• HDOJ1686 Oulipo KMP


    Oulipo

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2620    Accepted Submission(s): 1012


    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.

     
    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
     
     1 /* 功能Function Description:     HDOJ-1686
     2     开发环境Environment:          DEV C++ 4.9.9.1
     3     技术特点Technique:
     4     版本Version:
     5     作者Author:                   可笑痴狂
     6     日期Date:                      20120814
     7     备注Notes:
     8          查找模式字串在主串中出现的次数(可以有部分重叠)-------KMP算法
     9  */
    10  
    11  #include<stdio.h>
    12  #include<string.h>
    13  
    14  char s[1000005],t[10005];
    15  int next[10005];
    16  int k;
    17  
    18  void get_next(int len)
    19  {
    20      int i,j;
    21      i=0;
    22      j=-1;
    23      next[0]=-1;
    24      while(i<len)
    25      {
    26          if(j==-1||t[i]==t[j])
    27          {
    28              if(t[++i]==t[++j])
    29                  next[i]=next[j];
    30              else
    31                  next[i]=j;
    32          }
    33          else
    34              j=next[j];
    35      }
    36  }
    37  
    38  void KMP(int i,int len1,int len2)
    39  {
    40      int j=0;
    41      while(i<len1)
    42      {
    43          if(j==-1||s[i]==t[j])   //继续比较后续字符
    44          {
    45              ++i;
    46              ++j;
    47          }
    48          else                    //模式串向右移动
    49              j=next[j];
    50          if(j==len2)                 //匹配成功
    51          {
    52              ++k;
    53              j=next[j];
    54          }
    55      }
    56  }
    57  
    58  int main()
    59  {
    60      int len1,len2,n;
    61      scanf("%d",&n);
    62      while(n--)
    63      {
    64 
    65          k=0;
    66          scanf("%s",t);
    67          scanf("%s",s);
    68          len1=strlen(s);
    69          len2=strlen(t);
    70          get_next(len2);
    71          KMP(0,len1,len2);
    72          printf("%d\n",k);
    73      }
    74      return 0;
    75  }
     
    功不成,身已退
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  • 原文地址:https://www.cnblogs.com/dongsheng/p/2637525.html
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