二维数组trie pku2503

Babelfish

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12524		Accepted: 5522

Description

You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.

Input

Input consists of up to 100,000 dictionary entries, followed by a blank line, followed by a message of up to 100,000 words. Each dictionary entry is a line containing an English word, followed by a space and a foreign language word. No foreign word appears more than once in the dictionary. The message is a sequence of words in the foreign language, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

Output

Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".

Sample Input

dog ogday
cat atcay
pig igpay
froot ootfray
loops oopslay

atcay
ittenkay
oopslay

Sample Output

cat eh loops

Hint

Huge input and output,scanf and printf are recommended.

Source

用二维数组实现的trie树， 相对于用指针实现的trie，是不是能节约空间呢？

第四种解法 trie tree

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int index[1000100][26];
int length=0;
int value[1000100];
char english[100020][12];
inline bool myget(char *str)
{
     char ch=getchar();
     if(ch==' '||ch=='\n'||ch==EOF)
       return false;
     while(ch!=' '&&ch!='\n')
     {
           *str++=ch;
           ch=getchar();
     }
     *str=0;
     return true;
}

int main()
{
    int i=1;
    char foreign[12];
    while(myget(&english[i][0]))
    {
         myget(&foreign[0]);
         
         int j=0,start=0;
         while(foreign[j]!=0)
         {
               int next=foreign[j]-'a';
               if(index[start][next]==0)
                  index[start][next]=++length;
               start=index[start][next];
               ++j;
         }
         value[start]=i;
         ++i;
    }

    char forei[12];
    while(myget(forei))
    {
         int j=0,start=0;
         while(forei[j]!=0)
         {
               int next=forei[j]-'a';
               start=index[start][next];
               ++j;
         }
         
         start=value[start];
         if(start==0)
           printf("eh\n");
         else
           printf("%s\n",english[start]);
    }
    return 0;
}