poj1035Spell checker（串 暴力~~~）

Spell checker

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24417		Accepted: 8910

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

//就是单词查找，删除替换增加任何一个字母
解题思路：
  
输入词典中的单词，再输入需要查找的单词，查找到有四种方法：
  
1.  该单词存在词典中
  
2.该单词通过替换任何一个字母后存在于词典中（如果有多个，按词典序输出）
  
3.该单词通过删除任何一个字母后存在于词典中
  
4.该单词通过增加一个字母后存在于词典中

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

char str[1100][1100];//记录字典的单词
char re[1100][1100];//记录查询的单词
  int strl = 1;//记录字典单词的书目
    int rel  = 1;//记录需要查询单词的数目
int corr(int i)
{

    for(int j=1;j<=strl;j++)
    {
        if(strcmp(str[j],re[i])==0)
            return 1;
    }
    return 0;
}

int del(char *a,char *b)//删除或增加的匹配单词，a的字母数比b多
{
    int len1 = strlen(a);//最长的字母数目
    int i=0;//a的游动下表
    int j=0;//b的游动下表
    int mis=0;//记录不匹配数
    while(i<len1)
    {
        if(a[i]!=b[j])
        {
            i++;
            mis++;
            if(mis>1)//如果不匹配的数目超过了一 则肯定这俩单词不匹配
                return 0;
        }
        else
        {
            i++;
            j++;
        }
    }
    return 1;
}

int cha(char *a,char *b)//替换单词的数目
{
    int len1=strlen(a);
    int i=0,j=0,mis=0;
    for(i=0;i<len1;i++)
    {
        if(a[i]!=b[i])
            mis++;
    }
    if(mis>1)//如果超过了一就不匹配
        return 0;
    return 1;
}
int main()
{

    while(1)//读入
    {
        scanf("%s",str[strl]);
        if(str[strl][0]!='#')
        strl++;
        else break;
    }
   while(1)
    {
        scanf("%s",re[rel]);
        if(re[rel][0]!='#')
        rel++;
        else break;
    }
    rel--;
    strl--;
    for(int i=1;i<=rel;i++)
    {
          if(corr(i))
          {
                  printf("%s is correct\n",re[i]);
                  continue;
          }

          printf("%s:",re[i]);
          for(int j=1;j<=strl;j++)
          {
              int lenre = strlen(re[i]);
              int lenstr = strlen(str[j]);
              if(lenre - lenstr == 1)//这是判断删除一个字母是否匹配的情况
              {
                  if(del(re[i],str[j]))
                    printf(" %s",str[j]);
              }

              else if(lenstr - lenre == 1)//这是判断增加一个字母是否匹配的情况  就是删除的相反喽
              {
                  if(del(str[j],re[i]))
                    printf(" %s",str[j]);
              }

              else if(lenre - lenstr == 0)//这是判断替换的情况
              {
                  if(cha(re[i],str[j]))
                    printf(" %s",str[j]);
              }
          }
          printf("\n");



    }
return 0;
}