hdu 4300 Clairewd’s message （kmp）

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

Clairewd’s message

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3426    Accepted Submission(s): 1331

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

2abcdefghijklmnopqrstuvwxyzabcdabqwertyuiopasdfghjklzxcvbnmqwertabcde

 

Sample Output

abcdabcdqwertabcde

 

第一个Kmp题目=.=

题意：一个转换表，表示'a'转换成第一个字母，'b'转换成第二个字母；一个字符串表示完整的密文+可能不完整原文；输出完整密文+完整原文的字符串

思路：将给出的字符串全部按照转换表转换成另一个字符串，再从原字符串密文最短长度的地方进行kmp匹配查找；

#include <stdio.h>#include <stdlib.h>#include <string.h>char match[26];char trans[500];char str[100005];char p[100005];int len;int next[100005];void getNext(){    next[0]=-1;    int k=-1,j=0;    while(j<len)    {        if(k==-1 || p[k]==p[j])        {            k++;            j++;            next[j]==k;        }        else            k=next[k];    }}int kmp(){    int i,j=0;    if(len%2==1)  //保证密文长度大于等于给定字符串的一半        i=len/2+1;    else        i=len/2;    while(i<len&&j<len)    {        if(j==-1 || str[i] == p[j])        {            j++;            i++;        }        else            j=next[j];    }    return j;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%s",match);        int i;        for(i=0;i<26;i++) //转换规则        {            char ch = match[i];            trans[ch] = i;        }        scanf("%s",str);        len = strlen(str);        for(i=0;i<len;i++) //将明文和密文按照转换规则逐一转换        {            int temp = trans[str[i]];            p[i] = temp + 'a';        }        getNext();        int tmp = kmp();        if(tmp*2 == len)            printf("%s\n",str);        else        {            int tag = len-tmp; //如果后面缺明文，则用字符串长度减去给出的明文长度算出密文长度，则可得出明文的长度            printf("%s",str);            for(i=tmp;i<tag;i++)                printf("%c",p[i]);            printf("\n");        }    }    return 0;}