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
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:
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”。
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?”
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
[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?
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 }