HDOJ 题目1686 Oulipo（KMP）

最新推荐文章于 2019-07-30 21:44:05 发布

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最新推荐文章于 2019-07-30 21:44:05 发布

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本文介绍了一个基于文本匹配的比赛任务，旨在快速准确地计算给定单词在长文本中出现的次数，包括重叠情况。文章提供了问题描述、输入输出格式及样例，并附带了AC代码实现。

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Oulipo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5575 Accepted Submission(s): 2230

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

Source

华东区大学生程序设计邀请赛_热身赛

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题目意思就是问匹配了多少次

ac代码

#include<stdio.h>
#include<string.h>
char s1[1000005],s2[1000005];
int next1[1000005];
void getnext(char *s,int *next)
{
    int i=0,j=-1;
    int len=strlen(s);
    next[0]=-1;
    while(i<=len)
    {
        if(j==-1||s[i]==s[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
int kmp(char *s1,char *s2)
{
    int len1=strlen(s1);
    int len2=strlen(s2);
    int i=0,j=0,ans=0;
    memset(next1,0,sizeof(next1));
    //memset(next2,0,sizeof(next2));
    getnext(s1,next1);
    //getnext(s2,next2);
//    for(i=0;i<=len1;i++)
//        printf("%d ",next1[i]);
//    printf("\n");
//    for(i=0;i<=len2;i++)
//        printf("%d ",next2[i]);
//    printf("\n");
    while(j<=len2)
    {
        if(i==-1||s1[i]==s2[j])
        {
            i++;
            j++;
        }
        else
            i=next1[i];
        if(i==len1)
        {
            ans++;
            i=next1[i];
        //    j=next2[j];
        }
//        printf("%d %d\n",i,j);
    }
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",s1,s2);
        printf("%d\n",kmp(s1,s2));
    }
}