hdu 4300 Clairewd’s message (kmp)

最新推荐文章于 2019-01-23 18:42:43 发布

黎辰

最新推荐文章于 2019-01-23 18:42:43 发布

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文章标签： C++ kmp 字符串

本文链接：https://blog.youkuaiyun.com/qq_24451605/article/details/44211641

本文深入解析了FBI成员Clairewd在BUPT潜伏期间，通过预设转换表加密发送给ykwd的重要信息。面对GFW的监视，Clairewd巧妙停止了信息传输，而GFW则利用提供的转换表和截获的文本尝试还原密文与明文。文章指导如何通过特定算法分离出完整的信息串。

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Clairewd’s message

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3669 Accepted Submission(s): 1411

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

题目大意:第一格串是a-z字母对应的密文,
第二个串由完整的密文,和可能完整,可能不完整,可能不存在的明文组成
判断最短的完整密文+完整明文的串是什么
题目分析:直接算取next数组之后,判断当i >=(len-1)/2时,是否存在 next[len] >= len -i , 如果存在证明当前密文可以由后面的明文翻译出来

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#define MAX 100007

using namespace std;

int t;
char s1[MAX];
char s2[MAX];

char fmp[MAX];

int Next[MAX];

void get_next ( )
{
    int i = 0 , k = -1 , len = strlen(s2);
    Next[0] = -1;
    while ( i <= len )
        if ( k == -1 || s2[i] == fmp[s2[k]] )
            i++,k++,Next[i] = k;
        else k = Next[k];
}

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%s" , s1 );
        scanf ( "%s" , s2 );
        int len1 = strlen (s1);
        int len2 = strlen (s2);
        for ( int i = 0 ; i < len1 ; i++ )
            fmp[s1[i]] = (char)('a'+i);
        get_next ( );
        int len = len2;
        for ( int i = (len2-1)/2 ; i++ ; i <= len2 )
            if ( Next[len2] >= len2-i )
            {
                len = i;
                break;
            }
        for ( int i = 0 ; i < len ; i ++ )
            printf ( "%c" , s2[i] );
        for ( int i = 0 ; i < len ; i ++ )
            printf ( "%c" , fmp[s2[i]] );
        puts ( "" );
    }
}