• hdu 1686 Oulipo


    Oulipo

    http://acm.hdu.edu.cn/showproblem.php?pid=1686

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

    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
    题目大意:输出第一个字符串在第二个字符串中出现的次数
    KMP模板题
    初次写,1个错误:
    getnext函数中i的枚举要从1开始,不能从0,
    如果从0开始,j=f[0]=0,i=0 ,两者相等 f[1]=1
    然后j=f[1]=1,i=1,两者相等
    ……
    会导致一直相等
    因为f[i]表示的是0——i-1位的最大匹配长度
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    using namespace std;
    int n,len1,len2,f[1000001],ans;
    char s1[1000001],s2[1000001];
    void getnext()
    {
        for(int i=1;i<len1;i++)
        {
            int j=f[i];
            while(j&&s1[i]!=s1[j]) j=f[j];
            f[i+1]= s1[i]==s1[j] ? j+1 : 0;
        }
    }
    void getans()
    {
        int j=0;
        for(int i=0;i<len2;i++)
        {
            while(j&&s1[j]!=s2[i]) j=f[j];
            if(s1[j]==s2[i]) j++;
            if(j==len1) ans++;
        }
        printf("%d
    ",ans);
    }
    int main()
    {
        scanf("%d",&n);
        while(n--)
        {
            cin>>s1>>s2;
            len1=strlen(s1);len2=strlen(s2);
            memset(f,0,sizeof(f));
            getnext();
            getans();
            ans=0;
        }
    }
    //由于字符串位置从0开始存
    //所以f[i]在数值上=最大前缀子串=i的最大后缀子串=下一个要检验的位置 
    //也就是保证了 0——f[i]-1是相等的,再匹配i与f[i] 
    #include<cstdio>
    #include<cstring>
    #include<iostream>
    using namespace std;
    int n,len1,len2,f[1000001],ans;
    char s1[1000001],s2[1000001];
    void getnext()
    {
        for(int i=1;i<len1;i++)//注意这里从1开始 
        {
            int j=f[i];//j=下一个要匹配的位置 
            while(j&&s1[i]!=s1[j]) j=f[j];
            f[i+1]= s1[i]==s1[j] ? j+1 : 0;//j+1:字符串从0开始,所以长度+1 
        }
    }
    void getans()
    {
        int j=0;
        for(int i=0;i<len2;i++)//这里从1开始 
        {
            while(j&&s1[j]!=s2[i]) j=f[j];
            if(s1[j]==s2[i]) j++;
            if(j==len1) ans++;
        }
        printf("%d
    ",ans);
    }
    int main()
    {
        scanf("%d",&n);
        while(n--)
        {
            cin>>s1>>s2;
            len1=strlen(s1);len2=strlen(s2);
            memset(f,0,sizeof(f));
            getnext();
            getans();
            ans=0;
        }
    }
    一点儿注释
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  • 原文地址:https://www.cnblogs.com/TheRoadToTheGold/p/6483062.html
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