POJ 1035：Spell checker

Spell checker

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 22574		Accepted: 8231

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

题意是给你一个已知的字典，然后给你一个一个字符串，让你批改字符串，可能的情况是

1完全正确

2修改了一个字符

3增加了一个字符

4删除了一个字符

输出得到的结果。

简单题，对于字符串的枚举，就看其长度等于，大于1，小于1，之后比较相等的字符数量即可，符合的输出。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

vector<string>dic;
vector<string>modi;

void solve(string test)
{
	modi.clear();
	int i;
	int len=dic.size();

	for(i=0;i<len;i++)
	{
		if(dic[i]==test)
		{
			cout<<test<<" is correct"<<endl;
			return;
		}
		if(dic[i].length()==test.length())
		{
			int result=0,j;
			for(j=0;j<test.length();j++)
			{
				if(dic[i][j]==test[j])
					result++;
			}
			if(result==test.length()-1)
			{
				modi.push_back(dic[i]);
			}
		}
		else if(dic[i].length()+1==test.length())
		{
			int result=0,j=0,k=0;
			for(j=0;j<dic[i].length()&&k<test.length();k++)
			{
				if(dic[i][j]==test[k])
				{
					result++;
					j++;
				}
			}
			if(result==dic[i].length())
			{
				modi.push_back(dic[i]);
			}
		}
		else if(dic[i].length()-1==test.length())
		{
			int result=0,j=0,k=0;
			for(k=0;k<test.length()&&j<dic[i].length();j++)
			{
				if(dic[i][j]==test[k])
				{
					result++;
					k++;
				}
			}
			if(result==test.length())
			{
				modi.push_back(dic[i]);
			}
		}
	}

	if(modi.size())
	{
		cout<<test<<":";
		for(i=0;i<modi.size();i++)
		{
			cout<<" "<<modi[i];
		}
		cout<<endl;
	}
	else
		cout<<test<<":"<<endl;
}

int main()
{
	string test;
	dic.clear();
	while(cin>>test)
	{
		if(test=="#")
			break;
		dic.push_back(test);
	}
	while(cin>>test)
	{
		if(test=="#")
			break;
		solve(test);
	}
	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。