nyoj 5 Binary String Matching （KMP）

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

这题数据量较小，所以朴素算法也可以A掉，但目的是学习kMP字符串匹配算法，所以，本题我用的KMP算法求解，第一次接触kmp，写起来还是比较棘手…… 其他就不多说了，还是要多练习才有手感。
AC代码：

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=1000+10;
char a[12],b[maxn];
int next[12];

void get_next(){
next[0]=0; //next[i]表示对应0~i位置匹配的最长长度
int k=0; //前后缀匹配最长长度
int len=strlen(a);
for(int p=1;p<len;p++){
while(k>0 && a[p]!=a[k])
k=next[k-1];

if(a[p]==a[k]){
k++;
}
next[p]=k;
}
}

int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%s%s",a,b);
get_next();

int cnt=0;
int len1=strlen(a);
int len2=strlen(b);
int i=0,j=0;

while(j<len2){
if(a[i]==b[j]){
i++;
j++;
}
else if(i==0){ //表示从头开始，无法匹配到，j+1;
j++;
}
else {
i=next[i-1];
}

if(i==len1){
cnt++;
i=next[i-1];
}

}

printf("%d\n",cnt);
}
return 0;
}