HDU 1686.Oulipo

285


题目



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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

首先回顾一下KMP的步骤

ABABABBABBAABBAAB
ABBAAB
 00011

ABABABBABBAABBAAB
ABBAAB
00011

ABABABBABBAABBAAB
    ABBAAB
     00011

ABABABBABBAABBAAB
    ABBAAB
     00011

ABABABBABBAABBAAB
        ABBAAB
         00011


当遇到不匹配的时候,跳转到其对应的next指针位置(从开始位置到该位置之前的内容与当前位置前的内容相同)

当匹配结束后,我们可以"假装"匹配失败,这样就会紧接着往下匹配
记一下数即可

与正常的kmp的区别仅在于
++ans;
--i;
j = next[j - 1];

代码


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#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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