Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22615 Accepted Submission(s): 8746
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22615 Accepted Submission(s): 8746
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:
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.
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.
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
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
1
3
0
C/C++:
1 #include <map> 2 #include <queue> 3 #include <cmath> 4 #include <vector> 5 #include <string> 6 #include <cstdio> 7 #include <cstring> 8 #include <climits> 9 #include <iostream> 10 #include <algorithm> 11 #define INF 0x3f3f3f3f 12 using namespace std; 13 const int my_max = 1e6 + 10; 14 15 char s1[my_max], s2[my_max]; 16 int my_next[my_max]; 17 18 int my_kmp() 19 { 20 int len1 = strlen(s1), len2 = strlen(s2), ans = 0; 21 for (int i = 1, j = 0; i < len1; ++ i) 22 { 23 while (j > 0 && s1[i] != s1[j]) j = my_next[j]; 24 if (s1[i] == s1[j]) ++ j; 25 my_next[i + 1] = j; 26 } 27 for (int i = 0, j = 0; i < len2; ++ i) 28 { 29 while (j > 0 && s1[j] != s2[i]) j = my_next[j]; 30 if (s1[j] == s2[i]) ++ j; 31 if (j == len1) ++ ans, j = my_next[j]; 32 } 33 return ans; 34 } 35 36 int main() 37 { 38 int t; 39 scanf("%d", &t); 40 memset(my_next, 0, sizeof(my_next)); 41 while (t --) 42 { 43 scanf("%s", s1); 44 scanf("%s", s2); 45 printf("%d ", my_kmp()); 46 } 47 return 0; 48 }