hdoj 1686 Oulipo 【kmp(模板)】

Oulipo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7789 Accepted Submission(s): 3143



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

[align=left]Sample Output[/align]

1
3
0

[align=left]Source[/align]
华东区大学生程序设计邀请赛_热身赛

思路：
输入一个数T代表测试数据的个数，然后T组数据，每组数据包括两个字符串！第一个字符串代表子字符串（就是匹配字符串），第二个字符串代表文本字符串（也就是被匹配字符串），然后让你看看文本里面有多少个子字符串！

代码：

#include <stdio.h>
#include <string.h>
char a[10005];
char b[1000005];
int p[10005];
int lena,lenb;
int cnt;
void getp()
{
int i=0,j=-1;
p[0]=-1;
while(i<lena)
{
if(j==-1||a[i]==a[j])
{
i++;j++;
p[i]=j;
}
else
j=p[j];
}
}
void kmp()
{
getp();
int i=0,j=0;
while(i<lenb)
{
if(j==-1||b[i]==a[j])
{
i++;j++;
if(j==lena)
{
cnt++;
}
}
else
j=p[j];
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%s%s",a,b);
cnt=0;
lena=strlen(a);
lenb=strlen(b);
kmp();
printf("%d\n",cnt);
}
return 0;
}