• Oulipo(Kmp)


    Oulipo

    Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)
    Total Submission(s) : 67   Accepted Submission(s) : 44
    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
    题解:最基础的KMP,类似于看小说中的査找文字;注意abab abababab,出现的次数是3,而不是2,所以我们只需要在kmp中到p结尾时直接答案加一就好,而不是重新匹配,这样j跳到失配的上一个位置;重新匹配;
    代码:
     1 #include<stdio.h>
     2 #include<string.h>
     3 const int MAXN=10010;
     4 char str[MAXN],buf[MAXN*100];
     5 int len1,len2,cnt,p[MAXN];
     6 void getp(){int i=0,j=-1;
     7 p[0]=-1;
     8 while(i<len2){
     9     if(j==-1||str[j]==str[i]){
    10         i++,j++;
    11         p[i]=j;
    12     }
    13     else j=p[j];
    14 }
    15 }
    16 void kmp(){
    17     getp();
    18     int i=0,j=0;
    19     while(i<len1){
    20         if(j==-1||buf[i]==str[j]){
    21             i++,j++;
    22             if(j==len2)cnt++;
    23         }
    24         else j=p[j];
    25     }
    26 }
    27 int main(){
    28     int T;
    29     scanf("%d",&T);
    30     while(T--){cnt=0;
    31         scanf("%s%s",str,buf);
    32         len1=strlen(buf);
    33         len2=strlen(str);
    34         kmp();
    35         printf("%d
    ",cnt);
    36     }
    37     return 0;
    38 }
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace  std;
    typedef long long LL;
    #define mem(x,y) memset(x,y,sizeof(x))
    #define SI(x) scanf("%d",&x)
    #define PI(x) printf("%d",x)
    const int INF=0x3f3f3f3f;
    const int MAXN=1000010;
    int p[MAXN];
    
    void getp(char* s){
        int len=strlen(s);
        int i=0,j=-1;
        p[0]=-1;
        while(i<len){
            if(j==-1||s[i]==s[j]){
                i++;j++;
                p[i]=j;
            }
            else j=p[j];
        }
    }
    void kmp(char *s,char* m,int& ans){
        int len=strlen(m);
        getp(s);
        int len2=strlen(s);
        int j=0,i=0;
        while(i<len){
            if(j==-1||s[j]==m[i]){
                i++;j++;
                if(j==len2){
                    ans++;
                }
            }
            else j=p[j];
        }
    }
    int main(){
        char s[MAXN];
        char m[MAXN];
        int T;
        SI(T);
        while(T--){
            scanf("%s%s",s,m);
            int ans=0;
            kmp(s,m,ans);
            printf("%d
    ",ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/handsomecui/p/4710935.html
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