2013
12-21

# Oulipo

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.

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.

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.

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

1
3
0

http://acm.hdu.edu.cn/showproblem.php?pid=1686

其实这是一道简单的kmp  的算法     只要你在index_kmp()    将 i
的值一直小于N  就可以了    还是看看我具体代码吧     或许要看就知道了

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char w[10005],t[1000005];
int next[10005];
int N,M;
void get_next()
{
int i=0,j=next[0]=-1;
while(i<M)
{
if(j==-1||w[i]==w[j])
{
i++,j++;
next[i]=j;
}
else j=next[j];
}
}

int kmp()
{
int i=0,j=0;
int num=0;
while(i<N)
{
if(j==-1||t[i]==w[j])
{
i++,j++;
}
else j=next[j];
if(j==M)
{
num++;
j=next[j];
}

}
cout<<num<<endl;
}
int main()
{
int c,h,i;
while(cin>>c)
{
while(c--)
{
scanf("%s",w);
scanf("%s",t);
N=strlen(t);
M=strlen(w);
get_next();
kmp();
}
}
}