алгоритъм, позволяващ проследяване на дума, срещаща се две думи след дума с определен модел

Извинете ме, че не използвах псевдокод, вместо това използвах действителен код. Надявам се, че вече разбирам проблема достатъчно добре, че убеждението ми, че не изглежда много сложно, е точно.

# include <stdio.h>
# include <ctype.h>
# include <string.h>


# define BUFF_SIZE       1024
# define WORD_DELIM     " "
# define MATCH_PATT     "barf"


int main(  int ac ,  char *av[]  )
{
    __u_char    false = ( 1 == 0 ) ;
    __u_char    true = ( 1 == 1 ) ;

    __u_char    match_1_back = false ;
    __u_char    match
barf   barf  barf   healthy     feeling     better   barf  barf barf uh oh sick again
barf   barf  healthy     feeling     better   barf  barf uh oh sick again
barf   healthy     barf   feeling     better     barf   uh   barf   oh sick again
barf   healthy     feeling     better   barf uh oh sick again
back = false ;

    char        line_buff[  BUFF_SIZE  ] ;
    char        *buff_ptr ;
    char        *word_ptr ;


    while (  fgets( line_buff ,  BUFF_SIZE ,  stdin )  )
    {
        puts(  "\nInput line was:  "  ) ;
        puts(  line_buff  )  ;

        puts(  "Output line is:  "  ) ;

        buff_ptr = line_buff ;

        while (  ( word_ptr = strtok( buff_ptr ,  WORD_DELIM )  )  !=  NULL  )
        {
            buff_ptr = NULL ;

            if (  strcmp( word_ptr ,  MATCH_PATT  )  ==  0  )
            {
                // Set these to what they should be for next iteration.
                match
barf   barf  barf   healthy     feeling     better   barf  barf barf uh oh sick again
barf   barf  healthy     feeling     better   barf  barf uh oh sick again
barf   healthy     barf   feeling     better     barf   uh   barf   oh sick again
barf   healthy     feeling     better   barf uh oh sick again
back = match_1_back ;
                match_1_back = true ;

                // Don't output matched token.
            }
            else
            {
                // Don't output token if a token matched 2 tokens back.
                if (  ! match
barf   barf  barf   healthy     feeling     better   barf  barf barf uh oh sick again
barf   barf  healthy     feeling     better   barf  barf uh oh sick again
barf   healthy     barf   feeling     better     barf   uh   barf   oh sick again
barf   healthy     feeling     better   barf uh oh sick again
back  )
                    printf(  "%s " ,  word_ptr  ) ;

                // Set these to what they should be for next iteration.
                match
barf   barf  barf   healthy     feeling     better   barf  barf barf uh oh sick again
barf   barf  healthy     feeling     better   barf  barf uh oh sick again
barf   healthy     barf   feeling     better     barf   uh   barf   oh sick again
barf   healthy     feeling     better   barf uh oh sick again
back = match_1_back ;
                match_1_back = false ;
            }
        }

        printf(  "\n"  ) ;
    }
}

С този вход:

barf   barf  barf   healthy     feeling     better   barf  barf barf uh oh sick again
barf   barf  healthy     feeling     better   barf  barf uh oh sick again
barf   healthy     barf   feeling     better     barf   uh   barf   oh sick again
barf   healthy     feeling     better   barf uh oh sick again

Получих този резултат:

Input line was:  
barf   barf  barf   healthy     feeling     better   barf  barf barf uh oh sick again

Output line is:  
better sick again


Input line was:  
barf   barf  healthy     feeling     better   barf  barf uh oh sick again

Output line is:  
better sick again


Input line was:  
barf   healthy     barf   feeling     better     barf   uh   barf   oh sick again

Output line is:  
healthy feeling uh oh again


Input line was:  
barf   healthy     feeling     better   barf uh oh sick again

Output line is:  
healthy better uh sick again

Просто използвах просто сравнение, а не действителни регулярни изрази. Исках само да илюстрирам алгоритъма. Резултатът отговаря ли на изискванията?

Звучи като класическа употреба на регулярни изрази. Не посочвате език, но много езици поддържат RegEx.

Следният сайт е добра отправна точка/справка за RegEx, ако не сте запознати с него. http://www.regular-expressions.info/quickstart.html

Ще използвате израз, съставен от три части. Синтаксисът по-долу е от паметта, така че ще трябва да го проверите отново:

• (first) - Съвпада с думата "първо"
• [\w]*{1,3} - Всяка дума, която се повтаря например между 1 и 3 пъти -
• (second) - Съвпада с думата "втори"

Нещо като това?

for (i = 0; i<wordcount; i++)
{
    CurrentWord = Words[i]
    if (WordMatchesCritera(CurrentWord))
    {
        if (HavePrecedingWord)
        { 
             success !!!
        } 
        else
        {
            i ++;
            HavePrecedingWord = true
        }
    }
    else
    {
        HavePrecedingWord = false;
    }
}

Ще работи ли това?

    matchID = -1;
    eraseID = -1;


    for(i = 0; i < ... ; i++)
    {
         if( wordMatches ( word[i] ) )
         {                  
            matchID = i;     /* found the chosen one */
            eraseID = -1;         
         }
         else 
         {
            if( matchID != -1 ) /* chosen one was found ? */
            {
                 eraseID = i;   /* erase the next non-matching one */
                 break; /* ? done for now  ? */
            }

          }
    }