BNUOJ 3580 Oulipo

Oulipo

Time Limit: 1000ms

Memory Limit: 65536KB

 

This problem will be judged on PKU. Original ID: 3461
64-bit integer IO format: %lld      Java class name: Main

 

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

• One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
• One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.

 

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

 

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

 

Sample Output

1
3
0

 解题：KMP，看不懂，网上找了点代码，研究研究，这份代码写得很俊啊！

 

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 using namespace std;
 const int MaxN = ;
 char word[MaxN/], txt[MaxN];
 int next[MaxN/];
 void KMP_next(char b[], int pre[]) {
     int n = strlen(b), k;
     pre[] = -;
     k = -;
     for(int i = ; i < n; i++) {
         while(k > - && b[k+] != b[i]) k = pre[k];
         if(b[k+] == b[i]) k++;
         pre[i] = k;
     }
 }

 int main() {
     int n;
     scanf("%d%*",&n);
     while(n--) {
         gets(word);
         gets(txt);
         KMP_next(word, next);
         int cnt = , len = strlen(word);
         for(int i = , j = -; txt[i]; ++i) {
             while(j > - && word[j+] != txt[i]) j = next[j];
             if(word[j+] == txt[i]) j++;
             if(j == len-) {
                 cnt++;
                 j = next[j];
             }
         }
         printf("%d\n", cnt);
     }
     return ;
 }

Source

BAPC 2006 Qualification

 

貌似这样写。。。。更优化，但是对此题而言，无用

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 const int maxn = ;
 int fail[maxn];
 void getNext(const char *pStr, int *nextArr) {
     int i = , j = -, pLen = strlen(pStr);
     nextArr[i] = j;
     while (i < pLen) {
         if (pStr[++i] != pStr[++j]) {
             nextArr[i] = j;
             while (j != - && pStr[i] != pStr[j]) j = nextArr[j];
         } else nextArr[i] = nextArr[j];
     }
 }
 char word[maxn],text[maxn];
 int main(){
     int n,ret;
     scanf("%d",&n);
     while(n--){
         scanf("%s %s",word,text);
         getNext(word,fail);
         for(int i = ret = ,j = ; text[i]; ++i){
             while(j != - && word[j] != text[i]) j = fail[j];
             if(!word[++j]) ret++;
         }
         printf("%d\n",ret);
     }
     return ;
 }

整理以后的，可以选择开启所谓的优化

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 using namespace std;
 const int maxn = ;
 int fail[maxn];
 char word[maxn],text[maxn];
 void getFail() {
     fail[] = -;
     fail[] = ;
     for(int i = ,j = -; word[i]; ++i) {
         while(j != - && word[i] != word[j]) j = fail[j];
         fail[i+] = ++j;
         if(word[i+] == word[j]) fail[i+] = fail[j];//使用此句加优化，注释掉不加优化，都是正确的
     }
 }
 int main() {
     int n,ret;
     scanf("%d",&n);
     while(n--) {
         scanf("%s%s",word,text);
         getFail();
         for(int i = ret = , j = ; text[i] ; ++i) {
             while(j > - && word[j] != text[i]) j = fail[j];
             if(!word[++j]) {ret++;j = fail[j];}
         }
         printf("%d\n",ret);
     }
     return ;
 }