POJ1035-Spell checker

 

 

题目链接：点击打开链接

 

Spell checker

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 27197		Accepted: 9946

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

 

题目大意：从第二部分找第一部分的单词，3种情况。

 

思路：遍历，但是长度不等的不会处理。

cin超时

AC代码:

#include<iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;

char s[20];
char str[10005][20];//存储

int main() {
	int biao = 0;
	while(~scanf("%s", s)) {
		while(1) {
			if(s[0] != '#') {
				strcpy(str[biao++], s);
				scanf("%s", s);
			} else	break;
		}
		while(scanf("%s", s)) {
			int biao1 = 0;
			char ans[10005][20];
			int flag = 0;
			if(s[0] == '#') {
				break;
			}
			int i;
			for(i = 0; i < biao; i++) {
				if(strcmp(s, str[i]) == 0) {//相同
					flag = 2;
					break;
				}
				else {
					int len1 = strlen(s), len2 = strlen(str[i]);
					if(len1 == len2) {相等，只有一个不同才行
						int sum = 0;
						for(int k = 0; k < len1; k++) {
							if(s[k] != str[i][k])
								sum++;
						}
						if(sum == 1)
							strcpy(ans[biao1++], str[i]);
					} else if(len1 > len2) {//在长度大的字符串里找短的
						int sum = 0, k = 0;
						for(int z = 0; z < len1; z++, k++) {
							if(s[k] != str[i][z]) {
								z--;
								sum++;
							}
						}
						if(sum == 1)//只有一个不同，才是符合条件的
							strcpy(ans[biao1++], str[i]);
					} else {
						int sum = 0, k = 0;
						for(int z = 0; z < len2; z++, k++) {
							if(s[k] != str[i][z]) {
								k--;
								sum++;
							}
						}
						if(sum == 1)
							strcpy(ans[biao1++], str[i]);
					}
				}
			}
			if(flag == 2) {
				printf("%s is correct\n", s);
			} else {
				printf("%s: ", s);
				if(biao1 == 0)
					printf("\n");
				else {
					for(int k = 0; k < biao1; k++) {
						if(k != biao1-1)
							printf("%s ", ans[k]);
						else
							printf("%s\n", ans[k]);
					}
				}
			}
		}
		printf("\n");
	}
}