hdoj 1686 Oulipo(kmp)
阿新 • • 發佈：2017-08-08
should ace void lis col bsp oci get sam 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.
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.
Sample Output 1 3 0 題意：給定一個字符串，求它在另一串出現的次數
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 題意：給定一個字符串，求它在另一串出現的次數
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 const int maxn=1e6+10; 6 int Next[maxn]; 7 char str[maxn],mo[maxn]; 8 void getNext(){ 9 int i=0,j=-1,len=strlen(mo); 10 while(i<len) 11 { if (j==-1||mo[i]==mo[j]) 12 Next[++i]=++j; 13 else j=Next[j]; 14 } 15 } 16 int kmp(){ 17 int i=0,j=0,l1=strlen(str),l2=strlen(mo); 18 int ans=0; 19 while(i<l1) 20 { 21 if(j==-1||mo[j]==str[i]) 22 i++,j++; 23 else j=Next[j]; 24 if(j==l2) ans++; 25 } 26 return ans; 27 } 28 int main() 29 { int t; 30 scanf("%d",&t); 31 while(t--){ 32 scanf("%s%s",mo,str); 33 Next[0]=-1; 34 getNext(); 35 //for(int i=0;i<strlen(mo);i++) 36 //cout<<Next[i]<<" "; cout<<endl; 查看next數組 37 printf("%d\n",kmp()); 38 } 39 return 0; 40 } 41 /* 42 後人修改過後的KMP算法 43 針對aaaaabcde 模式串：aaaaax 44 int j=0,i=1; 45 next[1]=0; 46 while(i<t[0]) // t[0]存放模式串大小 47 { 48 if(j==0||T[i]==T[j]) 49 { 50 i++; 51 j++； 52 if(T[i]!=T[j]) 53 next[i]=j; 54 else next[i]=next[j]; 55 } 56 else j=next[j]; 57 } 58 */
hdoj 1686 Oulipo(kmp)