleb
>>>>
>>>> IEEE 2010 Student Chapter,
>>>> AC Active member
>>>> SPE 2009 Well Services Moderator
>>>> cairo.spe.org
>>>>
>>>>
>>>> Major: Electronics & Communications
>>>> Minor: Computer Engineering
>>>>
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>>>
>>>
>>>
>>> --
>>> Regards,
>>> Bharath B
>>>
>>> "When the going gets tough, the tough get going"
>>>
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>>
>>
>>
>> --
>> Paulo Cezar Pereira Costa
>> Graduando em Ciência da Computação
>> Universidade Federal de Goiás
>>
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>
>
>
> --
> Best regards, Дектярев Михаил
>
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