poj 2503 Babelfish(字典树·翻译)

poj 2503 Babelfish(字典树·翻译)

题目：​​class="data-table" data-id="t7a7e9d1-dZWS76iK" data-transient-attributes="class" data-width="802px" style="width: 100%; outline: none; border-collapse: collapse;">

Default

Babelfish

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 ogdaycat atcaypig igpayfroot ootfrayloops oopslayatcayittenkayoopslay

Sample Output

catehloops

Hint

Huge input and output,scanf and printf are recommended.

分析：这和hdu 1075 what are you talking about 很像，甚至比它还要简单。就是利用字典树进行单词翻译。

#include #include #include using namespace std;struct node{ bool cover; char word[15]; node *next[26]; node(){ cover=0; memset(word,0,sizeof(word)); memset(next,0,sizeof(next)); }};node *root;void insert(char *s1,char *s2){ node *p=root; int length=strlen(s2); for(int i=0;inext[s2[i]-'a']==0) p->next[s2[i]-'a']=new node(); p=p->next[s2[i]-'a']; } p->cover=1; strcpy(p->word,s1);}void del(node *p){ if(p==NULL){ return ; } for(int i=0;i<26;i++) del(p->next[i]); delete p;}char* query(char *s){ int length=strlen(s); node *p=root; for(int i=0;inext[s[i]-'a']==0) return 0; p=p->next[s[i]-'a']; } if(p->cover) return p->word; return 0;}char str[30],s1[15],s2[15];void strcpy(char *s,char *f,int start,int end){ for(int i=start;i<=end;i++){ s[i-start]=f[i]; } s[end-start+1]=0;}int main(){ //freopen("cin.txt","r",stdin); root=new node(); while(gets(str)){ int length1,slen=strlen(str); //slen=0意味着空行 if(slen==0) break; for(int i=0;i

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