POJ 3461 Oulipo（——KMP算法）

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
题意描述：
输入两个串s1和s2
计算并输出s1在s2中能够匹配多少次
解题思路：
KMP模板题，加深对next数组的理解，主要问题是更新i或者j的位置，其实不用更新，你会发现其实到最后j是会自动返回的。
代码实现:

#include<stdio.h>
#include<string.h>
char s[1000010],t[10010];
void get_next(char t[],int next[],int l2);
int kmp(char s[],char t[]);
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",t,s);
        printf("%d\n",kmp(s,t)); 
    }
    return 0;
}
int kmp(char s[],char t[])
{
    int i,j,l1,l2,c;
    
    int next[10010];//next[]数组中存的是左上匹配串前后缀的相似度，从0到l2 
    l1=strlen(s);
    l2=strlen(t);
    get_next(t,next,l2);
    
    c=0;
    i=0;
    j=0;
    while(i < l1)
    {
        if(j==-1 || s[i] == t[j])
        {
            i++;
            j++;
        }
        else//少了个else 
            j=next[j];
        if(j==l2)//i和j均不用改动
            c++;
    }
    return c;
}
void get_next(char t[],int next[],int l2)
{
    int i,j;
    i=0;
    j=-1;
    next[0]=-1;
    while(i < l2)
    {
        if(j==-1 || t[i] == t[j])
        {
            i++;
            j++;
            if(t[i] != t[j])
            next[i]=j;
            else
            next[i]=next[j];
        }
        else//少了个else 
            j=next[j];
    }
    /*for(i=0;i<=l2;i++)
        printf("%d ",next[i]);
    printf("\n");*/
}

易错分析：

1、next数组存储的结果是错位的，正好能够被下次利用。

2、代码能力，注意细节，像少个else，找了半天错。