8.5-Day1T1--Asm.Def 谈笑风生
题目大意
m个操作,
1:添加一个字符串
2:查询字符串s是否被添加过(中至多包含一个通配符“*”)
题解
trie树可以得部分分
用map映射
'*'就枚举26个英文字母来判断就可以了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<algorithm> #include<iostream> #include<map> #include<cstring> #include<string> using namespace std; inline int read() { int sum = 0,p = 1; char ch = getchar(); while(ch < '0' || ch > '9') { if(ch == '-') p = -1; ch = getchar(); } while(ch >= '0' && ch <= '9') { (sum *= 10) += ch - '0'; ch = getchar(); } return sum * p; } map<string,int> q; string s; int n,opt; int main() { n = read(); for(int i = 1;i <= n;i++) { opt = read(); cin>>s; if(opt == 1) q[s] = 1; if(opt == 2) { int len = s.length(); int o = 30; for(int i = 0;i < len;i++) if(s[i] == '*') { o = i; break; } if(o == 30) { if(q.count(s)) printf("YES\n"); else printf("NO\n"); } else { bool flag = 0; for(int i = 0;i < 26;i++) { char v = 'a' + i; s[o] = v; if(q.count(s)) { flag = 1; printf("YES\n"); break; } } if(!flag) printf("NO\n"); } } } return 0; }