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.
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.
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 be mre #
Sample Output
me is correct aware: award m: i my me contest is correct hav: has have yes: too or: i is correct fi: i mre: more me
题意就是给你一个字典，再给你一些字符，首先如果字典中有这个字符串，则直接输出，如果没有的话，那就找字符串加一个字符或少一个字符或者换一个字符是否可以在字典中找到相应的字符串
解题思路：我是用string类型的，比较方便看两个字符串是否相等，用char的话，就是strcmp函数也行。
如果找不到相等的，那么久分别在字典中找到与这个字符串的长度相差1的或者相等的。
然后匹配，如果匹配的结果相差一个则输出
#include<iostream>
#include<string>
#include<cmath>
using namespace std;
string s[10000];
string t[10000];
int main()
{
int k = 0;
int p = 0;
int d;
while (cin >> s[k] && s[k][0] != '#')
{
k++;
}
while (cin >> t[p] && t[p][0] != '#')
{
bool sign = false;
for (int i = 0; i < k; i++)
{
if (t[p] == s[i])
{
sign = true;
break;
}
}
if (sign)
{
cout << t[p] << " is correct" << endl;
continue;
}
cout << t[p] << ":";
for (int i = 0; i < k; i++)
{
int len1 = s[i].size();
int len2 = t[p].size();
if (abs(len1-len2) <= 1)
{
if (len2==1&& abs(len1 - len2) == 0)
{
cout << " " << s[i];
}
else
{
if (len1> len2)
{
int ans = 0;
d = 0;
for (int j = 0; j < len1; j++)
{
if (s[i][j] == t[p][d])
{
d++;
ans++;
}
}
if (ans == len2)
cout << " " << s[i];
}
else
{
int ans = 0;
d = 0;
for (int j = 0; j < len2; j++)
{
if (s[i][d] == t[p][j])
{
d++;
ans++;
}
}
if (ans == len1)
cout << " " << s[i];
}
if (len1 == len2)
{
int ts = 0;
for (int j = 0; j < len2; j++)
{
if (s[i][j] != t[p][j])
{
ts++;
if (ts > 1)
break;
}
}
if (ts <= 1)
cout << " " << s[i];
}
}
}
}
cout << endl;
p++;
}
return 0;
}