zoukankan      html  css  js  c++  java
  • POJ 1035.Spell checker

    Spell checker
    Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    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

    给一些合法单词和一些待测的单词,如果待测单词合法,则输出“xx is correct”。

    否则,输出删去、修改、增加一个字符后合法的单词

    输出按照输入顺序

    有个题解说用Trie树,写的一直WA、TLE

    换用暴力解

    AC代码:GitHub

      1 /*
      2 By:OhYee
      3 Github:OhYee
      4 HomePage:http://www.oyohyee.com
      5 Email:oyohyee@oyohyee.com
      6 Blog:http://www.cnblogs.com/ohyee/
      7 
      8 かしこいかわいい?
      9 エリーチカ!
     10 要写出来Хорошо的代码哦~
     11 */
     12 
     13 #include <cstdio>
     14 #include <algorithm>
     15 #include <cstring>
     16 #include <cmath>
     17 #include <string>
     18 #include <iostream>
     19 #include <vector>
     20 #include <list>
     21 #include <queue>
     22 #include <stack>
     23 #include <map>
     24 using namespace std;
     25 
     26 //DEBUG MODE
     27 #define debug 0
     28 
     29 //循环
     30 #define REP(n) for(int o=0;o<n;o++)
     31 
     32 int ok;
     33 int ldict,lstr;
     34 string temp;
     35 vector<string> dict;
     36 
     37 
     38 void replace(string dict) {
     39     int count = 0;
     40     for(size_t i = 0; i < dict.length(); i++) {
     41         if(temp[i] != dict[i]) count++;
     42         if(count > 1) return;
     43     }
     44     ok = 1;
     45     cout << " " + dict;
     46 }
     47 
     48 void insert(string dict) {
     49     string tmp;
     50     for(size_t i = 0; i < temp.length(); i++) {
     51         if(temp[i] != dict[i]) {
     52             tmp = dict;
     53             tmp.insert(i,1,temp[i]);
     54             if(tmp == temp) {
     55                 ok = 1;
     56                 cout << " " + dict;
     57             }
     58             return;
     59         }
     60     }
     61 }
     62 
     63 void del(string dict) {
     64     string tmp;
     65     for(size_t i = 0; i < dict.length(); i++) {
     66         if(temp[i] != dict[i]) {
     67             tmp = dict;
     68             tmp.erase(i,1);
     69             if(tmp == temp) {
     70                 ok = 1;
     71                 cout << " " + dict;
     72             }
     73             return;
     74         }
     75     }
     76 }
     77 
     78 bool Do() {
     79     while(1) {
     80         if(!(cin >> temp))
     81             return false;
     82         if(temp == "#") break;
     83         dict.push_back(temp);
     84     }
     85 
     86     while(1) {
     87         cin >> temp;
     88         if(temp == "#") break;
     89 
     90         ok = -1;
     91         lstr = temp.length();
     92 
     93         for(size_t i = 0; i < dict.size(); i++) {
     94             if(dict[i] == temp) {
     95                 cout << temp + " is correct";
     96                 ok = 0;
     97                 break;
     98             }
     99         }
    100 
    101 
    102         if(ok < 0) {
    103             cout << temp + ':';
    104             for(size_t i = 0; i < dict.size(); i++) {
    105                 ldict = dict[i].length();
    106                 if(lstr == ldict) {
    107                     replace(dict[i]);
    108                 } else if(lstr == ldict + 1) {
    109                     insert(dict[i]);
    110                 } else if(lstr == ldict - 1) {
    111                     del(dict[i]);
    112                 }
    113             }
    114         }
    115 
    116         cout << endl;
    117     }
    118 
    119     return true;
    120 }
    121 
    122 int main() {
    123     while(Do());
    124     return 0;
    125 }
  • 相关阅读:
    hdu 5119 Happy Matt Friends
    hdu 5128 The E-pang Palace
    hdu 5131 Song Jiang's rank list
    hdu 5135 Little Zu Chongzhi's Triangles
    hdu 5137 How Many Maos Does the Guanxi Worth
    hdu 5122 K.Bro Sorting
    Human Gene Functions
    Palindrome(最长公共子序列)
    A Simple problem
    Alignment ( 最长上升(下降)子序列 )
  • 原文地址:https://www.cnblogs.com/ohyee/p/5510784.html
Copyright © 2011-2022 走看看