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

題目大意：前面輸入一系列字符串，以「#」結束，後面在輸入一系列字符串，一樣以「#」結束，問後面的字符串能夠由前面哪些字符串經過添加，刪除，替換一個字母得來，按照字典序輸出，若是後面的和前面的相同，則直接輸出xx is correct。
解題方法：先將前面的字符串保存到一顆字典樹當中，以此來判斷後面的字符串在前面是否存在，若是不存在再另做判斷，看後面的字符串能夠由前面哪些字符串經過添加，刪除，替換一個字母得來。

#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;
}

//字典序判斷後面的字符串是否在前面出現過
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)//若是兩個字符串長度不相等
    {
        while(i < nLen1 && j < nLen2)
        {
            if (str1[i] != str2[j])//碰到字符不相等時，長度大的字符串下邊增長一
            {
                nCount++;//遇到不相等的字符，統計數加一
                if (nLen1 > nLen2)
                {
                    i++;
                }
                else
                {
                    j++;
                }
            }
            else//若是兩個字符相等，則下標同時增長一
            {
                i++;
                j++;
            }
        }
        if (nCount > 1)//不相等的字符數大於一，返回false
        {
            return false;
        }
        else
        {
            return true;
        }
    }
    else//當兩個字符串相同時，直接判斷相同位置不一樣字符串的數目是否大於一
    {
        while(i < nLen1)
        {
            if (str1[i] != str2[i])
            {
                nCount++;//不相同，計數加一
            }
            i++;
        }
        if (nCount == 1)//計數等於一，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\n", 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("\n");
        }
    }
    return 0;
}