201612-3 CCF permission query full score solution (detailed notes + clear logic) + problem-solving idea: STL simulation

Problem Description

problem solving ideas

unordered_map <string, int> p; //store all permission sets (including graded permissions and graded permissions)
unordered_map <string, unordered_map <string, int>> r; //store all permissions included in all roles (including permissions The maximum level value)
unordered_map <string, unordered_map <string, int>> u; // store all permissions contained in all users (the maximum level value of the contained permissions)

Store permissions (corresponding levels), roles (corresponding permissions), and users (corresponding permissions) according to read-in

Then perform permission query based on the stored collection

if (没有该用户) false
else if (该用户没有该权限) false
else(该用户拥有该权限)
{
    
    
	if (该权限是无等级权限) true
	else(该权限是有等级权限)
	{
    
    
		if (该查询是有等级查询)
		{
    
    
			if (拥有该查询等级权限)true
			else(不拥有该查询等级权限)false
		}
		else(该查询是无等级查询)
		{
    
    
			if (不具有该类的任何等级权限)false
			else (返回具有该类权限的等级)
		}
	}
}

Code

#include <iostream>
#include <cstring>
#include <algorithm>
#include <unordered_map>

using namespace std;

unordered_map <string, int> p; //存储所有权限集合（包括有等级权限和有等级权限）
unordered_map <string, unordered_map <string, int>> r; //存储所有角色包含的所有权限（所含权限的最大等级值）
unordered_map <string, unordered_map <string, int>> u; //存储所有用户包含的所有权限（所含权限的最大等级值）

void work(string &str, int &x) //将含有冒号的表达式进行名字和等级的划分
{
    
    
    int n = str.size();
    if (n >= 2 && str[n - 2] == ':') //因为题目中指出等级只有可能是一位数字，所以如果存在，则一定是最后一位
    {
    
    
        x = str[n - 1] - '0';
        str = str.substr(0, n - 2);
    }
    //cout << str << " " << x << endl;
}

int main()
{
    
    
    int n;
    cin >> n;

    //存储权限
    for (int i = 0; i < n; i ++)
    {
    
    
        string str = "";
        int x = -1;
        cin >> str;
        work(str, x);
        p[str] = x;
    }

    //存储角色
    cin >> n;
    for (int i = 0; i < n; i ++)
    {
    
    
        string name;
        cin >> name;
        int num;
        cin >> num;
        for (int j = 0; j < num; j ++)
        {
    
    
            string str = "";
            int x = -1;
            cin >> str;
            work(str, x);
            if (r[name].count(str) == 0) r[name][str] = x; //该角色还没有存储过该权限
            else if (r[name][str] < x) r[name][str] = x; //该角色已存储的该权限的权限值小，更新该权限值
        }
    }

    //存储用户
    cin >> n;
    for (int i = 0; i < n; i ++)
    {
    
    
        string name;
        cin >> name;
        int num;
        cin >> num;
        for (int j = 0; j < num; j ++)
        {
    
    
            string role = "";
            cin >> role;
            unordered_map <string, int > t = r[role]; //该用户拥有的角色所拥有的权限集合
            for (auto x : t) //遍历权限集合，将其更新加入用户对应的权限集合中
            {
    
    
                string privilege = x.first;
                int level = x.second;
                if (u[name].count(privilege) == 0) u[name][privilege] = level;
                else if (u[name][privilege] < level) u[name][privilege] = level;
            }
        }
    }

    //查询权限
    cin >> n;
    while (n --)
    {
    
    
        string uname;
        cin >> uname;
        string privilege;
        int level = -1;
        cin >> privilege;
        work(privilege, level);
        
        if (u.count(uname) == 0) puts("false"); //没有该用户
        else if (u[uname].count(privilege) == 0) puts("false"); //该用户没有该权限
        else //该用户拥有该权限
        {
    
    
            if (p[privilege] == -1) puts("true"); //该权限是无等级权限
            else //该权限是有等级权限
            {
    
    
                if (level != -1) //该查询是有等级查询
                {
    
    
                    if (u[uname][privilege] >= level) puts("true"); //拥有该查询等级权限
                    else puts("false");//不拥有该查询等级权限
                }
                else//该查询是无等级查询
                {
    
    
                    if (u[uname][privilege] == -1) puts("false"); //不具有该类的任何等级权限
                    else cout << u[uname][privilege] << endl; //返回具有该类权限的等级
                }
            }
        }
    }
    return 0;
}