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\n",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\n",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;
}
}
