Binary String Matching

题目链接  http://acm.nyist.net/JudgeOnline/problem.php?pid=5

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

31110011101101011100100100100011010110100010101011 

样例输出

303 

解题思路：

这道题就是数据结构中bf算法的应用。y 表示长的字符串，x 表示短的字符串，i 表示y的第 i 个元素，j 表示 x 的第 j 个元素。文字说不明白，大家看一下代码吧，很容易理解的。

代码：

#include<iostream>#include<cstring>using namespace std;int bf(string x,string y){    int i=0,j=0,sum=0;   //sum表示匹配的个数    while(i<y.size()&&j<x.size())    {        if(y[i]==x[j])   //如果对应的元素一样        {            i++;            j++;            if(j==x.size())   //如果已经匹配到了小字符串的最后一位后面的一位，那么说明这个子串存在于大字符串中            {                sum++;   //匹配的个数++                i=i-x.size()+1;   //让i指向 匹配成功 的子串的下一个字符                j=0;   //让j指向小字符串的第一个字符            }        }        else        {            i=i-j+1;   //这是一个规律，如果匹配失败i=i-j+1，记住即可            j=0;   //如果匹配失败，小字符串需要从头开始再次开始匹配        }    }    return sum;}int main(){    string x,y;    int t;    cin>>t;    while(t--)    {        cin>>x>>y;        cout<<bf(x,y)<<endl;    }    return 0;}