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

这是昨天多校联合赛的第一题，属于字符串处理，读题都发了我进一个小时，真心读不懂题，最后还是有道翻译的，各种弱爆了，本以为题意读懂后，这一题应该是囊中之物，却没想到，WA了近十次还是没过，真心伤不起啊，赛后交流的时候发现找最长明码的方法错 了，知道方法后有种想哭的感觉，，，，

题意：首先给你个密码转换规则，然后给你一段电文，已知电文是由明文和密文构成的，其中密文在前明文在后，让你求尽可能短的电文，其中前面是密文，后面是前面密文的明文。

AC代码：

#include<iostream>
#include<string.h>
#include<string>
#define N 100005
#include<cstdio>
using namespace std;
char s[N],s1[N];
char table[27];
char str[27];
int Next[N];
void _kmp(char *t)
{
	int n=strlen(t);
	int i=-1;
	Next[0]=-1;
	for(int j=1;j<n;++j)
	{
		if(t[i+1]!=t[j]&&i>-1) i=Next[i];
		if(t[i+1]==t[j]) i++;
		Next[j]=i;
	}
}
int _match(char *r,char *t)
{
	_kmp(t);
	int n=strlen(r);
	int m=strlen(t);
	int i=-1;
	for(int j=0;j<n;++j)
	{
		if(t[i+1]!=r[j]&&i>-1) i=Next[i];
		if(t[i+1]==r[j]) i++;
	}
	return i+1;//最长明文的长度
}
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%s%s",str,s);
		for(int i=0;i<26; ++i) table[str[i]-'a']='a'+i;
		strcpy(s1,s);
		int len=strlen(s);
		  for(int i=0;i<len;++i)//全部翻译成明文
			s[i]=table[s[i]-'a'];
		   int L=_match(s1+(len+1)/2,s);
		   printf("%s",s1);
		   for(int i=L;i<len-L;++i)//补全明文
			  printf("%c",s[i]);
		      printf("\n");
	}return 0;
}