KMP算法——字符串的匹配

String Matching——字符串匹配

题目描述
Finding all occurrences of a pattern in a text is a problem that arises frequently in text-editing programs. Typically,the text is a document being edited,and the pattern searched for is a particular word supplied by the user. We assume that the text is an array T[1…n] of length n and that the pattern is an array P[1…m] of length m<=n.We further assume that the elements of P and T are all alphabets(∑={a,b…,z}).The character arrays P and T are often called strings of characters. We say that pattern P occurs with shift s in the text T if 0<=s<=n and T[s+1…s+m] = P[1…m](that is if T[s+j]=P[j],for 1<=j<=m). If P occurs with shift s in T,then we call s a valid shift;otherwise,we calls a invalid shift. Your task is to calculate the number of vald shifts for the given text T and p attern P.
输入描述:
For each case, there are two strings T and P on a line,separated by a single space.You may assume both the length of T and P will not exceed 10^6.
输出描述:
You should output a number on a separate line,which indicates the number of valid shifts for the given text T and pattern P.

示例1
输入
abababab abab
输出
3

牛客网题目链接

题意：就是字符串的匹配，看第二个字符串在第一个字符串中出现了几次
KMP纯模板题，其中KMP最主要的就是找next数组，相关知识可自行学习（王道的《数据结构》有相关知识）

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
int nx[1000100];
string m,n;
int num;
void getnx(){   //next数组的构造函数
	nx[1]=0;
		int j;
		for(int i=2,j=1;i<=m.size();){
			nx[i]=j;
			while(j&&m[j-1]!=m[i-1])
				j=nx[j];
			j++;
			i++;
		}
}
void kmp(){    //匹配函数
	for(int i=1,j=1;i<=m.size();){
		while(j&&n[j-1]!=m[i-1])
			j=nx[j];
		if(j==n.size()){
			j=nx[j];
			num++;
		}
		else{
			j++;
			i++;
		}
	}
}
int main(){
	while(cin>>m>>n){
		num=0;	
		getnx();
		kmp();
		cout<<num<<endl;
	}
	return 0;
}

一个更简单的方法就是直接用C++的STL库函数find

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
int main(){
	string s,t;
	cin>>s>>t;
	int count=0;
	int i=s.find(t);
	while(i!=-1){
		count++;
		i=s.find(t,i+1);    //第二个变量指的是从s第i+1个位置上开始进行匹配
	}
	cout<<count<<endl;
	return 0;
}