Babelfish
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 41263 | Accepted: 17561 |
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.
思路：把字符串哈希成２６进制数，然后查找的复杂度就是线性的。字符串共有26^10种可能，约等于10^14，可以用long long存下。使用map，key为哈希值，value为该串的位置。
1 //2016.9.4
2 #include <iostream>
3 #include <cstdio>
4 #include <cstdlib>
5 #include <cstring>
6 #include <map>
7 #define N 100005
8
9 using namespace std;
10
11 char s1[N][12], s2[N][12], str[25];
12 map<long long, int> Hash;
13
14 long long F(char* s)//把字符串hash成一个26进制数
15 {
16 int len = strlen(s);
17 long long h = 0;
18 for(int i = 0; i < len; i++)
19 h = h*26+(s[i]-‘a‘);
20 return h;
21 }
22
23 int main()
24 {
25 int cnt = 0;
26 while(gets(str))
27 {
28 if(str[0] == ‘\0‘)break;
29 sscanf(str,"%s%s", s1[cnt], s2[cnt]);
30 long long h = F(s2[cnt]);
31 Hash[h] = cnt;
32 cnt++;
33 }
34 while(gets(str))
35 {
36 if(str[0] == ‘\0‘)break;
37 long long h = F(str);
38 if(Hash.find(h)!=Hash.end())//map按key查找，失败返回end
39 printf("%s\n", s1[Hash[h]]);
40 else printf("eh\n");
41 }
42
43 return 0;
44 }