HDU 4300 Clairewd's message KMP

The question is too difficult to understand.

The first line gives you a piece of ciphertext with 26 letters. The corresponding plaintext is from a-z.

The second line is followed by a plaintext string. The ciphertext must be complete, but either or not.

Ask for the shortest ciphertext + plaintext.

 

Example 1: abcdab

Shortest ciphertext: abcd, which corresponds to abcd.

Therefore, the shortest ciphertext + plaintext is abcdabcd.

Example 2: qwertabcde

Shortest ciphertext: qwert, whose plaintext is abcde

Therefore, the shortest ciphertext + plaintext is qwertabcde.

Difficult to understand.

 

Idea: the plaintext length must be smaller than or equal to len/2, and then match the later half with the corresponding plaintext

For example, in the first example, the maximum matching of dab and abcdab is t = 2, indicating that two are already in plain text, and the output from t to len-t is not displayed in plain text,

 

AC code:

 

# Include <iostream> # include <cstdio> # include <cstring> # include <string> # include <cstdlib> # include <cmath> # include <vector> # include <list> # include <deque> # include <queue> # include <iterator> # include <stack> # include <map> # include <set> # include <algorithm> # include <cctype> using namespace std; typedef _ int64 LL; const int N = 100090; const ll ii = 1000000007; const int INF = 0x3f3f3f3f; const double PI = acos (-1.0); char word [N], yi [26], mi [26], xh [N]; int wlen, next [N]; void getnext (char * p) {int j = 0, k =-1; next [0] =-1; while (j <wlen) {if (k =-1 | p [j] = p [k]) {j ++; k ++; next [j] = k;} else k = next [k] ;}} int kmp (char * text, char * word) {int I = 0, j = 0, tlen = strlen (text); while (I <tlen) {if (j =-1 | text [I] = word [j]) j ++, I ++; else j = next [j];} return j; // return the maximum number of overlapped prefixes between the suffix of the parent string and the pattern string} int main () {int I, j, T; scanf ("% d ", & T); while (T --) {scanf ("% s", yi, word); printf ("% s", word); for (I = 0; I <26; I ++) mi [yi [I]-'a'] = I + 'a'; // plaintext wlen = strlen (word) corresponding to the ciphertext ); strcpy (xh, word + (wlen + 1)/2); for (I = 0; I <wlen; I ++) word [I] = mi [word [I]-'a']; getnext (word); int t = kmp (xh, word ); // int p = wlen-t where the plain text overlaps; // the word [p] To be output = '\ 0'; printf ("% s \ n ", word + t); // from t to p} return 0 ;} # include <iostream> # include <cstdio> # include <cstring> # include <string> # include <cstdlib> # include <cmath> # include <vector> # include <list> # include <deque> # include <queue> # include <iterator> # include <stack> # include <map> # include <set> # include <algorithm> # include <cctype> using namespace std; typedef _ int64 LL; const int N = 100090; const ll ii = 1000000007; const int INF = 0x3f3f3f3f; const double PI = acos (-1.0); char word [N], yi [26], mi [26], xh [N]; int wlen, next [N]; void getnext (char * p) {int j = 0, k =-1; next [0] =-1; while (j <wlen) {if (k =-1 | p [j] = p [k]) {j ++; k ++; next [j] = k;} else k = next [k] ;}} int kmp (char * text, char * word) {int I = 0, j = 0, tlen = strlen (text); while (I <tlen) {if (j =-1 | text [I] = word [j]) j ++, I ++; else j = next [j];} return j; // return the maximum number of overlapped prefixes between the suffix of the parent string and the pattern string} int main () {int I, j, T; scanf ("% d ", & T); while (T --) {scanf ("% s", yi, word); printf ("% s", word); for (I = 0; I <26; I ++) mi [yi [I]-'a'] = I + 'a'; // plaintext wlen = strlen (word) corresponding to the ciphertext ); strcpy (xh, word + (wlen + 1)/2); for (I = 0; I <wlen; I ++) word [I] = mi [word [I]-'a']; getnext (word); int t = kmp (xh, word ); // int p = wlen-t where the plain text overlaps; // the word [p] To be output = '\ 0'; printf ("% s \ n ", word + t); // from t to p} return 0 ;}