3461 Oulipo 计算a中多少个与b匹配的子串 两个KMP模板

Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6169		Accepted: 2342

Description

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).

One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

1
3
0

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
const int maxn=2000000;
int next[maxn];
char a[maxn],b[maxn];
//s数组从0开始，但是next数组计算时是从1开始(next[i]--s[i-1]),计算结果也从1开始
void get_next(char *s,int next[])
{
int len=strlen(s);
int i=0,j=-1;
next[0]=-1;
while(i<len)
{
if(j==-1||s[i]==s[j]) i++,j++,next[i]=j;
else j=next[j];
}
}
int KMP(char *a,char *b)//a主串，b子串 返回值为a中与b匹配的第一个字母的位置(从0开始)
{
get_next(b,next);//计算next
int n=strlen(a),m=strlen(b);
int i=0,j=0;
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j]) i++,j++;
else j=next[j];
}
if(j>=m) return i-m;
return -1;
}
int KMP_Count(char *a,char *b)//a主串，b子串 返回值为共有a中多少个与b匹配的子串
{
get_next(b,next);//计算next
int n=strlen(a),m=strlen(b);
int i=0,j=0;
int cnt=0;
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j]) i++,j++;
else j=next[j];
if(j==m) cnt++,j=next[j];
}
return cnt;
}
int main()
{
int ci;scanf("%d",&ci);
while(ci--)
{
scanf("%s%s",b,a);
cout<<KMP_Count(a,b)<<endl;
}
return 0;
}

[/code]

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
using namespace std;
const int maxn=2000000;
int next[maxn];
char a[maxn],b[maxn];
//s数组从0开始，但是next数组计算时是从1开始(next[i]--s[i-1]),计算结果也从1开始
void get_next(char *s,int next[])
{
int len=strlen(s);
int i=0,j=-1;
next[0]=-1;
while(i<len)
{
if(j==-1||s[i]==s[j]) i++,j++,next[i]=j;
else j=next[j];
}
}
int KMP(char *a,char *b)//a主串，b子串 返回值为a中与b匹配的第一个字母的位置(从0开始)
{
get_next(b,next);//计算next
int n=strlen(a),m=strlen(b);
int i=0,j=0;
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j]) i++,j++;
else j=next[j];
}
if(j>=m) return i-m;
return -1;
}
int KMP_Count(char *a,char *b)//a主串，b子串 返回值为共有a中多少个与b匹配的子串
{
get_next(b,next);//计算next
int n=strlen(a),m=strlen(b);
int i=0,j=0;
int cnt=0;
while(i<n&&j<m)
{
if(j==-1||a[i]==b[j]) i++,j++;
else j=next[j];
if(j==m) cnt++,j=next[j];
}
return cnt;
}

/*void get_next(char *s,int next[])//next数组从0开始
{
int len=strlen(s);
next[0]=-1;
for(int j=1;j<len;j++)
{
int i=next[j-1];
while(s[j]!=s[i+1]&&i>=0) i=next[i];
if(s[j]==s[i+1]) next[j]=i+1;
else next[j]=-1;
}
}
int KMP(char *a,char *b)//a主串，b子串 返回值为a中与b匹配的第一个字母的位置(从0开始)
{
get_next(b,next);//计算next
int n=strlen(a),m=strlen(b);
int i=0,j=0;
while(i<m&&j<n)
{
if(b[i]==a[j]) i++,j++;
else if(i==0) j++;
else i=next[i-1]+1;
}
if(i<m) return -1;
return j-m;
}
int KMP_Count(char *a,char *b)
{
get_next(b,next);//计算next
int n=strlen(a),m=strlen(b);
int i=0,j=0,cnt=0;
while(i<m&&j<n)
{
if(b[i]==a[j]) i++,j++;
else if(i==0) j++;
else i=next[i-1]+1;
if(i==m) i=next[i-1]+1,cnt++;
}
return cnt;
}*/
int main()
{
int ci;scanf("%d",&ci);
while(ci--)
{
scanf("%s%s",b,a);
cout<<KMP_Count(a,b)<<endl;
}
return 0;
}