POJ 1035 Spell checker

Spell checker

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16826		Accepted: 6154

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


题目大意：前面输入一系列字符串，以“#”结束，后面在输入一系列字符串，同样以“#”结束，问后面的字符串可以由前面哪些字符串通过添加，删除，替换一个字母得来，按照字典序输出，如果后面的和前面的相同，则直接输出xx is correct。
解题方法：先将前面的字符串保存到一颗字典树当中，以此来判断后面的字符串在前面是否存在，如果不存在再另作判断，看后面的字符串可以由前面哪些字符串通过添加，删除，替换一个字母得来。

#include <stdio.h>
#include <iostream>
#include <math.h>
#include <string.h>
using namespace std;

typedef struct node
{
    bool isword;
    node * next[26];
    node()
    {
        isword = false;
        memset(next, 0, sizeof(next));
    }
}TreeNode;

int cmp(char *str1, char *str2)
{
    return strcmp(str1, str2);
}

void Insert(TreeNode *pRoot, char str[])
{
    int nLen = strlen(str);
    for (int i = 0; i < nLen; i++)
    {
        if (pRoot->next[str[i] - 'a'] == NULL)
        {
            pRoot->next[str[i] - 'a'] = new TreeNode;
        }
        pRoot = pRoot->next[str[i] - 'a'];
    }
    pRoot->isword = true;
}

//字典序判断后面的字符串是否在前面出现过
bool Find(TreeNode *pRoot, char str[])
{
    int nLen = strlen(str);
    for (int i = 0; i < nLen; i++)
    {
        if (pRoot->next[str[i] - 'a'] == NULL)
        {
            return false;
        }
        pRoot = pRoot->next[str[i] - 'a'];
    }
    if (pRoot->isword)
    {
        return true;
    }
    return false;
}

bool IsEqual(char str1[], char str2[])
{
    int nLen1 = strlen(str1);
    int nLen2 = strlen(str2);
    int nCount = 0;
    int i = 0, j = 0;
    if (nLen1 != nLen2)//如果两个字符串长度不相等
    {
        while(i < nLen1 && j < nLen2)
        {
            if (str1[i] != str2[j])//碰到字符不相等时，长度大的字符串下边增加一
            {
                nCount++;//遇到不相等的字符，统计数加一
                if (nLen1 > nLen2)
                {
                    i++;
                }
                else
                {
                    j++;
                }
            }
            else//如果两个字符相等，则下标同时增加一
            {
                i++;
                j++;
            }
        }
        if (nCount > 1)//不相等的字符数大于一，返回false
        {
            return false;
        }
        else
        {
            return true;
        }
    }
    else//当两个字符串相同时，直接判断相同位置不同字符串的数目是否大于一
    {
        while(i < nLen1)
        {
            if (str1[i] != str2[i])
            {
                nCount++;//不相同，计数加一
            }
            i++;
        }
        if (nCount == 1)//计数等于一，返回true
        {
            return true;
        }
        else
        {
            return false;
        }
    }
}
int main()
{
    int nCount1 = 0;
    char s1[10005][55], s2[55];
    TreeNode *pRoot = new TreeNode;
    while(scanf("%s", s1[nCount1]) != EOF && s1[nCount1][0] != '#')
    {    
        Insert(pRoot, s1[nCount1]);
        nCount1++;
    }
    while(scanf("%s", s2) != EOF && s2[0] != '#')
    {
        if (Find(pRoot, s2))
        {
            printf("%s is correct
", s2);
        }
        else
        {
            printf("%s:", s2);
            for (int i = 0; i < nCount1; i++)
            {
                int temp = strlen(s1[i]) - strlen(s2);
                if (temp <= 1 && temp >= -1)
                {
                    if (IsEqual(s1[i], s2))
                    {
                        printf(" %s", s1[i]);
                    }
                }
            }
            printf("
");
        }
    }
    return 0;
}