Spell checker

 Spell checker
Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
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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

题目大意：输入字典中的单词，以#结束，再输入要查找的单词，也以#结束
若在字典中找到了，则输出：x is correct；
若在字典中找到单词满足或者更改一个字母，或者删除一个字母，或者增加一个字母之后成为要查找的单词，就按照字典序输出；
若是以上两种都没有找到，则不输出；
代码：
  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<cstdlib>
  5 using namespace std;
  6 char dic[100000][82],word[100000][82];
  7 int dicnum=0,wordnum=0;
  8 void add()
  9 {
 10     while(scanf("%s",dic[dicnum])&&dic[dicnum][0]!='#')
 11     dicnum++;
 12     while(scanf("%s",word[wordnum])&&word[wordnum][0]!='#')
 13     wordnum++;
 14     dicnum--;
 15     wordnum--;
 16     //cout<<wordnum<<endl<<dicnum;
 17 }
 18 int change(char f[],char g[])//f是字典中的单词，g是查找的单词
 19 {
 20     int i=0;
 21     int t=0;
 22     for(i=0;f[i]!='\0';i++)
 23     {
 24         if(f[i]!=g[i])
 25         {
 26         t++;
 27         if(t==2)return 0;
 28         }
 29     }
 30     return 1;
 31 }
 32 int zeng(char f[],char g[])//g大
 33 {
 34     int s=0,t=0,zong=0;
 35     while(g[t]!='\0')
 36     {
 37         if(f[s]!=g[t])
 38         {
 39             zong++;
 40             t++;
 41             if(zong==2)return 0;
 42         }
 43         else
 44         {
 45             s++;
 46             t++;
 47         }
 48     }
 49     return 1;
 50 }
 51 int shan(char f[],char g[])//f大
 52 {
 53     int s=0,t=0,zong=0;
 54     while(f[s]!='\0')
 55     {
 56         if(f[s]!=g[t])
 57         {
 58             zong++;
 59             s++;
 60             if(zong==2)return 0;
 61         }
 62         else
 63         {
 64             s++;
 65             t++;
 66         }
 67     }
 68     return 1;
 69 }
 70 int main()
 71 {
 72     add();
 73     int diclen[100000]={0};
 74     int i,j;
 75     for(i=0;i<=dicnum;i++)
 76     diclen[i]=strlen(dic[i]);
 77     for(i=0;i<=wordnum;i++)
 78     {
 79         int f[100000],t=0,flag=0,s=0,zong=0;
 80         int len=strlen(word[i]);
 81         for( j=0;j<=dicnum;j++)
 82         {
 83             if(strcmp(dic[j],word[i])==0)
 84             {
 85                 flag=1;
 86                 break;
 87             }
 88             else if(diclen[j]==len)
 89             {
 90                 s=change(dic[j],word[i]);
 91                 if(s==1)
 92                 {
 93                     f[t++]=j;
 94                     zong++;
 95                 }
 96             }
 97             else if(diclen[j]-len==-1)
 98             {
 99                 s=zeng(dic[j],word[i]);
100                 if(s==1)f[t++]=j;
101             }
102             else if(diclen[j]-len==1)
103             {
104                 s=shan(dic[j],word[i]);
105                 if(s==1)
106                 {
107                     f[t++]=j;
108                     zong++;
109                 }
110             }
111         }
112         if(flag==1)
113         {
114             cout<<word[i]<<' '<<"is"<<' '<<"correct"<<endl;
115             continue;
116         }
117         else
118         {
119             //if(zong==0)continue;
120             cout<<word[i]<<": ";
121             int st=0;
122             for(j=0;j<=t-1;j++)
123             {
124                 if(st==0)
125                 {
126                     cout<<dic[f[j]];
127                     st=1;
128                 }
129                 else
130                 cout<<" "<<dic[f[j]];
131             }
132             cout<<endl;
133         }
134     }
135     return 0;
136 }
View Code

 

posted @ 2013-08-19 10:30  狂盗一枝梅  阅读(575)  评论(0)  编辑  收藏  举报