• <ins id="pjuwb"></ins>
    <blockquote id="pjuwb"><pre id="pjuwb"></pre></blockquote>
    <noscript id="pjuwb"></noscript>
          <sup id="pjuwb"><pre id="pjuwb"></pre></sup>
            <dd id="pjuwb"></dd>
            <abbr id="pjuwb"></abbr>
            posts - 195,  comments - 30,  trackbacks - 0
            Spell Checker
            Status In/Out TIME Limit MEMORY Limit Submit Times Solved Users JUDGE TYPE
            stdin/stdout 3s 8192K 135 46 Standard

            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

            This problem consists of several test cases, each of which is described below:

            The first part of each test case 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 test case 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 (words from the dictionary and words to be checked) consist only of small alphabetic characters and each one contains 15 characters at most.

            The first line of the input contains an integer N, which is the number of test cases in the input, followed by N test cases described above.

            Output

            For each test case, first write to the output 'Scenario #k:', where k is the number of test case of input. Then on the next line write to the output exactly one line for every checked word in the order of their appearance in the second part of the test case. If the word is correct (i.e. it exists in the dictionary) write the message: "<checked word> 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.

            Print a blank line after each test case.

            Sample Input

            1
            i
            is
            has
            have
            be
            my
            more
            contest
            me
            too
            if
            award
            #
            me
            aware
            m
            contest
            hav
            oo
            or
            i
            fi
            mre
            #
            

            Sample Output

            Scenario #1:
            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
            

            啟發(fā):map用于查找字符串還是很爽的。
            #include<iostream>
            #include
            <cstdlib>
            #include
            <map>
            #include
            <math.h>
            #include
            <string>
            using namespace std;
            map 
            <string,int> ans;
            bool solve(char *s,char *key)
            {
                
            int i=strlen(s);
                
            int j=strlen(key);
                
            int num=0;
                
            if(abs(i-j)>1)
                
            return false;
                
            else
                {
                    
            if(i-j==1)//比字典中多一個(gè) 
                    {
                        
            for(int k=0;k<j&&num<=2;k++)
                        {
                            
            if(s[k+num]!=key[k])
                            {
                                num
            ++;
                                k
            --;
                            }
                        }
                        
            if(num<=1)
                        
            return true;
                        
            else
                        
            return false;
                    }
                    
            if(j-i==1)//比字典中少一個(gè) 
                    {
                        
            for(int k=0;k<i&&num<=2;k++)
                        {
                            
            if(s[k]!=key[k+num])
                            {
                                num
            ++;
                              k
            --;
                            } 
                        }
                        
            if(num<=1)
                        
            return true;
                        
            else
                        
            return false;
                    }
                    
            if(i==j)
                    {
                        
            for(int k=0;k<i&&num<=2;k++)
                        {
                            
            if(s[k]!=key[k])
                            num
            ++;
                        }
                        
            if(num==1)
                        
            return true;
                        
            else
                        
            return false;
                    }
                }
            }
              
            int main()
              {
              freopen(
            "s.txt","r",stdin);
              freopen(
            "key.txt","w",stdout);
              
            int num,casetime=0;
              
            int i,j,k;
              cin
            >>num;
              
            char s[16];
              
            char str[10000][16];
              
            while(num--)
              {
                    casetime
            ++;
                    cout
            <<"Scenario #"<<casetime<<":"<<endl;
                    ans.clear();
                    i
            =0;
                    
            while(1)
                    {
                        scanf(
            "%s",&str[i]);
                        
            if(str[i][0]=='#')
                          
            break;
                        ans[str[i]]
            =1;
                        i
            ++;
                    }
                    
            while(1)
                    {
                        scanf(
            "%s",&s);
                        
            if(s[0]=='#')
                          
            break;
                        
            if(ans[s]==1
                        {
                            cout
            <<s<<" is correct"<<endl;

                        } 
                        
            else
                        {
                            cout
            <<s<<":";
                           
            for(int j=0;j<i;j++)
                          {
                            
            if(solve(s,str[j]))
                              cout
            <<" "<<str[j];
                          }
                          cout
            <<endl;
                        } 
                    }
                    cout
            <<endl;
                    
              }

              
            //system("PAUSE");
              return   0;
              }
            posted on 2009-07-06 20:13 luis 閱讀(276) 評(píng)論(0)  編輯 收藏 引用 所屬分類: 格式.輸入輸出.數(shù)據(jù)類型
            <2009年7月>
            2829301234
            567891011
            12131415161718
            19202122232425
            2627282930311
            2345678

            常用鏈接

            留言簿(3)

            隨筆分類

            隨筆檔案

            文章分類

            文章檔案

            友情鏈接

            搜索

            •  

            最新評(píng)論

            閱讀排行榜

            評(píng)論排行榜

            国内精品伊人久久久久妇| 国产精品久久波多野结衣| 久久久久亚洲精品无码网址 | 性做久久久久久久久| 久久精品国产欧美日韩99热| 99精品国产99久久久久久97| 99精品国产在热久久无毒不卡| 91精品婷婷国产综合久久| 亚洲精品成人久久久| 丰满少妇高潮惨叫久久久| 亚洲精品成人久久久| 伊人久久精品线影院| 99精品国产综合久久久久五月天| 91久久精品国产免费直播| 亚洲中文字幕无码久久2017| 99久久精品无码一区二区毛片| 亚洲综合日韩久久成人AV| 久久精品中文字幕一区| 777米奇久久最新地址| 亚洲va中文字幕无码久久| 久久久综合香蕉尹人综合网| 久久96国产精品久久久| 一本色道久久综合亚洲精品| 欧美日韩中文字幕久久久不卡| 99久久超碰中文字幕伊人| 一本一本久久A久久综合精品| 日本精品一区二区久久久| 精品久久久久久国产免费了| .精品久久久麻豆国产精品| 日日躁夜夜躁狠狠久久AV| 久久久亚洲欧洲日产国码是AV| 麻豆久久| 无码人妻久久一区二区三区蜜桃 | 国产免费久久精品99re丫y| 久久久久国色AV免费看图片| 91亚洲国产成人久久精品| 狠狠色丁香久久婷婷综| 99久久婷婷国产综合亚洲| 久久成人国产精品| 久久国产精品国产自线拍免费| 99精品久久精品一区二区|