POJ-3461 Oulipo (经典) kmp ʕ •ᴥ•ʔ

最新推荐文章于 2019-10-18 13:31:02 发布

henucm

最新推荐文章于 2019-10-18 13:31:02 发布

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本文链接：https://blog.youkuaiyun.com/henucm/article/details/81869812

本文介绍了一种高效的字符串匹配算法——KMP算法，并通过一个具体的编程实例详细解释了该算法的工作原理及实现步骤。文章首先定义了问题背景，然后介绍了如何构建next数组来加速匹配过程，最后展示了完整的C++代码实现。

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Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 49354		Accepted: 19605

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

题意：给你两个串子，第一为模板串第二为子串求子串中有多少个模板串

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#define ll long long
using namespace std;
string s,ss;
int ans;
int nexxt[1000010];
void getnext()
{
	int j=-1,i=0;
	nexxt[0]=-1;
	while(i<ss.size())
	{
		if(j==-1||ss[i]==ss[j])
		{
			i++,j++;
			nexxt[i]=j;
		}
		else
		j=nexxt[j];
	}
        //  A B C A B D
        // -1 0 0 0 1 2 0  nexxt数组的值 
        
     
}
void kmp()
{
	getnext();// 计算出nexxt数组的值  nexxt 也可以求循环串 
	int i=0,j=0;
	while(i<s.size())
	{
		if(s[i]==ss[j]||j==-1)
		{
			i++,j++;//值一样 跳过
		}
		else
			j=nexxt[j];// 看图
		if(j==ss.size())
		{
			ans++;
		}
	}

}
int main()
{
	ios::sync_with_stdio(false);
	int t;
	cin>>t;
	while(t--)
	{
		ans=0;
		cin>>ss>>s;
		kmp();
		cout<<ans<<endl;
	}
	return 0;
}