POJ 1035 Spell checker
2015-08-07 14:30
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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
Sample Output
Source
Northeastern Europe 1998
/*
刚开始想用字典树,方便查找来节约时间,等字典树敲完之后发觉这个题用字典树来操作似乎麻烦的很。题目大意就是先给你一些词,让你组成一个词典,然后以#号结束,之后再给你一些字符,问你是否可以通过增加、删除或改变词典中某个词的一个字符让两个字符相等,当然,如果词典里已经有这个词的时候就输出correct。
方法是像是字符串哈希,把所有的词先存起来,然后给予查询的字符串去词典中找,有相同的就输出correct否则就依次寻找有没有和目标词汇字符长度差1的词,找到之后判断能否操作就好了。注意有坑,当时用if去查fi的时候也是输出可以替换,其实要经过两步才可以。
*/
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
Source
Northeastern Europe 1998
/*
刚开始想用字典树,方便查找来节约时间,等字典树敲完之后发觉这个题用字典树来操作似乎麻烦的很。题目大意就是先给你一些词,让你组成一个词典,然后以#号结束,之后再给你一些字符,问你是否可以通过增加、删除或改变词典中某个词的一个字符让两个字符相等,当然,如果词典里已经有这个词的时候就输出correct。
方法是像是字符串哈希,把所有的词先存起来,然后给予查询的字符串去词典中找,有相同的就输出correct否则就依次寻找有没有和目标词汇字符长度差1的词,找到之后判断能否操作就好了。注意有坑,当时用if去查fi的时候也是输出可以替换,其实要经过两步才可以。
*/
#include <map> #include <set> #include <cmath> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <cstring> #include <string> #include <algorithm> const int INF = 0x3f3f3f3f; using namespace std; char a[10010][20],b[10010][20]; char str[20]; int main() { int top = 0,kep = 0; while(scanf("%s",&a[top]) && strcmp(a[top],"#")) top++; while(scanf("%s",str) && strcmp(str,"#")) { kep = 0; bool flag = true; for(int i=0; i<top; i++) { if(!strcmp(a[i],str)) { printf("%s is correct\n",str); flag = false; break; } else { int len1 = strlen(str); int len2 = strlen(a[i]); int s = 0; if(len1 == len2) { int ant = 0; for(int j=0; str[j]!='\0'; j++) { if(a[i][j] == str[j]) ant++; } if(ant == len2-1) strcpy(b[kep++],a[i]); } else if(len1+1 == len2)//len1 str { int ant = 0; for(int j=0; a[i][j]!='\0'; j++) { if(a[i][j] == str[s]) { ant++; s++; } } if(ant == len1) strcpy(b[kep++],a[i]); } else if(len1 == len2+1)//len1 str { int ant = 0; for(int j=0; str[j]!='\0'; j++) { if(a[i][s] == str[j]) { ant++; s++; } } if(ant == len2) strcpy(b[kep++],a[i]); } } } if(flag) { printf("%s:",str); for(int i=0; i<kep; i++) printf(" %s",b[i]); printf("\n"); } } return 0; }
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