538

# 题目

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

# 题解

KMP算法,需要理解next数组的意义


ABABABBABBAABBAAB
ABBAAB
00011

ABABABBABBAABBAAB
ABBAAB
00011

ABABABBABBAABBAAB
ABBAAB
00011

ABABABBABBAABBAAB
ABBAAB
00011

ABABABBABBAABBAAB
ABBAAB
00011

++ans;
--i;
j = next[j - 1];

# 代码

#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1000005;

char s1[maxn], s2[maxn];
int next[maxn];
void getNext(char *s) {
int len = strlen(s);
next[0] = -1;
for (int i = 1; i < len; ++i) {
int t = next[i - 1];
while (t != -1 && s[t] != s[i - 1])
t = next[t];
next[i] = (s[i - 1] == s[t] ? t + 1 : 0);
}
//for (int i = 0; i <= len; ++i)
//    printf("next[%d]=%d\n", i, next[i]);
}
int KMP(char *a, char *b) {
getNext(b);
int alen = strlen(a);
int blen = strlen(b);
int i = 0, j = 0;
int ans = 0;
while (i != alen) {
//printf("judge a[%d](%c) b[%d](%c)\n", i, a[i], j, b[j]);
while (j != -1 && a[i] != b[j])
j = next[j];
++i, ++j;
if (j == blen) {
//printf("match at %d\n", i - blen);
++ans;
--i;
j = next[j - 1];
}
}
return ans;
}

int main() {
int T;
scanf("%d", &T);
while (T--) {
scanf("%s%s", s1, s2);
printf("%d\n", KMP(s2, s1));
}
return 0;
}


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