HDU 1686 Oulipo

Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6885 Accepted Submission(s): 2746

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

解题思路：赤裸裸的KMP套板题。
简介一下KMP算法：有人叫做看毛片算法，这让我一下子就记住了……它的主要作用就是快速匹配字符串复杂度为O(m + n),m、n为匹配字符串与待匹配字符串的长度。它的原理就是利用之前搜索过的结果减少运算。

其实KMP最核心的就只有一个next数组，这个搞懂了，KMP也就搞懂了，next数组存储的是字符串的前后缀最大匹配度，含义也可引申为遍历不符时，下一个字符串应比较的位置。还没接触next数组的童靴先去把next数组的原理搞清楚吧。

这套模板是我自己有所修改的，呵呵O(∩_∩)O~，仔细领悟吧。

#pragma comment(linker, "/STACK:1024000000,1024000000") //黑科技
#include <stdio.h>
#include <map>
#include <vector>
#include <queue>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <string>
#include <math.h>
using namespace std;
typedef long long LL;
const double eqs=1e-9;
//freopen("data.in", "r", stdin);
#define maxd 1000005
char str1[maxd];
char str2[10005];  
int cnt = 0;
int Next[10005];
void get_Next()
{
    int len = strlen(str2);
    Next[0] = 0;
    int k = 0;
    for(int i = 1; i < len; i++)
    {
        while(k > 0 && str2[i] != str2[k]) k = Next[k - 1];//
        if (str2[i] == str2[k]) k++;
        Next[i] = k;
    }
}
void kmp()
{
    int len1 = strlen(str1);
    int len2 = strlen(str2);
    int now = 0;
    for(int i = 0; i < len1; i++)
    {
        while(now > 0 && (str1[i] != str2[now] || now >= len2)) now = Next[now - 1];//
        if (str1[i] == str2[now]) now++;
        if (now == len2) cnt++;
    }
}
int main()
{
    #ifndef ONLINE_JUDGE
    //freopen("data.in", "r", stdin);
    #endif
    int t;
    scanf("%d", &t);
    while(t--)
    {
        memset(Next, 0, sizeof(Next));
        cnt = 0;
        scanf("%s %s", str2, str1);
        get_Next();//
        kmp();
        printf("%d\n", cnt);
    }
    return 0;
}