AOJ Dictionary
作者:互联网
Your task is to write a program of a simple dictionary which implements the following instructions:
-
insert *str*: insert a string str in to the dictionary
-
find *str*: if the distionary contains str, then print 'yes', otherwise print 'no'
Input
In the first line n, the number of instructions is given. In the following n lines, n instructions are given in the above mentioned format.
Output
Print yes or no for each find instruction in a line.
Constraints
-
A string consists of 'A', 'C', 'G', or 'T'
-
1 ≤ length of a string ≤ 12
-
n ≤ 1000000
Sample Input 1
5
insert A
insert T
insert C
find G
find A
Sample Output 1
no
yes
Sample Input 2
13
insert AAA
insert AAC
insert AGA
insert AGG
insert TTT
find AAA
find CCC
find CCC
insert CCC
find CCC
insert T
find TTT
find T
Sample Output 2
yes
no
no
yes
yes
yes
首先考虑用set秒杀:
#include<bits/stdc++.h> using namespace std; set<string> S; int main(){ char ch[20]; string s; int n; cin >> n; for(int i = 0;i < n;i ++){ scanf("%s",ch); if(ch[0] == 'i'){ cin >> s; S.insert(s); } else if(ch[0] == 'f'){ cin >> s; if(S.find(s) != S.end()) cout << "yes" << endl; else cout << "no" << endl; } } return 0; }
嗯,确实可以秒了,但是这题的目的是为了应用一下散列法,所以还是老实地写了一下用散列法。
#include<bits/stdc++.h> #define ll long long using namespace std; //M是用来对哈希值取模的 const int N = 1e6 + 10,M = 999997; //采用拉链法来构造哈希表 list<string>L[N]; //哈希函数 int getchar(char ch){ if(ch == 'A')return 1; else if(ch == 'C')return 2; else if(ch == 'G')return 3; else if(ch == 'T')return 4; else return 0; } int getkey(char ch[]){ ll sum = 0,p = 1,i; for(int i = 0;i < strlen(ch);i ++){ sum += p*getchar(ch[i]); p *= 5; } return sum%M; } int main(){ char s[20]; int n; scanf("%d",&n); for(int i = 0;i < n;i ++){ cin >> s; char str[12]; //判断是insert还是find if(s[0] == 'i'){ cin >> str; ll key = getkey(str); //在对应哈希值的链表里插入这个元素 L[key].push_back(str); } else if(s[0] == 'f'){ cin >> str; int key = getkey(str); //在对于哈希值的链表里查找对应元素 list<string>::iterator iter = L[key].begin(); for(;iter != L[key].end();iter ++){ if(str == *iter){ cout << "yes" << endl; break; } } //迭代器指向链表end(),则没有找到相应元素 if(iter == L[key].end())cout << "no" << endl; } } return 0; }
原书上的开放地址法感觉没有拉链法来得简洁明了。
标签:ch,return,Dictionary,cin,int,AOJ,else,str 来源: https://www.cnblogs.com/Moniq/p/15557957.html