权限查询

题目链接

这道题一开始我想用map表存储数据，后来才意识到这是一种错误的想法。过多的map会大大增加编程的复杂度。参考题目中给出的最多只有一百个权限、一百个角色和一百个用户名，所以这道题最理想的做法是用遍历穷举的方式一个一个去找。

100分代码如下，存在一定的冗余但是并不影响最后的效果。

#include <iostream>
#include <map>
#include <vector>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cmath>
#include<set>
#include<bitset>
#include<stack>
#include<cstring>
#include<string>
using namespace std;
#define MAX 110
struct QX
{
	string name;
	int rank;
}qx[MAX];

struct Role
{
	string name;
	QX qxs[11];
	int cntquan;
}roles[MAX];
struct Name
{
	string name;
	Role roles[11];
	int cntrolt;
}user[MAX];
int c1, c2, c3,c4;

Role findrole(string role)
{
	for (int i = 1; i <= c2; i++)
		if (roles[i].name == role)
			return roles[i];
}



bool find_name(string name)
{
	for (int i = 1; i <= c3; i++)
		if (user[i].name == name)
			return true;
	return false;
}
//在权限表中寻找一个权限
bool find_qx(string qxname,int rank)
{

	for (int i = 1; i <= c1; i++)
	{
		if (qx[i].name == qxname&&qx[i].rank>=rank)
		{
			return true;
		}
	}
	return 0;
}



void find_name_qx(string username, string inpu)
{
	if (!find_name(username))
	{
		cout << "false" << endl;
		return;
	}

	int loc = inpu.find(':');

	string qxname;// = inpu.substr(0, loc);
	int rank = 10;
	if (loc != string::npos)
	{
		qxname = inpu.substr(0, loc);
		rank = atoi(inpu.substr(loc + 1).c_str());
	}
	else
	{
		qxname = inpu;
	}
	int finalans = -1;
	//cout << username << "  " << inpu << endl;

	//cout << qxname << "  " << rank << endl;
	for (int i = 1; i <= c3; i++)
	{
		if (user[i].name == username)
		{
			for (int j = 1; j <= user[i].cntrolt; j++)
			{
				Role role = user[i].roles[j];
				for (int k = 1; k <=role.cntquan; k++)
				{
					if (role.qxs[k].name == qxname)
					{
					//	cout<<"------" << role.qxs[k].name << endl;
						if (rank == 10&&role.qxs[k].rank==10)
						{
							cout << "true" << endl; 
							return;
						}
						else if(role.qxs[k].rank!=10)
						{
							finalans = max(finalans, role.qxs[k].rank);
						}
					}
				}
			}
		}
	}
	if (finalans == -1)
	{
	//	cout << "fina" << endl;
		cout << "false" << endl;
		return ;
	}
	else
	{
		if (find_qx(qxname, finalans))
		{
			if (rank == 10)
			{
				cout << finalans << endl;
			}
			else if (rank <= finalans)
			{
				cout << "true" << endl;
			}
			else {
				cout << "false" << endl;
			}
		}
		else
		{
			cout << "false" << endl;
		}
	}

}
QX find(string qx1)
{
	for(int i=1;i<=c1;i++)
		if (qx[i].name == qx1)
		{
			return qx[i];
		}
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("sb.txt", "r", stdin);
#endif // !ONLINE_JUDGE
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> c1;
	for (int i = 1; i <= c1; i++)
	{
		string t;
		cin >>t;
		int loc = t.find(':');
		if (loc != string::npos)
		{
			qx[i].name = t.substr(0, loc);
			qx[i].rank = atoi(t.substr(loc + 1).c_str());
		}
		else
		{
			qx[i].name = t;
			qx[i].rank = 10;
		}
	}
	cin >> c2;
	for (int i = 1; i <= c2; i++)
	{
		string rolename;
		int cnt;
		cin >> rolename >> cnt;
		roles[i].name = rolename;
		for (int j = 1; j <= cnt; j++)
		{
			string str;
			cin >> str;
			string qxname;
			int loc = str.find(':');
			int rank = 10;
			if (loc == string::npos)
			{
				qxname = str;
			}
			else {
				qxname = str.substr(0, loc);
				rank = atoi(str.substr(loc + 1).c_str());
			}
			roles[i].cntquan++;
			roles[i].qxs[roles[i].cntquan].name = qxname;
			roles[i].qxs[roles[i].cntquan].rank = rank;
		}
	}
	cin >> c3;
	for (int i = 1; i <= c3; i++)
	{
		string username;
		int cnt;
		cin >> username >> cnt;
		user[i].name = username;
		for (int j = 1; j <= cnt; j++)
		{
			string rolename;
			cin >> rolename;
			user[i].cntrolt++;
			user[i].roles[user[i].cntrolt]= findrole(rolename);
		}
	}
	cin >> c4;
	for (int i = 1; i <= c4; i++)
	{
		string name, item;
		cin >> name >> item;
		find_name_qx(name, item);
	}

	return 0;
}