Problem 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 <stdio.h>
#include <string.h>
int next[10005];
char str1[1000005],str2[10005];
int cnt;
void get_next(int len2)//生成next数组
{
int i = 0,j = -1;
next[0] = -1;
while (i<len2)
{
if(j == -1 || str2[i] == str2[j])
{
i++;
j++;
if (str2[i]!=str2[j])
next[i] = j;
else
next[i] = next[j];
}
else
j = next[j];
}
}
int kmp(int len1,int len2)//kmp算法
{
int i=0,j=0;
get_next(len2);
while(i<len1)
{
if(j==-1||str1[i]==str2[j])
{
++i;
++j;
}
else
j=next[j];
if(j == len2)
{
cnt++;
j = next[j];
}
}
}
int main()
{
int n;
int len1,len2;
scanf("%d",&n);
getchar();
while(n--)
{
gets(str2);
gets(str1);
len1 = strlen(str1);
len2 = strlen(str2);
cnt = 0;
kmp(len1,len2);
printf("%d\n",cnt);
}
return 0;
}
KMP算法 hdu4686 Oulipo
原文：http://www.cnblogs.com/wangmenghan/p/5697747.html
