Oulipo

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 3461
64-bit integer IO format: %lld      Java class name: Main

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.

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.

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.

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

1
3
0

 解题：KMP，看不懂，网上找了点代码，研究研究，这份代码写得很俊啊！

#include
#include
#include
using namespace std;
const int MaxN = ;
char word[MaxN/], txt[MaxN];
int next[MaxN/];
void KMP_next(char b[], int pre[]) {
int n = strlen(b), k;
pre[] = -;
k = -;
for(int i = ; i < n; i++) { while(k > - && b[k+] != b[i]) k = pre[k];
if(b[k+] == b[i]) k++;
pre[i] = k;
}
}

int main() {
int n;
scanf("%d%*",&n);
while(n--) {
gets(word);
gets(txt);
KMP_next(word, next);
int cnt = , len = strlen(word);
for(int i = , j = -; txt[i]; ++i) {
while(j > - && word[j+] != txt[i]) j = next[j];
if(word[j+] == txt[i]) j++;
if(j == len-) {
cnt++;
j = next[j];
}
}
printf("%d\n", cnt);
}
return ;
}

BAPC 2006 Qualification

貌似这样写。。。。更优化，但是对此题而言，无用

#include
#include
#include
using namespace std;
const int maxn = ;
int fail[maxn];
void getNext(const char *pStr, int *nextArr) {
int i = , j = -, pLen = strlen(pStr);
nextArr[i] = j;
while (i < pLen) {
if (pStr[++i] != pStr[++j]) {
nextArr[i] = j;
while (j != - && pStr[i] != pStr[j]) j = nextArr[j];
} else nextArr[i] = nextArr[j];
}
}
char word[maxn],text[maxn];
int main(){
int n,ret;
scanf("%d",&n);
while(n--){
scanf("%s %s",word,text);
getNext(word,fail);
for(int i = ret = ,j = ; text[i]; ++i){
while(j != - && word[j] != text[i]) j = fail[j];
if(!word[++j]) ret++;
}
printf("%d\n",ret);
}
return ;
}

整理以后的，可以选择开启所谓的优化

#include
#include
#include
using namespace std;
const int maxn = ;
int fail[maxn];
char word[maxn],text[maxn];
void getFail() {
fail[] = -;
fail[] = ;
for(int i = ,j = -; word[i]; ++i) {
while(j != - && word[i] != word[j]) j = fail[j];
fail[i+] = ++j;
if(word[i+] == word[j]) fail[i+] = fail[j];//使用此句加优化，注释掉不加优化，都是正确的
}
}
int main() {
int n,ret;
scanf("%d",&n);
while(n--) {
scanf("%s%s",word,text);
getFail();
for(int i = ret = , j = ; text[i] ; ++i) {
while(j > - && word[j] != text[i]) j = fail[j];
if(!word[++j]) {ret++;j = fail[j];}
}
printf("%d\n",ret);
}
return ;
}