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
Source
Recommend
代码:
1 #include <vector> 2 #include <map> 3 #include <set> 4 #include <algorithm> 5 #include <iostream> 6 #include <cstdio> 7 #include <cmath> 8 #include <cstdlib> 9 #include <string> 10 #include <cstring> 11 #include <queue> 12 using namespace std; 13 #define INF 0x3f3f3f3f 14 15 char p[10010],s[1000010]; 16 int nex[10010]; 17 18 void get(char *p) 19 { 20 int plen=strlen(p); 21 nex[0]=-1; 22 int k=-1,j=0; 23 while(j < plen){ 24 if(k==-1 || p[j] == p[k]){ 25 ++j; 26 ++k; 27 if(p[j] != p[k]) 28 nex[j]=k; 29 else 30 nex[j]=nex[k]; 31 } 32 else{ 33 k=nex[k]; 34 } 35 } 36 } 37 38 int kmp(char *s,char *p) 39 { 40 int i=0,j=0,ans=0; 41 int slen=strlen(s); 42 int plen=strlen(p); 43 while(i < slen && j< plen){ 44 if(j==-1 || s[i]==p[j]){ 45 ++i; 46 ++j; 47 } 48 else{ 49 j=nex[j]; 50 } 51 if(j == plen){ //重点注意，这里是为了回到当匹配完后，next[j]应该回到的位置 52 j=nex[j]; //例： a="aza" b="azazaza" 第一次结束后，next[j]应该所指的位置为a中的‘z’,然后继续匹配 53 ans++; 54 } 55 } 56 57 return ans; 58 } 59 60 int main() 61 { 62 int t; 63 scanf("%d",&t); 64 while(t--){ 65 scanf("%s",p); 66 scanf("%s",s); 67 get(p); 68 printf("%d ",kmp(s,p)); 69 } 70 }