Advanced Fruits(好题，LCS的模拟)

Advanced Fruits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2052    Accepted Submission(s): 1053
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

 题解：

这个题就是让找出这两个串的最长公共子序列，然后加上这两个串中减去公共子序列的字符，输出就行；

我的思路就是先求出最长公共子序列的dp数组，然后再倒过来，模拟dp数组走的路径倒着记录就行了；

代码：

 #include<stdio.h>
 #include<string.h>
 #define MAX(x,y) x>y?x:y
 const int MAXN=;
 int dp[MAXN][MAXN];
 char s1[MAXN],s2[MAXN],ans[MAXN*];
 int t1,t2,t;
 void LCS(){
     memset(dp,,sizeof(dp));
     t1=strlen(s1+);t2=strlen(s2+);
     for(int i=;i<=t1;i++){
         for(int j=;j<=t2;j++){
             if(s1[i]==s2[j])dp[i][j]=dp[i-][j-]+;
             else{
                 dp[i][j]=MAX(dp[i-][j],dp[i][j-]);
             }
         }
     }
 }
 void add(char a){
     ans[t++]=a;
     ans[t]='\0';
 }
 void print(){
     while(dp[t1][t2]){
             if(s1[t1]==s2[t2]){
                 add(s1[t1]);
                 t1--;t2--;
             }
             else{
                 if(dp[t1-][t2]>dp[t1][t2-]){
                     add(s1[t1]);
                     t1--;
                 }
                 else{
                     add(s2[t2]);
                     t2--;
                 }
             }
         }
         while(t1>)add(s1[t1--]);
     while(t2>)add(s2[t2--]);
 }
 int main(){
     while(~scanf("%s%s",s1+,s2+)){
         t=;
         LCS();
         print();
         for(int i=t-;i>=;i--)printf("%c",ans[i]);
         puts("");
     }
     return ;
 }