POJ 3461 Oulipo 深刻理解KMP算法next数组含义

秒速五厘米 2022-09-20 15:14 115阅读 0赞

题目大意：在 字符串2 中找 字符串1 出现的次数。

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;22537</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;8989</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

Source

[BAPC 2006 Qualification][]

#include <stdio.h>
    
    void getnext(char *ch, int next[]);
    int kmp(char *ch1, char *ch2, int next[]);
    
    char ch1[10010], ch2[1000010];
    int next[10010];
    
    int main()
    {
    	
    	int t;
    	
    	scanf("%d", &t);
    	while(t--)
    	{
    		
    		scanf("%s", ch1);
    		getnext(ch1, next);//获得next数组。
    		
    		scanf("%s", ch2);
    		printf("%d\n", kmp(ch1, ch2, next));
    	}
    	
    	return 0;
    }
    void getnext(char *ch, int next[])//对匹配串进行kmp匹配获得next数组值
    {
    	int i, j;
    	i=0;
    	j=-1;
    	next[0] = -1;//初始化，next[0] = -1;
    	while(ch[i])
    	{
    		if(j==-1 || ch[i]==ch[j])//匹配成功时，当前ch[i]对应的next值 变为 匹配成功的ch[j]的下标值。
    		{
    			i++;
    			j++;
    			next[i] = j;
    		}
    		else
    			j = next[j];//匹配失败则向左移动匹配串，重新寻找，即从next[j]对应的ch[next[j]]开始。
    	}
    }
    int kmp(char *ch1, char *ch2, int next[])
    {
    	int i, j, count=0;
    	
    	i=0;
    	j=0;
    
    	while(ch2[i])
    	{
    		if(j==-1 || ch2[i]==ch1[j])//与上方原理基本相同
    		{
    			i++;
    			j++;
    			if(ch1[j]==0)	//当匹配成功时
    			{
    				count++;	//计数的加1
    				j = next[j];//下一个匹配位置，不需要从匹配串的开始进行，只需从其对应的next数组里的值开始就行。				
    			}
    		}
    		else
    			j = next[j];
    	}
    	return count;
    }

[BAPC 2006 Qualification]: http://poj.org/searchproblem?field=source&key=BAPC+2006+Qualification