【POJ 3461】Oulipo

桃扇骨 2021-10-30 06:52 98阅读 0赞

Oulipo

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;55633</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;21703</td> 
  </tr> 
 </tbody> 
</table>

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

[BAPC 2006 Qualification][]

题目大意：给你两个字符串p和s，求出p在s中出现的次数，求的是主串和子串的匹配问题。

#include<cstdio>
    #include<iostream>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    typedef long long ll;
    using namespace std;
    const int N=100010;
    char s[N * 10], p[N];
    int nextval[N];
    int lens, lenp;
    void getnext(){
        int i=0,j=-1;
        nextval[0]=-1;
        while(i!=lenp){
            if(j==-1||s[i]==s[j])
                nextval[++i]=++j;
            else
                j=nextval[j];
        }
    }
     
    int KMP(){
        int i=0,j=0,count=0;
        while(i != lens && j != lenp){
            if(s[i] == p[j] || j == -1)
                ++i, ++j;
            else
                j = nextval[j];
            if(j == lenp){
                count++;
                j = nextval[j];
            }
        }
        return count;
    }
     
    int main(){
        freopen("3461.in","r",stdin);
        freopen("3461.out","w",stdout);
        int ncase;
        int len;
        int ans;
        scanf("%d", &ncase);
        while(ncase--){
            scanf("%s%s", p, s);
            lens=strlen(s);
            lenp=strlen(p);
            getnext();
            ans=KMP();
            printf("%d\n", ans);
        }
        return 0;
    }

转载于:https://www.cnblogs.com/wuhu-JJJ/p/11146327.html

[BAPC 2006 Qualification]: http://poj.org/searchproblem?field=source&key=BAPC+2006+Qualification