poj 3461 Oulipo ----KMP，字符串处理

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Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12706		Accepted: 5075

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:


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

题目大意： 赤裸裸的KMP。就是求 子串 在 主串中出现的次数。

题目思路： KMP 。KMP 重要的就是 next数组的建立。next数组就是对 子串 在与 主串匹配时加了预处理。

使得匹配效率提升了很多。避免了不必要的匹配。预处理也就是求next数组这里可能难理解一点

http://baike.baidu.com/view/574553.htm

#include;
#include;
int next[10050];
char p[10050],t[1001000];
void get_next(char *a,int len);
int main()
{
// freopen("in.txt","r",stdin);
int n,lenp,lent,i,j;
while(scanf("%d",&n)!=EOF)
{
while(n--)
{
int sum=0;
i=0,j=0;
scanf("%s%s",p,t);
lenp=strlen(p);
lent=strlen(t);
get_next(p,lenp);       //求next数组
// for(i=0;i<lenp;i++)
// printf("%d ",next[i]);
while(i<lent)             //匹配
{
if(j==-1|| t[i]==p[j])
{
i++; j++;
if(j==lenp)
{
sum++;
j=next[j];
}
}
else j=next[j];
}
printf("%d\n",sum);
}
}
return 0;
}
void get_next(char *a,int len)
{
int i=0,j=-1;
next[0]=-1;
while(i<len)        //很巧妙的方法。
{
if(j==-1 || a[i]==a[j])
{
i++; j++;
next[i]=j;
}
else j=next[j];
}
}