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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ 1035 Spell checker

POJ 1035 Spell checker

編輯:C++入門知識

Spell checker Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 15145 Accepted: 5504 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 Source   Northeastern Europe 1998 首先 理清思路,用最普通的方法做就可以; 然後 睡覺 [cpp]  #include <iostream>   #include <string>   using namespace std;   string s1[20000];   string s2;   int main()   {       int i,j,n,m,s,t,x,k;       int l1,l2;       i=0;       while(1)       {           getline(cin,s2);           if(s2=="#")           {               break;           }           s1[i]=s2; i++;       }       n=i;       while(1)       {           getline(cin,s2);           if(s2=="#")           {               break;           }           for(i=0;i<=n-1;i++)           {               if(s1[i]==s2)               {                   break;               }           }           if(i!=n)           {               cout<<s2<<" is correct"<<endl;           }else           {               cout<<s2<<":";               l2=s2.size();               for(i=0;i<=n-1;i++)               {                   l1=s1[i].size();                   k=0;                   if(l1==l2)                   {                       for(j=0;j<=l1-1;j++)                       {                           if(s1[i][j]!=s2[j]&&k==0)                           {                               k=1;                           }else if(s1[i][j]!=s2[j])                           {                               break;                           }                       }                       if(j==l1)                       {                           cout<<" "<<s1[i];                       }                   }else if(l1-l2==1)                   {                       for(j=0,x=0;j<=l1-1;j++)                       {                           if(s1[i][j]!=s2[x]&&k==0)                           {                               k=1;                           }else if(s1[i][j]!=s2[x])                           {                               break;                           }else if(s1[i][j]==s2[x])                           {                               x++;                           }                       }                       if(j==l1)                       {                           cout<<" "<<s1[i];                       }                   }else if(l2-l1==1)                   {                       for(j=0,x=0;j<=l2-1;j++)                       {                           if(s1[i][x]!=s2[j]&&k==0)                           {                               k=1;                           }else if(s1[i][x]!=s2[j])                           {                               break;                           }else if(s1[i][x]==s2[j])                           {                               x++;                           }                       }                       if(j==l2)                       {                           cout<<" "<<s1[i];                       }                   }               }               cout<<endl;           }       }       return 0;   }  

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