KMP算法匹配入门(Hdu 1686)
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.
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.
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.
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
1
3
0
#代码如下
#include
#include
#include
#include
#include
using namespace std;
string ps,ts;
const int N=10005;
int Next[N];
void make_next(){
int j=0,k=-1;
Next[0]=k;
for(;j<ps.length()-1;){
if(k==-1||ps[j]==ps[k])
Next[++j]=++k;
else k=Next[k];
}
}
int main(){
int t;scanf("%d",&t);
while(t--){
cin>>ps>>ts;
make_next();
int len=ts.length();
//cout<
int cnt=0,ans=0;
int Len=ps.length();
int i=0,j=0;
while(i<len){
if(j==-1||ts[i]==ps[j])i++,j++;
else j=Next[j];
if(Len==j)j--,i--,ans++,j=Next[j];
}
cout<<ans<<endl;
}
}