hdu 1686 Oulipo

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


kmp，看样例就可以看懂题目。
每次两个串，问第二个串了又几个第一个串，可以重叠，只要是连续的即可。
每找到一个串，j退到前缀下一位。

code:

#include <stdio.h>
#include <string.h>
int Next[10005],n;
char w[10005],t[1000005];
void getnext(int n)
{
    int i = 0,j = -1;
    Next[0] = -1;
    while(i < n) {
        if(j == -1 || w[j] == w[i]) {
            Next[++ i] = ++ j;
        }
        else j = Next[j];
    }
}
int kmp(int n,int m) {
    int c = 0;
    getnext(n);
    int i = -1,k = -1;
    while(i < m) {
        if(k == -1 || t[i] == w[k]) {
            i ++;k ++;
        }
        else k = Next[k];
        if(k == n)c ++,k = Next[k];
    }
    return c;
}
int main() {
    scanf("%d",&n);
    while(n --) {
        scanf("%s%s",w,t);
        printf("%d
",kmp(strlen(w),strlen(t)));
    }
}

 后来又做了一遍。

代码：

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
using namespace std;
void getNext(int *Next,char *w) {
    Next[0] = -1;
    int i = -1,j = 0;
    while(w[j]) {
        if(i == -1 || w[i] == w[j]) {
            Next[++ j] = ++ i;
        }
        else i = Next[i];
    }
}
int Kmp(char *w,char *t) {
    int Next[10001];
    getNext(Next,w);
    int i = 0,j = 0,c = 0;
    while(t[j]) {
        if(i == -1 || w[i] == t[j]) {
            i ++;
            j ++;
            if(!w[i]) {
                c ++;
                i = Next[i];
            }
        }
        else i = Next[i];
    }
    return c;
}
int main() {
    int n;
    scanf("%d",&n);
    char w[10001],t[1000001];
    for(int i = 0;i < n;i ++) {
        scanf("%s%s",w,t);
        printf("%d
",Kmp(w,t));
    }
}