CCF CSP 201612-3权限查询（c 100）

CCF CSP 201612-3权限查询

按题意模拟，一层层嵌套，有注释

#include<iostream>#include<map>#include<string>#include<algorithm>using namespace std;map<string,int>category,cy;map<string,map<string, int> >role;map<string,map<string, int> >user;int main(){int n;cin>>n;//存储权限 for(int i = 0; i < n; i  ){string s;cin>>s;int f = -1;//分有权限和无权限两种 if((f = s.find(':'))!=-1){int num = 0;for(int j = f 1; j < s.size(); j  ){num = num * 10   s[j]-'0';}category[s.substr(0,f)] = num;}else category[s] = -1;}cin>>n;//存储权限 for(int i = 0; i < n; i  ){string str,s;int m;cin>>str>>m;cy.clear();for(int k = 0; k < m; k  ){cin>>s;int f = -1;//带权限的存储最大权限，无权限的存储为-1 if((f = s.find(':'))!=-1){int num = 0;for(int j = f 1; j < s.size(); j  ){num = num * 10   s[j]-'0';}if(cy.count(s.substr(0,f))){cy[s.substr(0,f)] = max(cy[s.substr(0,f)],num);}else{cy[s.substr(0,f)] = num;}}else cy[s] = -1;}role[str] = cy;}//存储角色cin>>n;for(int i = 0; i < n; i  ){string str,s;int m;cin>>str>>m;cy.clear();for(int k = 0; k < m; k  ){cin>>s;int f = -1;//遍历角色中的权限，替换到用户中 map<string,int> mp = role[s];for(map<string,int>::iterator it = mp.begin(); it != mp.end(); it  ){if(cy.count(it->first)){cy[it->first] = max(cy[it->first],it->second);}else{cy[it->first] = it->second;}}}user[str] = cy;}//for(map<string,map<string,int> > ::iterator it = user.begin(); it != user.end(); it  ){//cout<<it->first<<":";//for(map<string,int>::iterator it1 = it->second.begin(); it1 != it->second.end(); it1  ){//cout<<it1->first<<"  "<<it1->second<<" ";//}//cout<<endl;//}cin>>n;for(int i = 0; i < n; i  ){string name,privilege;cin>>name>>privilege;//是否有该用户 if(user.count(name)){int f = -1;//是否是带权限查询 if((f = privilege.find(':'))!=-1){int num = 0;for(int j = f 1; j < privilege.size(); j  ){num = num * 10   privilege[j]-'0';}//权限是否存在 if(user[name].count(privilege.substr(0,f))){if(user[name][privilege.substr(0,f)] >= num){cout<<"true";}else{cout<<"false";}}else{cout<<"false";}}else{if(user[name].count(privilege)){if(user[name][privilege] == -1)cout<<"true";else{cout<<user[name][privilege];}}else{cout<<"false";}}}else{cout<<"false";}cout<<endl;}return 0;}

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