nyoj5 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 

题意：求出主串中模式串出现的次数.

朴素算法code:

#include <cstdio>
#include <cstring>
int main()
{
  int n,count;
  char a[200],b[1200];
  scanf("%d",&n);
  getchar();
  while(n--)
  {
    count=0;
    int i=0,j=0,len;
    scanf("%s\n%s",a,b);
    len=strlen(b);
    while(i<=len)
    {
      if(a[j]=='\0')
      {
        count++;
        i=i-j+1;
        j=0;
      }
      else if(a[j]==b[i])
      {
        i++;
        j++;
      }
      else
      {
        i=i-j+1; //关键在于回溯
        j=0;
      }
    }
    printf("%d\n",count);
  }
  return 0;
}

KMP算法:

#include<cstdio>
#include<cstring>
int nextval[200];
void get_next(char a[])//得到next数组；
{
  int len;
  int i=0,j=-1;
  nextval[0]=-1;
  len=strlen(a);
  while(i<=len)
  {
    if(j==-1 || a[i]==a[j])
    {
      ++i;
      ++j;
      if(a[i]==a[j])
      nextval[i] = nextval[j];  //把回溯的内容全换成是next数组；
      else
      nextval[i] = j;
    }
    else
      j=nextval[j];
  }
}
int kmp(char a[],char b[])//kmp的主体函数
{
  int i=0,j=0,count=0;
  int lena,lenb;
  lena=strlen(a);
  lenb=strlen(b);
  get_next(a);
  while(i<=lenb)
  {
    if(j==-1 || a[j]==b[i])
    {
      ++i;
      ++j;
    }
    else
      j=nextval[j];
    if(j>=lena)
    {
      count++;
      j=nextval[j];
    }
  }
  return count;
}
int main()
{
  int n;
  char a[20],b[1200];
  scanf("%d",&n);
  while(n--)
  {
    scanf("%s\n%s",a,b);
    printf("%d\n",kmp(a,b));
  }
  return 0;
}