题目描述
blablabla
样例
/*
对于70%数据的读取没有想到好的方法,所以目标是做40%
*/
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int n, m;
const int N = 2510, M = 510, null = 0x3f3f3f3f;//null for N/A
char ch[N];
struct node{
int dn;
int attr[M];// 0x3f3f3f3f or val
}p[N];
bool cmp(node s1, node s2){
return s1.dn < s2.dn;
}
bool isSuitable(int i, int a, char op, int c){
if(op == ':'){
if(p[i].attr[a] == c) return true;
}else{// ~
if(p[i].attr[a] != null && p[i].attr[a] != c) return true;
}
return false;
}
int main(){
cin >> n;
int t;
for(int i = 0; i < n; i++){
cin >> p[i].dn >> t;//输入多个需要用scanf,这里较少
memset(p[i].attr, 0x3f, sizeof p[i].attr);
while(t--){
int data, val;
cin >> data >> val;
p[i].attr[data] = val;
}
}
sort(p, p + n, cmp);//sorted by dn
cin >> m;
while(m--){
scanf("%s", ch);
if(ch[0] == '&' || ch[0] == '|'){ //easy expr
int a, b, c, d;
char op, op1, op2;
sscanf(ch, "%c(%d%c%d)(%d%c%d)", &op, &a, &op1, &b, &c, &op2, &d);
if(op == '&'){
for(int i = 0; i < n; i++){
if(isSuitable(i, a, op1, b) && isSuitable(i, c, op2, d)) cout << p[i].dn << ' ';
}
}else{// |
for(int i = 0; i < n; i++){
if(isSuitable(i, a, op1, b) || isSuitable(i, c, op2, d)) cout << p[i].dn << ' ';
}
}
}else{ //base expr
int a, c;
char op;
sscanf(ch, "%d%c%d", &a, &op, &c);
for(int i = 0; i < n; i++){
if(isSuitable(i, a, op, c)) cout << p[i].dn << ' ';
}
}
puts("");
}
return 0;
}```
----------
### 算法1
##### (暴力枚举) $O(n^2)$
blablabla
#### 时间复杂度
#### 参考文献
#### C++ 代码
blablabla
----------
### 算法2
##### (暴力枚举) $O(n^2)$
blablabla
#### 时间复杂度
#### 参考文献
#### C++ 代码
blablabla
```