poj 3461 Oulipo (KMP入门）

DescriptionThe 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: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 textT, 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.InputThe 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.OutputFor 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

题意：在第二个字符串中查找第一个字符串出现的个数。
以第一个字符串建立next函数，然后在第二个字符串中查找。

#include<cstdio>
#include<cstring>
using namespace std;
int next[100000];
char s1[100000],s2[2000000];
void get_next()
{
int i=0,j=-1;
next[0]=-1;
while (s1[i])
{
if (j==-1||s1[i]==s1[j])
{
i++;
j++;
next[i]=j;
}
else j=next[j];
}
}
int kmp()
{
int i=0,j=0,k=0,len;
len=strlen(s1);
while (s2[i])
{
if (j==-1||s1[j]==s2[i])
{
j++;
i++;
}
else j=next[j];
if (j==len)
{
k++;
j=next[j];
}
}
return k;
}
int main()
{
int n;
scanf("%d",&n);
getchar();
while (n--)
{
gets(s1);
gets(s2);
memset(next,0,sizeof(next));
get_next();
int ans=kmp();
printf("%d\n",ans);
}
}