HDU 1711 Oulipo

Oulipo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8375    Accepted Submission(s): 3380

Problem Description

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.

 

Input

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.

 

Output

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.

 

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

 

Sample Output

1
3
0

 

继续全裸套KMP轻松过，但是不知道为什么G++老是超时但是C++就在62ms内妙过。有点奇怪。

但是还是想说，模版真好用。。。

代码如下：

/*************************************************************************
	> File Name: Oulipo.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Tue 24 Nov 2015 01:27:59 AM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

void kmp_pre(char x[], int m, int nexta[]){
    int i , j;
    j = nexta[0] = -1;
    i = 0;
    while(i < m){
        while(j != -1 && x[j] != x[i])
            j = nexta[j];
        nexta[++ i] = ++ j;
    }
}
void preKMP(char x[], int m, int kmpNext[]){
    int i, j;
    j = kmpNext[0] = -1;
    i = 0;
    while(i < m){
        while(j != -1 && x[i] != x[j])
            j = kmpNext[j];
        if(x[++ i] == x[++ j])
            kmpNext[i] = kmpNext[j];
        else 
            kmpNext[i] = j;
    }
}

int nexti[10010];
int KMP_Count(char x[], int m, char y[], int n){
    int i , j;
    int ans = 0;
    kmp_pre(x, m, nexti);
    i = j = 0;
    while(i < n){
        while(j != -1 && y[i] != x[j])
            j = nexti[j];
        i ++;
        j ++;
        if(j >= m){
            ans ++;
            j = nexti[j];
        }
    }
    return ans;
}

char a[10010];
char b[1000010];
int T;
int main(void){
    cin >> T;
    getchar();
    while(T --){
        gets(a);
        gets(b);
        printf("%d
", KMP_Count(a, strlen(a), b, strlen(b)));
    }
}