hdu1075

What Are You Talking About

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others)
Total Submission(s): 6842    Accepted Submission(s): 2086


Problem Description
Ignatius is so lucky that he met a Martian yesterday. But he didn't know the language the Martians use. The Martian gives him a history book of Mars and a dictionary when it leaves. Now Ignatius want to translate the history book into English. Can you help him?
 

Input
The problem has only one test case, the test case consists of two parts, the dictionary part and the book part. The dictionary part starts with a single line contains a string "START", this string should be ignored, then some lines follow, each line contains two strings, the first one is a word in English, the second one is the corresponding word in Martian's language. A line with a single string "END" indicates the end of the directory part, and this string should be ignored. The book part starts with a single line contains a string "START", this string should be ignored, then an article written in Martian's language. You should translate the article into English with the dictionary. If you find the word in the dictionary you should translate it and write the new word into your translation, if you can't find the word in the dictionary you do not have to translate it, and just copy the old word to your translation. Space(' '), tab('\t'), enter('\n') and all the punctuation should not be translated. A line with a single string "END" indicates the end of the book part, and that's also the end of the input. All the words are in the lowercase, and each word will contain at most 10 characters, and each line will contain at most 3000 characters.
 

Output
In this problem, you have to output the translation of the history book.
 

Sample Input
START from fiwo hello difh mars riwosf earth fnnvk like fiiwj END START difh, i'm fiwo riwosf. i fiiwj fnnvk! END
 

Sample Output
hello, i'm from mars. i like earth! 好开心,125ms,statistic里面第三,happying……
第二个trie,虽然调了老半天,是些字符串处理的错误
wa了两边,原因是count是用来判断有无该单
#include <cstdio>
#include 
<cstdlib>
#include 
<cstring>
#include 
<cmath>
#include 
<ctime>
#include 
<cassert>
#include 
<iostream>
#include 
<sstream>
#include 
<fstream>
#include 
<map>
#include 
<set>
#include 
<vector>
#include 
<queue>
#include 
<algorithm>
#include 
<iomanip>
#define pp printf("here\n")
using namespace std;
struct node
{
    
int next[26];
    
char name[11];
    
int count;
    
void init()
    
{
        memset(next,
-1,sizeof(next));
        count
=0;
    }

}
 s[5000005];
int ind,sind;
void cas_init()
{
    s[
0].init();
    sind
=1;
}

void ins(char str[],char str1[])
{
    
int len=strlen(str);
    
int i,j,ind;
    
for(i=ind=0; i<len; i++)
    
{
        j
=str[i]-'a';
        
if(s[ind].next[j]==-1)
        
{
            s[sind].init();
            s[ind].next[j]
=sind++;
        }

        
//s[ind].count++;
        ind=s[ind].next[j];
    }

    s[ind].count
++;
    len
=strlen(str1);
    
for(i=0; i<len; i++)  s[ind].name[i]=str1[i];
    s[ind].name[len]
='\0';
}

int search(char str[])
{
    
int ind,i,j,len=strlen(str);
    ind
=0;
    
for(i=0; i<len; i++)
    
{
        j
=str[i]-'a';
        
if(s[ind].next[j]==-1)
            
return 0;
        
else ind=s[ind].next[j];
    }

    
return ind;
}

int main()
{
    
int i,j,len,tmp,ai,k;
    cas_init();
    
char str[36],str11[15],strn[15];
    
char str1[1005],str2[10050];
    gets(str);
    
while(gets(str)&&!(strcmp(str,"END")==0)) 
    
{
        len
=strlen(str);
        i
=0;
        
while(i<len&&str[i]!=' '{str11[i]=str[i];i++;}str11[i]='\0';
        i
++;
        j
=i;
        
while(j<len){strn[j-i]=str[j];j++;}strn[j-i]='\0';
    
//    puts(strn);puts(str11);
        ins(strn,str11);

    }

    gets(str);
    
while(gets(str1)&&(strcmp(str1,"END")!=0))
    
{
        len
=strlen(str1);
        i
=0;
        j
=0;
        ai
=0;
        
while(j!=len)
        
{
            
while((str1[i]<'a'||str1[i]>'z')&&(i<len))
            
{
                str2[ai]
=str1[i];
                ai
++;i++;
            }

            j
=i;
            
while(str1[j]<='z'&&str1[j]>='a'&&j<len)
            
{
                strn[j
-i]=str1[j];
                j
++;
            }

            strn[j
-i]='\0';
            i
=j;
            tmp
=search(strn);
            
if(tmp!=0&&s[tmp].count>0)
            
{
                
for(k=0; k<strlen(s[tmp].name); k++)
                
{
                    str2[ai]
=s[tmp].name[k];
                    ai
++;
                }

            }

            
else 
            
{
                
for(k=0;k<strlen(strn);k++)
                
{
                    str2[ai]
=strn[k];
                    ai
++;
                }

            }

        }

        str2[ai]
='\0';
        puts(str2);
    }

    
return 0;
}


posted on 2012-07-16 00:38 jh818012 阅读(179) 评论(0)  编辑 收藏 引用


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