HDU 4300 Clairewd’s message (next函数的应用)

题意：给你一个明文对密文的字母表，在给你一段截获信息，截获信息前半段是密文，后半段是明文，但不清楚它们的分界点在哪里，密文一定是完整的，明文可能是残缺的，求完整的信息串（即完整的密文+明文串）。

题解：KMP next函数的应用。

 

#include <cstdio>

#include <cstring>

#include <cstdlib>



const int MAXN = 100010;



char table[32];

char extable[32];

char ori[MAXN];

char aft[MAXN];

int next[MAXN];

int  len;



void init()

{

    for ( int i = 0; i < 26; ++i )

        extable[ table[i]-'a' ] = 'a' + i;



    len = strlen(ori);

    for ( int i = 0; i < len/2; ++i )

        aft[i] = ori[i];



    for ( int i = len/2; i < len; ++i )

        aft[i] = table[ ori[i] - 'a' ];



    aft[len] = '\0';



    return;

}



void getNext( char* s, int* next )

{

    int length = len;

    int i = 0, j = -1;

    next[0] = -1;

    while ( i < length )

    {

        if ( j == -1 || s[i] == s[j] )

        {

            ++i, ++j;

            next[i] = j;

        }

        else j = next[j];

    }

}



int main()

{

    int T;

    scanf( "%d", &T );

    while ( T-- )

    {

        scanf( "%s", table );

        scanf( "%s", ori );

        init();

        getNext( aft, next );



        int ans = len;

        //printf("next[%d] = %d\n", ans, next[ans] );

        while ( next[ans] > len/2 ) ans = next[ans];

        ans = len - next[ans];

        //printf( "ans = %d\n", ans );

        for ( int i = 0; i < ans; ++i )

            printf( "%c", ori[i] );

        for ( int i = 0; i < ans; ++i )

            printf( "%c", extable[ ori[i] - 'a' ] );

        puts("");

    }

    return 0;

}