HDU - 1686 Oulipo (KMP)

Oulipo

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

BAPC

AZA

AZAZAZA

VERDI

AVERDXIVYERDIAN

Sample Output

题意描述：
求出字符串A在字符串B中出现的次数。

解题思路：

利用kmp算法；先求出字符串A的next[]数组，用kmp算法在B当找到A数量加一，将j更新为next[j-1]继续查找。

KMP算法讲解：https://www.bilibili.com/video/av3246487?from=search&seid=17974959124889101719

#include<stdio.h>
#include<string.h>
int main()
{
    char a[1000005],b[10005];
    int i,j,next[10005],lena,lenb,num,t;
    while(scanf("%d\n",&t)!=EOF)
    {
        while(t--)
        {
            gets(b); 
            gets(a);
            lena=strlen(a);
            lenb=strlen(b);
            j=0;
            next[0]=0;
            //子串next数组求法 
            for(i=1;i<=lenb-1;)
            {
                if(j==0&&b[j]!=b[i])//当j等于0，b[i]与b[j]不相等时，i向后查，next[i]=0 
                {
                    next[i]=0;
                    i++;
                }
                if(b[i]==b[j])//相等时，i，j都向后走 
                {
                    next[i]=j+1;
                    i++;
                    j++;
                }
                if(j!=0&&b[i]!=b[j])//当j大于0时，b[i]b[j]不相等，j更新为前一个的next值 
                {
                    j=next[j-1];
                }
            }
            num=0;
            for(i=0,j=0;i<=lena-1;)
            {
                if(a[i]==b[j])
                {
                    j++;
                    if(j==lenb)//找到时数量加一，j更新为前一个的next值 
                    {
                        num++;
                        j=next[j-1];
                    }    
                    i++;
                }
                if(j==0&&a[i]!=b[j])//j等于0，a[i]与b[j]不相等时i向后查找 
                    i++;
                if(j>0&&a[i]!=b[j])//j大于0，不相等j更新 
                {
                    j=next[j-1];
                }
            }
            printf("%d\n",num);    
        }    
    }
    return 0;
}