hdu4300 Clairewd’s message 扩展KMP

B - Clairewd’s message(EKMP)

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 4300

Appoint description:  System Crawler  (2016-01-22)

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 
      
      
      
      

 

Hint

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

     
     
     
     

      
      
      
      
          
      
     
     
     
     

这道题有一个很操蛋的地方，就是你读一遍之后根本不知道它在说什么。鉴于智商问题，我读了七八遍之后仍然不能理解，最终百度的，不过，其中GFW的梗还挺好笑。。。

题目大意：给出一个加密字典，即一个字母映射为另一个字母，类似于函数关系；给出一个字符串，它是由 完整的加密串+不一定完整的解密串 相连而成，要求将这个串补充完整，即将不完整的部分补充完整，并且这个完整的串还是最短的。

解题思路：1. 将原始串再做一次加密，加密+解密 就会变成 XX+加密

                  2. 设原始串为s（主串），新串为t（模式串），对两串进行扩展KMP运算

  3. 从中间开始扫描extend数组，如果遇到i+extend[i]==len&&i>=extend[i]的情况，表明找到解。

解释：为了找到最短的串，我们要尽量让加密串与解密串一样长，也就是要从中间开始找，如果找到一个解，那么从i到len的串必定全部匹配，i既表示解密串的起始位置，也表示加密串的长度，extend[i]则表示后一半长度，这就是i+extend[i]==len，而且，要保证两部分不重合，这就是i>=extend[i]。

代码

#include <iostream>
#include <stdio.h>
#include <string.h>
#define MAX 100010
using namespace std;

void get_next(char x[],int m,int next_[])
{
    next_[0]=m;
    int j=0;
    while(j+1<m&&x[j]==x[j+1]) j++;
    next_[1]=j;
    int k=1;
    for(int i=2; i<m; i++)
    {
        int p=next_[k]+k-1;
        int L=next_[i-k];
        if(i+L<p+1) next_[i]=L;
        else
        {
            j=max(0,p-i+1);
            while(i+j<m&&x[i+j]==x[j]) j++;
            next_[i]=j;
            k=i;
        }
    }
}


void ekmp(char x[],int m,char y[],int n,int next_[],int extend[])
{
    get_next(x,m,next_);
    int j=0;
    while(j<n&&j<m&&x[j]==y[j]) j++;
    extend[0]=j;
    int k=0;
    for(int i=1; i<n; i++)
    {
        int p=extend[k]+k-1;
        int L=next_[i-k];
        if(i+L<p+1) extend[i]=L;
        else
        {
            j=max(0,p-i+1);
            while(i+j<n&&j<m&&y[i+j]==x[j]) j++;
            extend[i]=j;
            k=i;
        }
    }
}




char s[MAX],t[MAX],dic[27];
char dc[200];
int len;
int Next[MAX],extend[MAX];


int main()
{
    int T,i,j,bg;
    char tmp;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",dic,t);
        len=strlen(t);

        for(i=0;dic[i]!='\0';i++)//计算解密字典
        {
            dc[dic[i]-'a']='a'+i;
        }

        for(i=0; i<len; i++)//二次加密
        {
            s[i]=dic[t[i]-'a'];
        }
        ekmp(t,len,s,len,Next,extend);
        for(i=len/2; i<len; i++)
        {
            if(i+extend[i]==len && i>=extend[i])//重要，需要长度相等且不重合
                break;
        }
        for(j=0; j<i; j++)
            printf("%c",t[j]);
        for(j=0; j<i; j++)//输出解密内容
            printf("%c",dc[t[j]-'a']);
        printf("\n");
    }
    return 0;
}