poj 2503 Babelfish
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Babelfish
Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 34169 Accepted: 14666
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.
题意:
给出字典的英文与外语的对照,然后给出外语,输出英语
题解:
裸的字典树
代码:
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int Max=27;const int maxn=811000;struct node{ int sign; char str[20]; int next[Max];}a[maxn];int cur=0;void insert(char *s,char *t){ int len,ans; int p=0; len=strlen(s); for(int i=0;i<len;i++) { ans=s[i]-'a'; if(a[p].next[ans]!=0) { p=a[p].next[ans]; } else { a[p].next[ans]=++cur; a[cur].sign=0; p=a[p].next[ans]; } } strcpy(a[p].str,t); a[p].sign=1;}void find(char *s){ int len,ans; int p=0; len=strlen(s); for(int i=0;i<len;i++) { ans=s[i]-'a'; if(a[p].next[ans]==0) { printf("eh\n"); return; } p=a[p].next[ans]; } if(a[p].sign) printf("%s\n",a[p].str); else printf("eh\n");}int main(){ char str[30]; while(gets(str)&&str[0]) { int pos=0; while(str[pos++]!=' '); str[pos-1]=0; insert(str+pos,str); } while(gets(str)) find(str); return 0;}
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