POJ 3461 Oulipo(KMP详解)

最新推荐文章于 2023-06-25 08:42:03 发布

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最新推荐文章于 2023-06-25 08:42:03 发布

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本文链接：https://blog.youkuaiyun.com/czh199638/article/details/52880260

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本文详细介绍KMP算法原理及其实现过程，包括next数组的意义和作用，通过具体实例讲解如何快速查找一个字符串在另一字符串中出现的次数，适合初学者理解和掌握。

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Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 36962		Accepted: 14918

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

题解：一道KMP模板题，通过这道题可以很好地理解KMP

KMP很重要的一点就是原字符串的下标i永不回退，回退的是匹配的字符串的下标j

还有一点重要的的是next[k]数组，next[k]表示的是下标为k之前的子字符串中最长的前缀后缀相同的长度，那next数组有什么用那，用处就是在匹配字符失败的时候直接移动j来进行下一次匹配。

在这里推荐几篇能够讲清楚kmp的文章：

从头到尾彻底理解KMP：http://blog.youkuaiyun.com/v_july_v/article/details/7041827

KMPnext数组详解：http://www.cnblogs.com/tangzhengyue/p/4315393.html

接下来是AC code(不过跑得好慢):

import java.io.*;
import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		sc.nextLine();
		for(int o=1;o<=n;o++){
			String pat=sc.nextLine();
			String txt=sc.nextLine();
			int count=0;
			int[]next=makeNext(pat);//创建匹配字符串的next数组
			int t_length=txt.length();
			int p_length=pat.length();
			int j=0;
			for(int i=0;i<t_length;i++){
				while(j>0&&txt.charAt(i)!=pat.charAt(j)){
					j=next[j];//当匹配失败时，回退j
				}
				if(txt.charAt(i)==pat.charAt(j)){
					j++;//如果当前字符匹配，j++往后找
				}
				if(j==p_length){
					count++;
					j=next[j];//找到了，count++过以后将当作没有匹配到，回退j
				}
			}
			System.out.println(count);
		}

	}
	public static int[] makeNext(String s){
		int s_length=s.length();
		int[] next=new int [s_length+1];
		int j=0;
		int k=-1;
		next[0]=-1;
		while(j<s_length){
			if(k==-1||s.charAt(j)==s.charAt(k)){
				next[++j]=++k;
			}
			else
				k=next[k];
		}
		return next;
	}

}
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