Oulipo POJ - 3461

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 wordW and a textT, count the number of occurrences ofW inT. All the consecutive characters of W must exactly match consecutive characters ofT. 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 wordW in the textT.

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

1
3
0

KMP问题。。。。不过和KMP稍有不同，这个题要求计算次数而不是匹配成功就可以，所以next数组多求一位。

AC代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
int lent,lenp,next[10005];
char p[10005],t[1000005];
void getnext()
{
    int i=0,j=-1;
    next[0]=-1;
    while(i!=lenp)
    {
        if(j==-1||p[i]==p[j])
        {
            next[++i]=++j;
        }
        else
        {
            j=next[j];
        }
    }
}
int KMP()
{
    int i=0,j=0,c=0;
    while(i!=lent&&j!=lenp)
    {
        if(t[i]==p[j]||j==-1)
        {
            ++i,++j;
        }
        else
        {
            j=next[j];
        }
        if(j==lenp)
        {
            c++;
            j=next[j];
        }
    }
    return c;

}
int main()
{
    int n,i,k,j,ans;
    scanf("%d",&n);
    getchar();
    while(n--)
    {
        scanf("%s%s",p,t);
        lent=strlen(t);
        lenp=strlen(p);
        getnext();
        ans=KMP();
        printf("%d\n",ans);
    }
    return 0;
}