hdu 4300 Clairewd’s message(扩展KMP)

Clairewd’s message

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2305    Accepted Submission(s): 905

Problem Description

Clairewd is a member of FBI. After several years concealing in BUPT, she intercepted some important messages and she was preparing for sending it to ykwd. They had agreed that each letter of these messages would be transfered to another one according to a conversion table.
Unfortunately, GFW(someone's name, not what you just think about) has detected their action. He also got their conversion table by some unknown methods before. Clairewd was so clever and vigilant that when she realized that somebody was monitoring their action, she just stopped transmitting messages.
But GFW knows that Clairewd would always firstly send the ciphertext and then plaintext(Note that they won't overlap each other). But he doesn't know how to separate the text because he has no idea about the whole message. However, he thinks that recovering the shortest possible text is not a hard task for you.
Now GFW will give you the intercepted text and the conversion table. You should help him work out this problem.

 

Input

The first line contains only one integer T, which is the number of test cases.
Each test case contains two lines. The first line of each test case is the conversion table S. S[i] is the ith latin letter's cryptographic letter. The second line is the intercepted text which has n letters that you should recover. It is possible that the text is complete.

Hint

Range of test data:
T<= 100 ;
n<= 100000;

 

Output

For each test case, output one line contains the shorest possible complete text.

 

Sample Input

2
abcdefghijklmnopqrstuvwxyz
abcdab
qwertyuiopasdfghjklzxcvbnm
qwertabcde

 

Sample Output

abcdabcd
qwertabcde

 

Author

BUPT

 

Source

2012 Multi-University Training Contest 1

思路：用扩展KMP 求出明文串的匹配串长。输出。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int mm=1e5+9;
char s[mm],t[mm];
int has[155];
int next[mm],ex[mm];
void getnext(const char*t)
{ int len=strlen(t);
 next[0]=len;
 int a=0;
 while(a<len-1&&t[a]==t[a+1])++a;
 next[1]=a;
 int l,le;a=1;
 for(int i=2;i<len;++i)
 {
   le=a+next[a]-1;///a影响断范围
   l=next[i-a];///已经确定断匹配
   if(le<=i-1+l)
   { int j=le-i+1>0?le-i+1:0;
     while(i+j<len&&t[i+j]==t[j])++j;
     next[i]=j;a=i;
   }else next[i]=l;
 }
}
void ekmp(const char*s,const char*t)
{ getnext(t);
 int a=0,ls,lt;ls=strlen(s);lt=strlen(t);
 while(a<ls&&a<lt&&s[a]==t[a])++a;
 ex[0]=a;a=0;
 int l,le;
 for(int i=1;i<ls;++i)
 {
   le=ex[a]+a-1;l=next[i-a];///已经确定的匹配
   if(le<=l+i-1)
   {
     int j=le-i+1>0?le-i+1:0;
     while(i+j<ls&&j<lt&&s[i+j]==t[j])++j;
     ex[i]=j;a=i;
   }
   else ex[i]=l;
 }
}
int main()
{ int cas;
  while(~scanf("%d",&cas))
  { while(cas--)
    {scanf("%s",s);
     memset(has,0,sizeof(has));
     for(int i=0;s[i];++i)
     has[s[i]]=i+'a';
     scanf("%s",s);
     for(int i=0;s[i];++i)
     t[i]=has[s[i]];
     int len=strlen(s);
     t[len]='\0';
     ekmp(s,t);
     ///for(int i=0;i<len;++i)
     //cout<<ex[i]<<" ";
     ///cout<<endl;
     bool yes=0;
     for(int i=0;s[i];++i)
     {
      if(i+ex[i]==len&&i>=ex[i])
      { //cout<<s<<endl;
        //cout<<i<<endl;
        yes=1;
       for(int j=0;j<i;++j)printf("%c",s[j]);
       for(int j=0;j<i;++j)printf("%c",has[s[j]]);
       puts("");
      break;
      }
     }
     if(!yes)
     {
       for(int j=0;j<len;++j)printf("%c",s[j]);
       for(int j=0;j<len;++j)printf("%c",has[s[j]]);
       puts("");
     }
    }
  }
}