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
裸的KMP,直接上代码
#include <iostream> #include <cstdio> #include <cstring> #include <vector> #include <string> #include <queue> using namespace std; char s2[10010],s1[1000010]; int sum,next[10010]; void get_next(int len) { int i=0,j=-1; next[0]=-1; while(i<len) { if(j==-1||s2[i]==s2[j]) { ++i; ++j; // if(s2[i]!=s2[j]) next[i]=j; // next[i]=next[j]; } else j=next[j]; } } void kmp(int len1,int len2) { int i=0,j=0; get_next(len2); while(i<len1) { if(j==-1||s1[i]==s2[j]) { ++i; ++j; } else j=next[j]; if(j==len2) { sum++; j=next[j]; } } } int main() { int t; scanf("%d",&t); while(t--) { getchar(); scanf("%s",s2); scanf("%s",s1); // cout<<s1<<" "<<s2<<endl; int len1=strlen(s1); int len2=strlen(s2); sum=0; kmp(len1,len2); printf("%d\n",sum); } return 0; }
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