Poj3461 Oulipo

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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

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3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

1
2
3
1
3
0

Analysis

KMP板题….

给出两个字符串，其中一个是另一个的子串，求子串P在母串T中出现了多少次。

显然暴力做法是从左到右一个个匹配，如果有某个字符不匹配，就跳回去，把字串右移。当然这是肯定会超时的。

观察发现，当匹配失败时，我们已经知道前面有一些字符是匹配的了，这时仅仅右移一位就显得很笨拙。而KMP的思想是：利用已经部分匹配这个有效信息，保持i指针不回溯，通过修改j指针，让模式串尽量地移动到有效的位置。

所以整个KMP的重点在于：当某一个字符与主串不匹配时，我们应该知道j指针要移动到哪？

如下图，C和B不匹配了，可以把j移到第2位，因为前面有两个字母是一样的：

可看出，当匹配失败时，j要移动的下一个位置k有这样的性质：最前面的k个字符和j之前的最后k个字符是一样的。

如上图，有公式：

那么

因为P的每一个位置都可能发生不匹配，所以用一个next数组保存每一个位置j对应的k，next[j]=k，表示当$T[i]!=P[j]$时，j的下一个位置。函数getnext()即为其过程。

对比这两个图发现规律：

证明：

当$P[k]!=P[j]$时，如下图

此时应为$k=next[k]$，如下图

然而上述算法还有点缺陷

如图，这一步移动毫无意义，因为后面的B已经不匹配，那么前面的B肯定也不匹配，原因在于$P[j]==P[next[j]]$。所以在getnext()中特判一下这种情况就ok了。

Code

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#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 1000100;
char s1[maxn], s2[maxn];
int Next[maxn];
int lens1, lens2;
int KMP()
{
	int i = 0, j = 0, ans = 0;
	while (i < lens1&&j < lens2)
	{
		if (i == -1 || s1[i] == s2[j])
		{
			i++; j++;
		}
		else
		{
			i = Next[i];
		}
		if (i == lens1)
		{
			ans++;
			i = Next[i];
		}
	}
	return ans;
}
void getnext()
{
	Next[0] = -1;
	int i = 0, j = -1;
	while (i < lens1)
	{
		if (j == -1 || s1[i] == s1[j])
		{
			i++; j++;
			if (s1[i] == s1[j])Next[i] = Next[j];
			else Next[i] = j;
		}
		else
		{
			j = Next[j];
		}
	}
}
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%s", s1);
		scanf("%s", s2);
		lens1 = strlen(s1);
		lens2 = strlen(s2);
		getnext();
		int ans = KMP();
		printf("%d\r\n", ans);
	}
	return 0;
}