[POJ1035]-Spell checker

Spell checker

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24409		Accepted: 8905

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

题目描述：给你不超过10000行的word（word长度不超过15），然后给你字符让你判断能匹配到他给出的哪些word,如果能匹配到相同的

输出s: is correct 如果没有相同的，就按照指定的匹配方式，匹配时有三种操作：

1：去掉一个字符。

2：增加一个字符。

3：覆盖一个字符。

解题思路：

直接暴力，使用STL中的string+vector，先判断word是否相等，在判断其长度差Len：

if(Len==0)

    那么就肯定为替换，这种情况直接一个一个一次比较就行了。

else if(Len==-1||)

    这种情况就是增加或者删除一个字符了。

温情提示：别用string s[]，这种字串组很耗时，我第一次就用的string s[]，直接TLE,换vector就不会超时。

//head.h
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <string>
#include <memory>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <vector>
#include <stack>
#include <utility>
#include <set>
#include <map>
#include <iterator>
#include <fstream>
#define LOCAL

#ifndef LOCAL
#include "head.h"
#endif
using namespace std;
const int maxn = 100000;
int main()
{
	vector<string> s;
	string tmp;
	// freopen("data1035.in","r",stdin);
	while(true)
	{
		cin >> tmp;
		if(tmp=="#")
			break;
		s.push_back(tmp);
	}
	while(true)
	{
		bool flag=false;
		cin >> tmp;
		if(tmp=="#") break;
		for(int i = 0;i < s.size();++ i)
		{
			if(tmp == s[i])
			{
				flag=true;
				break;
			}
		}
		if(flag)
		{
			cout << tmp << " is correct\n";
			continue;
		}
		cout << tmp << ":";
		int len = 0;
		for(int i = 0;i < s.size();++ i)
		{
			len = tmp.length() - s[i].length();
			if(len > 1|| len < -1)
				continue;
			int c=0;
			if(len == 0)//长度差相等
			{
				for(int j = 0;j < tmp.length();++ j)
					if(tmp[j]!=s[i][j])
						c++;
				if(c==1)
					cout << " " << s[i];
			}
			else if(len == -1)//删除一个字符，
			{
				c=0;
				for(int j = 0;j < s[i].length();++ j)
					if(tmp[c]==s[i][j])
					{
						c++;
					}
				if(c == tmp.length())
					cout << " " << s[i];
			}
			else if(len == 1)//增加一个字符
			{
				c=0;
				for(int j = 0;j < tmp.length();++ j)
					if(tmp[j]==s[i][c])
					{
						c++;
					}
				if(c == s[i].length())
					cout << " " << s[i];
			}
		}
		cout << endl;

	}
	return 0;
}