K - Oulipo

题目描述：

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：给出两个字符串，求第二个字符串中有多少个第一个字符串。

分析：

本题的第一反应，依次匹配。但是看到字符串的长度，便可知道这种朴素的匹配方式一定是会超时的。 所以在考虑就只能是KMP匹配了。
对于KMP匹配而言，只要找出next数组就不成问题了。
具体代码如下：

#include"stdio.h"
#include"string.h"
//得到next数组的函数
void Get_next(int next[],char word[])
{
    int i,len,j;
    len=strlen(word);
    next[0]=-1;
    i=0;
    j=-1;
    while(i<len)
    {
        if(j==-1||word[i]==word[j])
        {
            i++;j++;
            if(word[i]==word[j])
                next[i]=next[j];
            else
                next[i]=j;
        }
        else
            j=next[j];
    }

}
//KMP匹配
int Index_KMP(char text[],char word[],int next[])
{
    int i,j,len,lenw;
    int count=0;
    len=strlen(text);
    lenw=strlen(word);
    i=0;j=0;
//这次不是单纯的匹配成功就好。所以一定要匹配到text数组的最后一个元素
    while(i<len)
    {

            if(j==-1||text[i]==word[j])
            {
                i++;j++;
                if(j==lenw)//j到了word字符下的最后一个下标。即匹配成功
                {
                    count++;//匹配成功次数++ 
                    j=next[j];//在这里我们假设最后一个为匹配成功，从而将j回溯

                }
            }
            else
                j=next[j];

    }
  return count;
}
int main()
{
    int T;
    char word[10001];
    char text[1000001];
    int next[10001];
    int i,j,count;
    while(~scanf("%d",&T))
    {
        while(T--)
        {
            scanf("%s",word);
            scanf("%s",text);
            Get_next(next,word);
            count=Index_KMP(text,word,next);
            printf("%d\n",count);
        }
    }
}