POJ 3461 Oulipo（KMP模板题）

阳光穿透心脏的1/2处 2024-02-18 14:45 28阅读 0赞

Oulipo

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;43141</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;17358</td> 
  </tr> 
 </tbody> 
</table>

Description

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

AC代码：

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn = 1e4+10;
    const int max2 = 1e6+10;
    char W[maxn];
    char T[max2];
    int next[maxn];
    
    void get_next(int len){
    	int k = 0;
    	next[0] = 0;
    	for(int p = 1; p < len; p++){
    		while(k > 0 && T[p]!=T[k]){
    			k = next[k-1];
    		}
    		if(T[k] == T[p])k++;
    		next[p] = k;
    	}
    }
    
    int main(){
    	int n ;
    	scanf("%d",&n);
    	while(n--){
    		scanf("%s%s",W,T);
    		int len1 = strlen(W);
    		int len2 = strlen(T);
    		get_next(len1);
    		int cnt = 0;
    		int k = 0, j = 0;
    		while(k<len1 && j<len2){
    			if(W[k] == T[j]){
    				if(k == len1-1){
    					k = next[k];
    					j++;
    					cnt++;
    				}
    			    else k++,j++;
    			}
    			else if(k == 0) j++;
    			else k = next[k-1];
    		}
    		printf("%d\n",cnt);
    	}
    	return 0;
    }