HDOJ 1686 POJ 3461 Oulipo kmp

Oulipo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7326    Accepted Submission(s): 2939

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

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

 

 

Sample Output

130

 

很基础的kmp。其实题目本身很好做的，就算不理解kmp也能够做的。不过要注意的是，getnext函数求的是next，是str的失败数组。而kmp求的是str与buf的对称性。所以第一个函数的i<len1而第二个i<len2，并且第一次是i=0，j=-1，第二次的时候i=0，j=0。

HDOJ  AC代码

#include<stdio.h>#include<string.h>#define max 10010char str[max],buf[max * 100];int next[max],len1,len2;void getnext() {int i = 0,j = -1;next[i] = j;while(i < len1){if(j == -1 || str[i] == str[j]){++i;++j;next[i] = j;}else j = next[j];}}int kmp() { int i = 0,j = 0,cnt = 0;getnext();while(i < len2){if(j == -1 || buf[i] == str[j]){++i;++j;if(j == len1) cnt++;}else j = next[j];}return cnt;}int main(){int t;scanf("%d",&t);while(t--){scanf("%s%s",str,buf);len1 = strlen(str);len2 = strlen(buf);printf("%d\n",kmp());}return 0;}

 

注意，宏定义max的时候不要用10000+10的形式，会RE。
POJ  AC代码 (2015 08 10)

#include<stdio.h>#include<string.h>#define maxn 10010char str[maxn],buf[maxn*100];int next[maxn],len1,len2,cnt;void getnext(){int i=0,j=-1;next[i]=j;while(i<len1){if(j==-1||str[i]==str[j]){++i;++j;next[i]=j;}else j=next[j];}}int kmp(){int i=0,j=0;getnext();while(i<len2){if(j==-1||buf[i]==str[j]){++i;++j;if(j==len1) cnt++;}else j=next[j];}return cnt;}int main(){int t;scanf("%d",&t);while(t--){scanf("%s%s",str,buf);cnt=0;len1=strlen(str);len2=strlen(buf);printf("%d\n",kmp());}return 0;}