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.
InputThe 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.
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。
1 #include<stdio.h>
2 #include<string.h>
3 int Next[10010],n,m,_,tlen,plen;
4 char t[1000010],p[10010];
5
6 void prekmp() {
7 tlen=strlen(t);
8 plen=strlen(p);
9 int i,j;
10 j=Next[0]=-1;
11 i=0;
12 while(i<plen) {
13 while(j!=-1&&p[i]!=p[j]) j=Next[j];
14 if(p[++i]==p[++j]) Next[i]=Next[j]; //这里判断的是模式串
15 else Next[i]=j;
16 }
17 }
18
19 int kmp() {
20 prekmp();
21 int i,j,ans=0;
22 i=j=0;
23 while(i<tlen) {
24 while(j!=-1&&t[i]!=p[j]) j=Next[j];
25 i++;j++;
26 if(j>=plen) { //主要区别
27 ans++;
28 j=Next[j];
29 }
30 }
31 return ans;
32 }
33
34 int main() {
35 for(scanf("%d",&_);_;_--) {
36 scanf("%s",p);
37 scanf("%s",t);
38 printf("%d\n",kmp());
39 }
40 }
原文地址：https://www.cnblogs.com/ACMerszl/p/10263472.html