【POJ3461】Oulipo

Oulipo Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 35904 Accepted: 14492

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

[Submit]   [Go Back]   [Status]   [Discuss]

这个是子串匹配，和模板的模式串匹配有一点不同，就那一点next数组值的存储范围包括第lens个，因为你判断玩一个子串后要判断是不是前面有已经重合的部分，就像第二个输入输出样例。

#include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; const int N = 1e6+20; const int M = 1e5+10; char p[N],s[M]; int next[M]; int lens,lenp; void getNext() { int j,k; next[0]=-1; j=0,k=-1; while(j!=lens) { if(k==-1||s[k]==s[j]) { k++; j++; next[j]=k; } else k=next[k]; } } int KMPMatch() { int i=0,j=0,count=0; while(i<lenp) { if(j==-1||p[i]==s[j]) { i++; j++; } else j=next[j]; if(j==lens) { count++; j=next[j]; } } return count; } int main() { int T; scanf("%d",&T); while(T--) { scanf("%s%s",s,p); lens=strlen(s); //加了这两句就能ac，不加直接strlen（s）就超时，我也是醉了 lenp=strlen(p); //不造为啥，以后要规范细节啊，否则容易吃大亏， //尽量避免意向不到的错误而导致改都不知道怎么改 getNext(); int ans=KMPMatch(); printf("%d\n",ans); } return 0; } 题目地址：http://poj.org/problem?id=3461