Oulipo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8574 Accepted Submission(s): 3459

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

       题意是要求输入一个单词和一段话，找出单词在这段话中出现的次数。典型的单匹配KMP算法，在这里简单回顾一下KMP算法的核心思想：
       KMP是一种非常高效的字符串匹配算法，其核心是计算字符串f每一个位置之前的字符串的前缀和后缀公共部分的最大长度（不包括字符串本身，否则最大长度始终是字符串本身）。获得f每一个位置的最大公共长度之后，就可以利用该最大公共长度快速和字符串O比较。当每次比较到两个字符串的字符不同时，我们就可以根据最大公共长度将字符串f向前移动(已匹配长度-最大公共长度)位，接着继续比较下一个位置。事实上，字符串f的前移只是概念上的前移，只要我们在比较的时候从最大公共长度之后比较f和O即可达到字符串f前移的目的。
       理解了kmp算法的基本原理，下一步就是要获得字符串f每一个位置的最大公共长度。这个最大公共长度在算法导论里面被记为next数组。在这里要注意一点，next数组表示的是长度，下标从1开始；但是在遍历原字符串时，下标还是从0开始。假设我们现在已经求得next[1]、next[2]、……next[i]，分别表示长度为1到i的字符串的前缀和后缀最大公共长度，现在要求next[i+1]。由上图我们可以看到，如果位置i和位置next[i]处的两个字符相同（下标从零开始），则next[i+1]等于next[i]加1。如果两个位置的字符不相同，我们可以将长度为next[i]的字符串继续分割，获得其最大公共长度next[next[i]]，然后再和位置i的字符比较。这是因为长度为next[i]前缀和后缀都可以分割成上部的构造，如果位置next[next[i]]和位置i的字符相同，则next[i+1]就等于next[next[i]]加1。如果不相等，就可以继续分割长度为next[next[i]]的字符串，直到字符串长度为0为止。

以下是KMP有参函数核心代码：

void makeNext(const char P[],int next[])
{
     int q,k;
     int m = strlen(P);
     next[0] = 0;
     for (q = 1,k = 0; q < m; ++q)
     {
         while(k > 0 && P[q] != P[k])
             k = next[k-1];
         if (P[q] == P[k])
        {
             k++;
        }
         next[q] = k;
     }
 }
 
 void kmp(const char T[],const char P[],int next[])
 {
     int n,m;
     int i,q;
     n = strlen(T);
     m = strlen(P);
     makeNext(P,next);
     for (i = 0,q = 0; i < n; ++i)
     {
         while(q > 0 && P[q] != T[i])
             q = next[q-1];
         if (P[q] == T[i])
         {
             q++;
         }
         if (q == m)
         {
             printf("Pattern occurs with shift:%d\n",(i-m+1));
         }
     }    
 }

以下是KMP无参函数核心代码：

void makeNext()
{
    int i = 0,j = -1;   
    next[0] = -1;  
    while (i<n)  
    {  
        if(j == -1 || a[i] == a[j])  
        {  
            i++;  
            j++;  
            next[i] = j;  
        }  
        else  
        j = next[j];  
    }  
 }

理解了以上内容，这个题就不难做出来了。
以下是这道题的AC代码：

#include<stdio.h>
#include<string.h>
#define max1 10005
#define max2 1000005
int next[max1]={0},count;
char str1[max1],str2[max2];
void makeNext(char P[max1],int next[max1])
{
     int q,k;
     int m = strlen(P);
     next[0] = 0;
     for (q = 1,k = 0; q < m; ++q)
     {
         while(k > 0 && P[q] != P[k])
             k = next[k-1];
         if (P[q] == P[k])
        {
             k++;
        }
         next[q] = k;
     }
}
 
 void kmp(char T[max2],char P[max1],int next[max1])
 {
     int n,m;
     int i,q;
	 count=0;
     n = strlen(T);
     m = strlen(P);
     makeNext(P,next);
     for (i = 0,q = 0; i < n; ++i)
     {
         while(q > 0 && P[q] != T[i])
             q = next[q-1];
         if (P[q] == T[i])
         {
             q++;
         }
         if (q == m)
         {
			 count++;
          
         }
     }
	 printf("%d\n",count);
 }
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		while(n--)
		{
		   scanf("%s",str1);
		   scanf("%s",str2);
		   kmp(str2,str1,next);
		}
	}
	return 0;
}

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