ACM-Binary String Matching

最新推荐文章于 2018-08-08 16:08:13 发布

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最新推荐文章于 2018-08-08 16:08:13 发布

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分类专栏： Acm

本文链接：https://blog.youkuaiyun.com/dally2/article/details/80159061

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本文介绍了一种针对仅由'0'和'1'组成的两个字符串A和B的匹配问题，目标是找出字符串A作为子串在B中出现的次数。提供了三种不同的实现方法：KMP算法、暴力匹配及使用C++标准库string的功能。

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Binary String Matching

时间限制： 3000 ms | 内存限制： 65535 KB

难度： 3

描述

Given two strings A and B, whose alphabet consist only ‘0’ and ‘1’. Your task is only to tell how many times does A appear as a substring of B? For example, the text string B is ‘1001110110’ while the pattern string A is ‘11’, you should output 3, because the pattern A appeared at the posit

输入

The first line consist only one integer N, indicates N cases follows. In each case, there are two lines, the first line gives the string A, length (A) <= 10, and the second line gives the string B, length (B) <= 1000. And it is guaranteed that B is always longer than A.

输出

For each case, output a single line consist a single integer, tells how many times do B appears as a substring of A.

样例输入

3
11
1001110110
101
110010010010001
1010
110100010101011

样例输出

3
0
3

三种方式：吐舌头

1.KMP

#include<iostream>
#include<cstring>
using namespace std;
const int Max=1005;
char T[Max],S[Max]; 
int nex[Max];
int lenT,lenS;
void getNex(){
   	int j=nex[0]=-1;
   	int i=0;
	while(i<lenT){
		while(j!=-1&& T[i]!=T[j]) j=nex[j];
		if(T[++i]==T[++j]) 
		nex[i]=nex[j];
		else
		nex[i]=j;
	} 
}
int  kmp(){
  
  int ans=0;
  getNex();
  int j=0;
  int i=0;
    while(i<lenS){
    	while (j!=-1&&S[i]!=T[j]) j=nex[j];
    	i++,j++;
    	if(j>=lenT){
		ans++;
		j=nex[j]; 
		} 
    } 
  return ans; 
}
int main()
{
	int n;
	cin>>n;
	while(n--){
	   memset(nex,0,sizeof(nex));
	   cin>>T>>S;
	   lenT=strlen(T);
	   lenS=strlen(S);
	   int ans=kmp();
	   cout<<ans<<endl;
	}
	return 0;
}

2.暴力

#include<iostream>
#include<cstring>
#include<string>
using namespace std;
char S[1005],T[1005];
int main()
{
	int n;
	cin>>n;
	while(n--){
	cin>>T>>S;
	int lenT=strlen(T);
	int lenS=strlen(S);
	int flag=0;
	int ans=0;
	for(int i=0;i<lenS;i++){
		flag=0;
		for(int j=0;j<lenT;j++){
			if(S[i+j]!=T[j]){
				flag=1;
				break;
			}
		}
		if(!flag){
			ans++;
		}
	}
	cout<<ans<<endl;	
	}
	return 0; 
}

3.string库：

#include<iostream>
#include<string>
using namespace std;
string s,t;
int main()
{
	int n;
	cin>>n;
	while(n--){
	
	cin>>t>>s;
	int len=0;
	int ans=0;
	while(len=s.find(t.c_str(),len),len!=string::npos){
	len++;
	ans++;	
	}
	cout<<ans<<endl;	
	}
	return 0;
}