Spell checker dictionary tree && normal lookup (counting)

Problem Description

You, as a member of a development team for a new spell checking program, are to write a module that will check the correctness of given words using a known dictionary of all correct words in all their forms. 
If the word is absent in the dictionary then it can be replaced by correct words (from the dictionary) that can be obtained by one of the following operations: 
?deleting of one letter from the word; 
?replacing of one letter in the word with an arbitrary letter; 
?inserting of one arbitrary letter into the word. 
Your task is to write the program that will find all possible replacements from the dictionary for every given word. 

 

 

Input

The first part of the input file contains all words from the dictionary. Each word occupies its own line. This part is finished by the single character '#' on a separate line. All words are different. There will be at most 10000 words in the dictionary. 
The next part of the file contains all words that are to be checked. Each word occupies its own line. This part is also finished by the single character '#' on a separate line. There will be at most 50 words that are to be checked. 
All words in the input file (words from the dictionary and words to be checked) consist only of small alphabetic characters and each one contains 15 characters at most. 

 

 

Output

Write to the output file exactly one line for every checked word in the order of their appearance in the second part of the input file. If the word is correct (i.e. it exists in the dictionary) write the message: " is correct". If the word is not correct then write this word first, then write the character ':' (colon), and after a single space write all its possible replacements, separated by spaces. The replacements should be written in the order of their appearance in the dictionary (in the first part of the input file). If there are no replacements for this word then the line feed should immediately follow the colon.

 

 

Sample Input

i is has have be my more contest me too if award # me aware m contest hav oo or i fi mre #

 

 

Sample Output

me is correct aware: award m: i my me contest is correct hav: has have oo: too or: i is correct fi: i mre: more me

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Trie

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  1 #include<iostream>
  2 #include<string>
  3 #include<cstring>
  4 #include<cstdio>
  5 #include<cmath>
  6 using namespace std;
  7 char str1[10101][55],str2[55];
  8 int i,j;
  9 class node
 10 {
 11     public:
 12     node *p[26];
 13     int it;
 14     bool de;
 15     node()
 16     {
 17         for(it=0;it<26;it++)
 18           p[it]=NULL;
 19         de=false;;
 20     }
 21 };
 22 node *root;
 23 void get_num(char*s)
 24 {
 25     node*r=root;
 26     int len1=strlen(s);
 27     int it;
 28     for(it=0;it<len1;it++)
 29     {
 30         if(r->p[s[it]-'a']==NULL)
 31         {
 32             r->p[s[it]-'a']=new node();
 33         }
 34         r=r->p[s[it]-'a'];
 35     }
 36     r->de=true;
 37 }
 38 bool find(char *s)
 39 {
 40     int len2=strlen(s);
 41     int it;
 42     node *r=root;
 43     for(it=0;it<len2;it++)
 44     {
 45         if(r->p[s[it]-'a']==NULL)
 46             return false;
 47         r=r->p[s[it]-'a'];
 48     }
 49     if(r->de)
 50       return true;
 51     return false;
 52 }
 53 bool solve(char s1[],char s2[])
 54 {
 55    int len1=strlen(s1);
 56    int len2=strlen(s2);
 57    int ncount=0;
 58    if(len1!=len2)
 59    {
 60        int it=0,jt=0;
 61        while(it<len1&&jt<len2)
 62        {
 63            if(s1[it]!=s2[jt])
 64            {
 65                ncount++;
 66                if(len1>len2)
 67                {
 68                    it++;
 69                }
 70                else
 71                 jt++;
 72            }
 73            else
 74            {
 75                it++;
 76                jt++;
 77            }
 78        }
 79        if(ncount>1)
 80         return false;
 81        else
 82         return true;
 83    }
 84    else
 85    {
 86        int it=0;
 87        while(it<len1)
 88        {
 89            if(s1[it]!=s2[it])
 90             ncount++;
 91            it++;
 92        }
 93        if(ncount==1)
 94         return true;
 95        else
 96         return false;
 97    }
 98 }
 99 int main()
100 {
101     int st=0;
102     root=new node();
103     while(scanf("%s",str1[st])&&str1[st][0]!='#')
104     {
105         get_num(str1[st]);
106         st++;
107     }
108     while(scanf("%s",str2)&&str2[0]!='#')
109     {
110         if(find(str2))//判断是否有
111          printf("%s is correct\n",str2);
112         else
113         {
114             printf("%s:",str2);
115             for(i=0;i<st;i++)
116             {
117                 int temp=strlen(str1[i])-strlen(str2);
118                 if(temp<=1&& TEMP> = - . 1 )
 119                    IF (Solve (str1 [I], str2)) // determines whether there is a different or alternative 
120                     the printf ( " % S " , str1 [I]);
 121              }
 122              the printf ( " \ n- " );
 123          }
 124      }
 125      return  0 ;
 126  
127 }

View Code

 

Reproduced in: https: //www.cnblogs.com/sdau--codeants/p/3397536.html