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kmp算法(Oulipo)
http://www.cnblogs.com/dolphin0520/archive/2011/08/24/2151846.html
http://www.matrix67.com/blog/archives/115
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 22468 | Accepted: 8962 |
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
bf算法RE=。=
#include<iostream> #include<string> using namespace std; int main() { int n,c; cin>>n; for(c=1;c<=n;c++) { char s[100005],t[10005]; cin>>t; cin>>s; int slen,tlen; tlen=strlen(t); slen=strlen(s); int i,j; int sum=0; for(i=0;i<slen;i++) { j=0; while(t[j]==s[i]&&j<tlen) { i++; j++; } if(j==tlen) sum=sum+1; i=i-j+1; } cout<<sum<<endl; } return 0; }
kmp算法求next[]数组~
#include<iostream> #include<string> using namespace std; int next[10005]; char s[1000005],t[10005]; void getnext(char *p,int *next) { int j=0,k=-1; next[0]=-1; while(!j || t[j]!='\0') { if(k==-1 || t[j]==t[k]) { j++; k++; if(t[j]!=t[k]) next[j]=k; else next[j]=next[k]; } else k=next[k]; } } int main() { int n,c; cin>>n; for(c=1;c<=n;c++) { cin>>t>>s; int slen,tlen; slen=strlen(s); tlen=strlen(t); getnext(t,next); int i=0,j=0,sum=0; while(i<slen) { if(j==-1||s[i]==t[j]) { i++; j++; } else { j=next[j]; } if(j==tlen) { sum=sum+1; j=next[j]; } } cout<<sum<<endl; } return 0; } /* 3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN */