An LCS problem (Spoj : advanced fruits ADFRUITS)

Правка en2, от shengdebao, 2016-03-02 18:54:10

The company "21st Century Fruits" has specialized in creating new sorts of fruits by transferring genes from one fruit into the genome of another one. Most times this method doesn't work, but sometimes, in very rare cases, a new fruit emerges that tastes like a mixture between both of them.

A big topic of discussion inside the company is "How should the new creations be called?" A mixture between an apple and a pear could be called an apple-pear, of course, but this doesn't sound very interesting. The boss finally decides to use the shortest string that contains both names of the original fruits as sub-strings as the new name. For instance, "applear" contains "apple" and "pear" (APPLEar and apPlEAR), and there is no shorter string that has the same property. A combination of a cranberry and a boysenberry would therefore be called a "boysecranberry" or a "craboysenberry", for example.

Your job is to write a program that computes such a shortest name for a combination of two given fruits. Your algorithm should be efficient, otherwise it is unlikely that it will execute in the alloted time for long fruit names. Input Specification

Each line of the input file contains two strings that represent the names of the fruits that should be combined. All names have a maximum length of 100 and only consist of alphabetic characters. Input is terminated by end of file. Output Specification

For each test case, output the shortest name of the resulting fruit on one line. If more than one shortest name is possible, any one is acceptable. Sample Input

apple peach ananas banana pear peach

Sample Output

appleach bananas pearch

----------------- This is my solution but unfortunately I cant get ACCEPT ,

I simply make two vectors v and u indicating the index of the lcs in strings s and t (first string and the second string). I make these vectors by running a dp LCS algorithm that I think its correct at least I have not found any test case that proves sth else, after I've made these two vector I simply use three pointers one on the first char of s , another on the first char of string t, and another indicating the first index of LCS we will meet(on the vectors) while any of the pointers 1 or 2 point a character not on the LCS I simply add those characters to my result, else I add all the pointers and add that character of the LCS to my result

heres my code I'd be thankful if you could help me:

http://pastebin.com/raw/ZgJt6hak

P.S The reason why I'm not getting accept might be because of my input method :)

Теги lcs, dynamic programming, #dp, #strings

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en2 Английский shengdebao 2016-03-02 18:54:10 910 my own solution is added...
en1 Английский shengdebao 2016-03-02 14:05:51 1719 Initial revision (published)