POJ 1035 Spell checker （模擬）

題目連接

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

分析：
哇，個人天。剛看見題目的時候確實被測試數據給嚇到了，這都是什麼鬼？ 仔細讀一下題目就會發現沒有看着這麼的可怕，徹底靠着模擬來就好了。

首先給定一個單詞字典，裏面有一系列的單詞，以「#」做爲單詞字典輸入結束的標誌。 而後會給定一個須要查詢的單詞，若是這個單詞在上面給定的單詞字典裏面，就能夠直接輸出「該單詞 is correct」，否的的話輸出「該單詞： 與該單詞類似的單詞」。

所謂的類似單詞要求兩個單詞的長度差不能超過1
1：若是兩個單詞想等的話，要求這兩個單詞只能有一個對應位置的字符不同；
2：兩個單詞的長度相差1，較長的單詞能夠在任意位置比較短的單詞多出一個字符，若是去掉這個多出來的字符後，要求兩個單詞徹底相同。

理解題意以後主要的就是看代碼怎麼寫了，徹底的是一個模擬的過程。

代碼：

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;
string str[10009];
string str1[10009];
bool alike(string str1,string str2)
{
    string temp;
    if(str1.length()>str2.length())//要求必定是str2的長度比較大,這樣便於控制後面的字符串進行比較
    {
        temp=str1;
        str1=str2;
        str2=temp;
    }
    int cnt=0;
    if(str1.length()<str2.length())//考慮兩個串長度不相等的狀況，並且str1比較短
    {
        for(int i=0; i<str2.length()&&cnt<str1.length(); i++)//注意兩個串下表的變化
            if(str2[i]==str1[cnt])
                cnt++;
        if(cnt==str1.length())//能夠匹配，str1串中的任何一個字符均可以與str2串匹配上
            return true;
    }
    else//兩串長度相等,可是保證只更改了其中的一個值
    {
        cnt=0;
        for(int i=0; i<str1.length(); i++)
            if(str2[i]==str1[i])
                cnt++;
        if(cnt==str1.length()-1)//至關於兩個長度相同的單詞裏面，只有一個位置的字母是不同的
            return true;
    }
    return false;
}

int main()
{
    int len_dic=0;
    string s;
    while(cin>>s)//輸入單詞字典
    {
        if(s=="#")  break;
        str[len_dic]=s;
        str1[len_dic]=s;
        len_dic++;
    }
    sort(str,str+len_dic);//將字典中的單詞按照字典序排列，方便下面的查找是否存在時的二分算法，也只在這裏用到了
    while(cin>>s)//輸入要查找的單詞
    {
        if(s=="#") break;
        int len=s.length();
        //可以在單詞字典裏面找到這個單詞
        if(binary_search(str,str+len_dic,s)) //從a開始到a+size找和v相同的,//用二分查找（折半查找和v相同的）一、若存在，則輸出* is correct
        {
            cout << s << " is correct\n" ;
            continue;
        }
        cout<<s<<": ";
        for(int i=0; i<len_dic; i++) //遍歷整個單詞字典，仍是原來輸進去的順序，而不是按照字典序排列以後的
        {
            //與所須要查找的單詞的位數差大於1的話，就確定不是類似的單詞
            if(str1[i].length()<len-1||str1[i].length()>len+1)
                continue;
            //與所須要查找的單詞長度相等，或者位數只差一位，纔有可能爲類似的單詞
            if(alike(str1[i],s))//
                cout<<str1[i]<<" ";
        }
        printf("\n");
    }
    return 0;
}