Binary String Matching

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

輸入

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.

輸出

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

樣例輸入

3
11
1001110110
101
110010010010001
1010
110100010101011 

樣例輸出

3
0
3 

【分析】串的模式匹配

        題意：輸入串A（模式串）和串B（主串），問串A在串B中出現了幾次。

        這裡使用更高效的KMP演算法實現，稍作修改即可實現串A出現次數的統計。

#include <stdio.h>
#include <string.h>
#define maxlenA 15
#define maxlenB 1005
int N;
//next陣列-模式串部分匹配資訊
//next[i]表示模式串的第i個字元與主串B中相應字元"失配"時，
//模式串A中需重新和主串B中該字元進行比較的字元位置 
int next[maxlenA]; 
char A[maxlenA],B[maxlenB];  //A-模式串 B-主串 
//求模式串A的部分匹配資訊 
void getNext(char *s,int *next)
{
	int i,j,len;
	i=0,j=-1;
	len=strlen(s);
	//規定next[0]=-1 
	next[i]=j; 
	//打表 
	while(i<len)
	{
		if(j==-1 || s[i]==s[j])
		{
			i++;
			j++;
			next[i]=j;
		} 
		else
			j=next[j];
	}
}
//主串s 模式串t，統計模式串t在主串s中的出現次數 
void KMP(char *s,char *t)
{
	int i,j;
	int lens,lent;//主串s和模式串t的長度 
	int ans;      //模式串t在主串s中的出現次數 
	i=j=ans=0;
	lens=strlen(s);
	lent=strlen(t);
	getNext(A,next);
	while(i<lens && j<lent)
	{
		//j在模式串第一個字元前 或 主串和模式串對應位置字元相同，則向後考察 
		if(j==-1 || s[i]==t[j])
		{
			i++;
			j++;
		}
		//否則，j回到重新和主串s匹配的位置 
		else
			j=next[j];
		//匹配成功，i不動，j從next[j]位置開始繼續與主串s匹配 
		if(j>=lent)
		{
			ans++;
			j=next[j];
		} 
	}
	printf("%d\n",ans);
}
int main()
{
	int i;
	scanf("%d",&N);
	while(N--)
	{
		scanf("%s",A);
		scanf("%s",B);
		KMP(B,A);
	}
	return 0;
}