Oulipo

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30723		Accepted: 12349

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

 

题意：给若干组数据，每组两个字符串，问第一个字符串在第二个字符串中出现的次数

思路：KMP算法来判断长串中的子串是否和短串相等，

注意：如果发现一组匹配，让 ANS++，j=next[j]+1，i++，因为next[i]=x的意思是以1为起始的前缀和以i为结束的后缀的长度为x，这样可以看出此时P[next[i]]==P[1]，这样就避免了让 j=1,i=i-lenP+2，避免了时间的浪费。由KMP的next构造可知，P[10]={00AAAA}的next={0,0,1,2,3,}，P的前缀和后缀是可以重叠的，而且会让每个的next[i]的值尽量大，所以可以保证不会漏掉某一种情况

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
char T[1000010],P[10010];
int N,next[10010],lenP,lenT,ans;
int main(){
    scanf("%d",&N);
    while(N--){
        ans=0;
        scanf("%s%s",P+1,T+1);
        lenP=strlen(P+1); lenT=strlen(T+1);
        //get next[]
        for(int j=0,i=2;i<=lenP;i++){
            while(P[j+1]!=P[i]&&j) j=next[j];
            if(P[j+1]==P[i]) j++;
            next[i]=j;
        }
        
        for(int i=1,j=1;i<=lenT;){
            if(P[j]==T[i]){
                if(j==lenP){
                    ans++;
                    j=next[j]+1;
                    i++;
                }
                else i++,j++;
                }
            else{
                if(j==1) i++;
                else j=next[j-1]+1;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}

还可以用hash

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std ;
typedef unsigned long long uLL ;
const int N=1000005 ;
const uLL Base=1234567 ;
uLL base[N]={1},hs,hA[N] ;
inline uLL ask(int l,int r)
{return hA[r]-hA[l-1]*base[r-l+1] ;}
 
char A[N],s[N] ;
int T,ans,n,m ;
int main()
{
    for(int i=1;i<N;i++) 
        base[i]=base[i-1]*Base ;
    scanf("%d",&T);
    while( T-- )
    {
        ans=0 ; hs=0 ;
        memset(hA,0,sizeof(hA)) ;
         
        scanf("%s%s",s+1,A+1);
        m=strlen(A+1) ; n=strlen(s+1) ;
         
        for(int i=1;i<=m;i++)
        {
            hA[i]=hA[i-1]*Base+(A[i]-'A'+1) ;
            if(i<=n) hs=hs*Base+(s[i]-'A'+1) ;
        }
        for(int i=m-n+1; i; i--)
            if(ask(i,i+n-1)==hs) ans++ ;
             
        printf("%d
",ans) ;
    }
    return 0;
}