hdoj 1686 Oulipo【求一个字符串在另一个字符串中出现次数】

Oulipo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7141    Accepted Submission(s): 2835

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>
#define MAX 1000100
char p[10010],str[MAX];
int f[10010];
void getfail()
{
	int i,j,len;
	len=strlen(p);
	f[0]=f[1]=0;
	for(i=1;i<len;i++)
	{
		j=f[i];
		while(j &&p [i]!=p[j])
		j=f[j];
		f[i+1]=p[i]==p[j]?j+1:0;
	}
}
int main()
{
	int n,m,j,i,t;
	int l1,l2;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%s",p,str);
		getfail();
		l1=strlen(p);
		l2=strlen(str);
		int s=0,j=0;
		for(i=0;i<l2;i++)
		{
			while(j&&str[i]!=p[j])
            j=f[j];
            if(str[i]==p[j])
            j++;
            if(j>=l1)
            {
                s++;
                j=f[j];//注意此处 
            }
		}
		printf("%d
",s);
	}
	return 0;
}

失配函数优化写法：

#include<stdio.h>
#include<string.h>
#define MAX 10010
int ans,len1,len2;
char s[MAX],str[MAX*100];
int f[MAX];
void huang()
{
	int i=0,j = -1;
	f[i]=j;
	while(i < len1)
	{
		if(j == -1||s[i]==s[j])
		{
			i++;
			j++;
			f[i]=j;
		}
		else
		j=f[j];
	}
}
void kmp()
{
    huang();
	int i=0,j=0;
	for(i=0;i<len2;i++)
	{
		while(j && str[i] != s[j])
		j = f[j];
		if(str[i]==s[j])
			j++;       
		if(j >= len1)
		{
			ans++;
			j = f[j];
		}
	}
}
int main()
{
	int i,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s%s",s,str);
		len1=strlen(s);
		len2=strlen(str);
		huang();
		ans=0;
		kmp();
		printf("%d
",ans);
	}
	return 0;
}

　　

　

　借着此题，写下求失配函数的模板：

/*
*  f[]为失配函数数组 
*  
*/
getfail()
{
	int i,j;
    int len = strlen(p);
    f[0]=f[1]=1;
    for(i = 1; i < len; i++)
    {
    	j = f[i];
    	while(j && p[i] != p[j])
    	j = f[j];
    	f[i+1]= p[i] ==p[j]?j+1:0;
    }
}