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  • POJ——T 3461 Oulipo

    http://poj.org/problem?id=3461

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 42698   Accepted: 17154

    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

    Source

     
     
    ①:hash 搞   对于y的任意子串x   hash(x,y)=hash[y]-hash[x]*p^(y-x+1)
     1 #include <algorithm>
     2 #include <iostream>
     3 #include <cstring>
     4 #include <cstdio>
     5 
     6 using namespace std;
     7 
     8 #define p 147
     9 #define ull unsigned long long
    10 char s1[10005],s2[1000005];
    11 ull n,ss1,ss2,P;
    12 ull lens,hs[1000005],ans;
    13 
    14 inline void Get_hash()
    15 {
    16     lens=0;
    17     for(ull i=1;i<=ss1;i++)
    18         lens=lens*p+s1[i]-'A';
    19 }
    20 inline void Get_hs()
    21 {
    22     for(ull i=1;i<=ss2;i++)
    23         hs[i]=hs[i-1]*p+s2[i]-'A';
    24 }
    25 inline ull Q_pow(ull a,ull b)
    26 {
    27     ull ret=1,base=a;
    28     for(;b;b>>=1,base*=base)
    29         if(b&1) ret*=base;
    30     return ret;
    31 }
    32 inline ull Check(ull x,ull y)
    33 {
    34     return hs[y]-hs[x-1]*P;
    35 }
    36 
    37 int main()
    38 {
    39     for(cin>>n;n--;ans=0)
    40     {
    41         scanf("%s%s",s1+1,s2+1);
    42         ss1=strlen(s1+1),ss2=strlen(s2+1);
    43         Get_hash(); Get_hs(); P=Q_pow(p,ss1);
    44         for(ull i=ss1;i<=ss2;i++)
    45             if(lens==Check(i-ss1+1,i)) ans++;
    46         cout<<ans<<endl;
    47     }
    48     return 0;
    49 }
        ② kmp
     1 #include <algorithm>
     2 #include <iostream>
     3 #include <cstring>
     4 #include <cstdio>
     5 
     6 using namespace std;
     7 
     8 int n,ss1,ss2;
     9 char s1[10005],s2[1000005];
    10 int lens,next[1000005],ans;
    11 
    12 inline void Get_next()
    13 {
    14     int j=0;next[1]=0;
    15     for(int i=2;i<=ss1;i++)
    16     {
    17         for(;j>0&&s1[i]!=s1[j+1];) j=next[j];
    18         if(s1[i]==s1[j+1]) j++;
    19         next[i]=j;
    20     }
    21 }
    22 
    23 inline void kmp()
    24 {
    25     int j=0;
    26     for(int i=1;i<=ss2;i++)
    27     {
    28         for(;j>0&&s2[i]!=s1[j+1];) j=next[j];
    29         if(s1[j+1]==s2[i]) j++;
    30         if(j==ss1) ans++,j=next[j];
    31     }
    32 }
    33 
    34 int main()
    35 {
    36     for(cin>>n;n--;ans=0)
    37     {
    38         scanf("%s%s",s1+1,s2+1);
    39         ss1=strlen(s1+1),ss2=strlen(s2+1);
    40         Get_next(); kmp();
    41         printf("%d
    ",ans);
    42     }
    43     return 0;
    44 }
    ——每当你想要放弃的时候,就想想是为了什么才一路坚持到现在。
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  • 原文地址:https://www.cnblogs.com/Shy-key/p/7356040.html
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