Advanced Fruits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3958    Accepted Submission(s): 2081

Special Judge

Problem Description

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

Each line of the input 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

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

 

题意：就是求两个字符串最长公共子串再输出，

公共子串输出一次即可。

 

分析：对每个字符标记，公共子串输出一次，递归输出。

 

#include
        
         #include
         
          #include
          
           using namespace std;char a[1000],b[1000];int alen,blen;int dp[1000][1000];void LCS(){    memset(dp,0,sizeof(dp));    for(int i=1;i
           
            =dp[i][j-1]) dp[i][j]=dp[i-1][j];            else dp[i][j]=dp[i][j-1];        }    }}void Print(int i,int j){    if(i==0&&j==0) return;    if(i<0||j<0) return;    if(i>=1&&j>=1&&dp[i][j]==dp[i-1][j-1]+1&&dp[i-1][j]==dp[i][j-1]&&dp[i-1][j-1]==dp[i][j-1])    {
       //公共子串判断条件        Print(i-1,j-1);        printf("%c",a[i]);    }    else if(i>=1&&dp[i][j]==dp[i-1][j])    {        Print(i-1,j);        printf("%c",a[i]);    }    else if(j>=1)    {        Print(i,j-1);        printf("%c",b[j]);    }}int main(){    a[0]=12;b[0]=12;//使数组a,b的第一个字符不为0,以免影响alen,blen     while(scanf("%s%s",a+1,b+1)!=-1)    {        alen=strlen(a);        blen=strlen(b);        LCS();        Print(alen-1,blen-1);        printf("\n");    }    return 0;}