hdu 1880 魔咒词典(双hash)

魔咒词典
Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16775    Accepted Submission(s): 3963

Problem Description
哈利波特在魔法学校的必修课之一就是学习魔咒。据说魔法世界有100000种不同的魔咒,哈利很难全部记住,但是为了对抗强敌,他必须在危急时刻能够调用任何一个需要的魔咒,所以他需要你的帮助。
给你一部魔咒词典。当哈利听到一个魔咒时,你的程序必须告诉他那个魔咒的功能;当哈利需要某个功能但不知道该用什么魔咒时,你的程序要替他找到相应的魔咒。如果他要的魔咒不在词典中,就输出“what?”
 
Input
首先列出词典中不超过100000条不同的魔咒词条,每条格式为:
[魔咒] 对应功能
其中“魔咒”和“对应功能”分别为长度不超过20和80的字符串,字符串中保证不包含字符“[”和“]”,且“]”和后面的字符串之间有且仅有一个空格。词典最后一行以“@END@”结束,这一行不属于词典中的词条。
词典之后的一行包含正整数N(<=1000),随后是N个测试用例。每个测试用例占一行,或者给出“[魔咒]”,或者给出“对应功能”。
 
Output
每个测试用例的输出占一行,输出魔咒对应的功能,或者功能对应的魔咒。如果魔咒不在词典中,就输出“what?”
 
Sample Input
[expelliarmus] the disarming charm
[rictusempra] send a jet of silver light to hit the enemy
[tarantallegra] control the movement of one's legs
[serpensortia] shoot a snake out of the end of one's wand
[lumos] light the wand
[obliviate] the memory charm
[expecto patronum] send a Patronus to the dementors
[accio] the summoning charm
@END@
4
[lumos]
the summoning charm
[arha]
take me to the sky
 
Sample Output
light the wand
accio
what?
what?

 

C/C++:

 1 #include <map>
 2 #include <queue>
 3 #include <cmath>
 4 #include <vector>
 5 #include <string>
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <climits>
 9 #include <iostream>
10 #include <algorithm>
11 #define INF 0x3f3f3f3f
12 using namespace std;
13 const int Mod = 1e9 + 7;
14 
15 char str[300], s1[100], s2[200];
16 map<pair<int, int>, string> m0;
17 map<pair<int, int>, string> m1;
18 
19 int get_hash0(char *s)
20 {
21     int sum = 0;
22     for (int i = 0; s[i]; ++ i)
23         sum = (sum * 117 % Mod + s[i]) % Mod;
24     return sum;
25 }
26 
27 int get_hash1(char *s)
28 {
29     int sum = 0;
30     for (int i = 0; s[i]; ++ i)
31         sum = (sum * 131 % Mod + s[i]) % Mod;
32     return sum;
33 }
34 
35 int main()
36 {
37     int n, a, b;
38     m0.clear();
39     m1.clear();
40     ios::sync_with_stdio(false);
41     while (gets(str), str[0] != '@')
42     {
43         a = 1, b = 0;
44         for (a, b; str[a] != ']'; ++ a, ++ b)
45             s1[b] = str[a];
46         s1[b] = 0;
47         b = 0, a += 2;
48         for (a, b; str[a]; ++ a, ++ b)
49             s2[b] = str[a];
50         s2[b] = 0;
51         a = get_hash0(s1), b = get_hash1(s1);
52         m0[make_pair(a, b)] = s2;
53         a = get_hash0(s2), b = get_hash1(s2);
54         m1[make_pair(a, b)] = s1;
55     }
56     scanf("%d", &n);
57     getchar();
58     while (n --)
59     {
60         gets(str);
61         if (str[0] == '[')
62         {
63             str[strlen(str) - 1] = 0;
64             a = get_hash0(str + 1), b = get_hash1(str + 1);
65             if (m0.find(make_pair(a, b)) != m0.end()) cout <<m0[make_pair(a, b)] <<endl;
66             else puts("what?");
67         }
68         else
69         {
70             str[strlen(str)] = 0;
71             a = get_hash0(str), b = get_hash1(str);
72             if (m1.find(make_pair(a, b)) != m1.end()) cout <<m1[make_pair(a, b)] <<endl;
73             else puts("what?");
74         }
75     }
76     return 0;
77 }

 

posted @ 2018-08-21 09:59  GetcharZp  阅读(161)  评论(0编辑  收藏  举报