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

題目連結

求模式串在主串中的出現次數，KMP演算法模板題。

AC程式碼:

#include <iostream>
#include <vector>
#include <string>
using namespace std;

void KMPPre(string Pattern, vector<int> &Next) {
    int i = 0, j = -1;
    Next[0] = -1;
    int Len = int(Pattern.length());
    while (i != Len) {
        if (j == -1 || Pattern[i] == Pattern[j]) {
            Next[++i] = ++j;
        }
        else {
            j = Next[j];
        }
    }
}

void PreKMP(string Pattern, vector<int> &Next) {
    int i, j;
    i = 0;
    j = Next[0] = -1;
    int Len = int(Pattern.length());
    while (i < Len) {
        while (j != -1 && Pattern[i] != Pattern[j]) {
            j = Next[j];
        }
        if (Pattern[++i] == Pattern[++j]) {
            Next[i] = Next[j];
        }
        else {
            Next[i] = j;
        }
    }
}

int KMPCount(string Pattern, string Main, vector<int> &Next) {
    //PreKMP(Pattern, Next);
    KMPPre(Pattern, Next);
    int PatternLen = int(Pattern.length()), MainLen = int(Main.length());
    int i = 0, j = 0;
    int Ans = 0;
    while (i < MainLen) {
        while (j != -1 && Main[i] != Pattern[j]) {
            j = Next[j];
        }
        i++; j++;
        if (j >= PatternLen) {
            Ans++;
            j = Next[j];
        }
    }
    return Ans;
}

int main(int argc, char *argv[]) {
    int T;
    cin >> T;
    for (int Case = 1; Case <= T; ++Case) {
        string Main, Pattern;
        cin >> Pattern >> Main;
        vector<int> Next(int(Pattern.length()) + 1, 0);
        cout << KMPCount(Pattern, Main, Next) << endl;
    }
    return 0;
}