CCF-CSP 202303-3 LDAP（待更）_ccf202303星际网络

题目链接

题目非常长，但还没有到完全看不懂的地步，即给你一些匹配表达式，让你输出对应的用户DN即可

该题，我就想奔着40分去的，因为40分只是符合EASY_EXPR即简单的匹配表达式，它只包含一个逻辑符号和两个括号内的简单式子如

&(1:2)(2:3)

数据结构上我选择vector与pair进行存储如下

 

因为如果只是满足EASY_EXPR，那么可能的情况会比较少，因此完全可以使用if将所有情况列出来，具体实现如下（不知为何0分）

#include<iostream> 
#include<vector>
#include<string>
#include<set>
using namespace std;
 
int n, m;
 
typedef struct Node {
	int attr, value;
}Node;
vector<pair<int, vector<Node> > >arr;
 
vector<int> fun1(string exp) {//得到:符号匹配用户的DN
	int key = exp[0] - 48, v = exp[2] - 48;
	vector<int>res;A
	for (int i = 0; i < n; i++) {
		pair<int, vector<Node> >per = arr[i];
		int curDN = per.first;
		vector<Node>temp = per.second;
		int len = temp.size();
		for (int j = 0; j < len; j++) {
			if (temp[j].attr == key && temp[j].value == v) {
				res.push_back(curDN);
				break;
			}
		}
	}
	return res;
}
 
vector<int> fun2(string exp) {//得到~符号匹配用户的DN
	int key = exp[0] - 48, v = exp[2] - 48;
	vector<int>res;
	for (int i = 0; i < n; i++) {
		pair<int, vector<Node> >per = arr[i];
		int curDN = per.first;
		vector<Node>temp = per.second;
		int len = temp.size();
		for (int j = 0; j < len; j++) {
			if (temp[j].attr == key && temp[j].value != v) {
				res.push_back(curDN);
				break;
			}
		}
	}
	return res;
}
 
int main() {
	cin >> n;
	vector<Node>temp;
	for (int i = 1; i <= n; i++) {
		int DN, count_attr, attr, value;
		temp.clear();
		cin >> DN >> count_attr;
		for (int j = 1; j <= count_attr; j++) {
			cin >> attr >> value;
			temp.push_back({ attr,value });
		}
		arr.push_back(make_pair(DN, temp));
	}
	cin >> m;
	while (m--) {
		vector<int>ans;
		ans.clear();
		string exp;
		cin >> exp;
		if (exp[0] == '&') {
			string sub1 = exp.substr(2, 3);
			string sub2 = exp.substr(7, 3);
			vector<int>res1, res2;
			set<int>st;
			if (sub1[1] == ':') {
				res1 = fun1(sub1);
			}
			else if (sub1[1] == '~') {
				res1 = fun2(sub1);
			}
			if (sub2[1] == ':') {
				res2 = fun1(sub2);
			}
			else if (sub2[1] == '~') {
				res2 = fun2(sub2);
			}
			//找到res1与res2共同拥有的部分
			int len1 = res1.size(),len2=res2.size();
			for (int i = 0; i < len1; i++) {
				st.insert(res1[i]);
			}
			for (int j = 0; j < len2; j++) {
				if (st.count(res2[j])) {
					ans.push_back(res2[j]);
				}
			}
		}
		else if (exp[0] == '|') {
			string sub1 = exp.substr(2, 3);
			string sub2 = exp.substr(7, 3);
			vector<int>res1, res2;
			set<int>st;
			if (sub1[1] == ':') {
				res1 = fun1(sub1);
			}
			else if (sub1[1] == '~') {
				res1 = fun2(sub1);
			}
			if (sub2[1] == ':') {
				res2 = fun1(sub2);
			}
			else if (sub2[1] == '~') {
				res2 = fun2(sub2);
			}
			//找到res1与res2拥有的部分去重加入到ans中
			int len1 = res1.size(), len2 = res2.size();
			for (int i = 0; i < len1; i++) {
				st.insert(res1[i]);
			}
			for (int j = 0; j < len2; j++) {
				st.insert(res2[j]);
			}
			for (auto i = st.begin(); i != st.end(); i++) {
				ans.push_back(*i);
			}
		}
		else if(exp.length()==3){
			if (exp[1] == ':') {
				ans = fun1(exp);
			}
			else if (exp[1] == '~') {
				ans = fun2(exp);
			}
		}
		int len = ans.size();
		for (int i = 0; i < len; i++) {
			if (i == 0) {
				cout << ans[i];
			}
			else {
				cout << " " << ans[i];
			}
		}
		cout << endl;
	}
	return 0;
}