kuangbin专题十六 KMP&&扩展KMP HDU1686 Oulipo
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.
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 在主串中匹配到j>=plen时,j继续回溯。
#include<stdio.h>
#include<string.h>
int Next[],n,m,_,tlen,plen;
char t[],p[]; void prekmp() {
tlen=strlen(t);
plen=strlen(p);
int i,j;
j=Next[]=-;
i=;
while(i<plen) {
while(j!=-&&p[i]!=p[j]) j=Next[j];
if(p[++i]==p[++j]) Next[i]=Next[j]; //这里判断的是模式串
else Next[i]=j;
}
} int kmp() {
prekmp();
int i,j,ans=;
i=j=;
while(i<tlen) {
while(j!=-&&t[i]!=p[j]) j=Next[j];
i++;j++;
if(j>=plen) { //主要区别
ans++;
j=Next[j];
}
}
return ans;
} int main() {
for(scanf("%d",&_);_;_--) {
scanf("%s",p);
scanf("%s",t);
printf("%d\n",kmp());
}
}
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