Oulipo-白红宇

Oulipo

阅读量：5849 次

发布时间：2019-06-19

本文共 3782 字，大约阅读时间需要 12 分钟。

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　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

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

Sample Output

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]={\000AAAA}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

1 #include
     
       2 #include
      
        3 #include
       
         4 #include
        
          5 #include
         
           6 #include
          
            7 using namespace std; 8 char T[1000010],P[10010]; 9 int N,next[10010],lenP,lenT,ans;10 int main(){11 scanf("%d",&N);12 while(N--){13 ans=0;14 scanf("%s%s",P+1,T+1);15 lenP=strlen(P+1); lenT=strlen(T+1);16 //get next[]17 for(int j=0,i=2;i<=lenP;i++){18 while(P[j+1]!=P[i]&&j) j=next[j];19 if(P[j+1]==P[i]) j++;20 next[i]=j;21 }22 23 for(int i=1,j=1;i<=lenT;){24 if(P[j]==T[i]){25 if(j==lenP){26 ans++;27 j=next[j]+1;28 i++;29 }30 else i++,j++;31 }32 else{33 if(j==1) i++;34 else j=next[j-1]+1;35 }36 }37 printf("%d\n",ans);38 }39 return 0;40 }

还可以用hash

1 #include
     
       2 #include
      
        3 #include
       
         4 #include
        
          5 #include
         
           6 #include
          
            7 #include
           
             8 using namespace std ; 9 typedef unsigned long long uLL ;10 const int N=1000005 ;11 const uLL Base=1234567 ;12 uLL base[N]={ 1},hs,hA[N] ;13 inline uLL ask(int l,int r)14 { return hA[r]-hA[l-1]*base[r-l+1] ;}15  16 char A[N],s[N] ;17 int T,ans,n,m ;18 int main()19 {20     for(int i=1;i

转载于:https://www.cnblogs.com/CXCXCXC/p/4936469.html

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