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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:
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
#include<iostream> #include<cstring> #include<cstdio> #include<cmath> using namespace std; char s1[11000],s2[1100000]; int next[11000]; int match[1100000]; int main() { int n,i,j,k; cin>>n; getchar(); for(i = 0;i<n;i++) { int num = 0; scanf("%s",s1+1); scanf("%s",s2+1); next[1] = 0; int l1 = strlen(s1+1); for(j = 2;j<l1+1;j++) { int t = next[j-1]; while(t&&s1[j]!=s1[t+1]) t = next[t]; if(s1[j] == s1[t+1])t++; next[j] = t; } int l2 = strlen(s2+1); match[0] = 0; for(j = 1;j<l2+1;j++) { int t = match[j-1]; while(t&&s2[j]!=s1[t+1]) t = next[t]; if(s2[j] == s1[t+1]) t++; match[j] = t; if(match[j] == l1) num++; } cout<<num<<endl; } }
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原文地址：http://www.cnblogs.com/wos1239/p/4388923.html
