KMP算法之Oulipo
Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8137 Accepted Submission(s):
3280
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.
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.
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
首先,这道题属于一道KMP算法, 只是将此算法逆着用了一下, 求子串在母串中出现的次数。定义一个count记录次数,然后 利用平常的算法就OK。 若子串没在母串中出现过,
最后的count就还等于0啦!
1 #include<stdio.h> 2 #include<string.h> 3 4 const int MAXN = 1e6+1; 5 6 char Mum[MAXN], Son[10001]; 7 int next[MAXN]; 8 int count; 9 void GetNext(int l2) 10 { 11 int i = 0, j = -1; 12 next[0] = -1; 13 while(i < l2) 14 { 15 if(j == -1 || Son[i]==Son[j]) 16 next[++i] = ++j; 17 else 18 j = next[j]; 19 } 20 } 21 void KMP(int l1, int l2) 22 { 23 int i=0, j=0; 24 GetNext(l2); 25 while(i < l1) 26 { 27 if(j==-1 || (Mum[i]==Son[j]&&i<l1&&j<l2) ) 28 i++, j++; 29 else 30 j = next[j]; 31 if(j == l2) 32 { 33 count++; 34 j = next[j]; 35 } 36 } 37 } 38 39 int main() 40 { 41 int T; 42 43 scanf("%d", &T); 44 45 while(T--) 46 { 47 int Mn, Sn; 48 scanf("%s", Son); 49 scanf(" %s", Mum); 50 Sn = strlen(Son); 51 Mn = strlen(Mum); 52 count = 0; 53 KMP(Mn, Sn); 54 printf("%d\n", count); 55 } 56 return 0; 57 }