Clairewd’s message

http://acm.hdu.edu.cn/showproblem.php?pid=4300

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
   
   
   
   

제목: 한 단락의 전문을 캡처했는데 전문은 적어도 밀문(화명문)을 포함하고 밀문이 완전하며 명문이 완전하지 않을 수도 있고 심지어 전혀 없을 수도 있다.현재 요구는 제시된 일치 규칙에 따라 완전한 전문을 출력합니다.

#include<cstdio>
#include<cstring>
using namespace std;
char s[100005],t[100005];
char match[30],table[30];
int next[100005],slen,tlen;
char cop[30];
void get_next()
{
    int i=0,j=-1;
    next[0]=-1;
    for(;t[i];)
        if(j==-1||t[i]==t[j])
        {
            ++i;++j;
            next[i]=j;
        }
        else j=next[j];
}
int kmp()
{
    int i=0,j=0;
    for(;i<slen&&j<tlen;)
    {
        if(j==-1||s[i]==table[t[j]-'a'])
        {
            i++;j++;
            if(i==slen)
                return j;
        }
        else
            j=next[j];
    }
    return 0;
}
int main()
{
    int cas;
    scanf("%d",&cas);
    for(;cas--;)
    {
        scanf("%s%s",match,t);
        strcpy(s,t+(strlen(t)+1)/2);
        slen=strlen(s);
        tlen=strlen(t);
        for(int i=0;i<26;++i)
            table[match[i]-'a']='a'+i;
        get_next();
        int k=kmp();
        printf("%s",t);
        for(int i=k;i<tlen-k;++i)
           printf("%c",table[t[i]-'a']);
        printf("
");
    }
    return 0;
}

출처:http://blog.csdn.net/ACM_Ted/article/details/7823851