【kmp】POJ-3461 Oulipo
Time Limit: 1000MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
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
不多说,一道裸的kmp。。。只不过是求个数而不是求位置,于是乎刚开始时被光荣的坑了。。。找到一个后不是j=0;而是j = next[j],因为要充分利用好next这个资源,next是最大前缀后缀相同的长度。。。。也就是说w串只需移动j - next[j]个长度就够了,因为前面是相同的,不需重新比较。。
【代码】:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 using namespace std; 5 6 char w[10000 + 5]; 7 char t[1000000 + 5]; 8 int next[10000 + 5]; 9 int T; 10 int ans = 0; 11 12 void getnext(int &w_len) 13 { 14 int k = -1,j = 0; 15 next[0] = -1; 16 while(j < w_len) 17 { 18 if(k == -1 || w[k] == w[j]) 19 { 20 j++;k++; 21 if(w[j] != w[k]) next[j] = k; 22 else next[j] = next[k]; 23 } 24 else k = next[k]; 25 } 26 } 27 void kmp(int &w_len,int &t_len) 28 { 29 int i = 0, j = 0; 30 while(i < t_len && j < w_len) 31 { 32 if(t[i] == w[j] || j == -1) 33 { 34 i++;j++; 35 } 36 else j = next[j]; 37 if(j == w_len){ 38 ans++; 39 j = next[j]; 40 } 41 } 42 } 43 int main() 44 { 45 scanf("%d", &T); 46 while(T--) 47 { 48 ans = 0; 49 scanf("%s",w); 50 scanf("%s",t); 51 int w_len = strlen(w); 52 int t_len = strlen(t); 53 getnext(w_len); 54 kmp(w_len, t_len); 55 printf("%d\n",ans); 56 } 57 return 0; 58 }