[STL]IP and QQ (eden)

Descrption

An ip can login several qqs, and a qq can be logined by several ips.

Your task is to find which qqs have been logined by the ip and which ips have logined the qq.

output format : qq ==> [ ip1 ip2 … ] and ip ==> [ qq1 qq2 … ]

if no such qq or ip, output “no qq” and “no ip”

First input n, then n+2 lines follow..

n lines:

qq ip

2 lines:

ip // find which qqs have been logined by the ip

qq // find which ips have logined the qq.

sample input:

5  10258279649 192.168.1.45  10258279649 192.168.1.45  10258279643 192.168.1.40  10258279640 192.168.1.45  10258279641 192.168.1.30  192.168.1.45  10258279649  

sample output:

192.168.1.45 ==> [ 10258279640 10258279649 ]  10258279649 ==> [ 192.168.1.45 ]  

Hint

map

set

From: 黄浩然

Submission

#include <iostream>#include <map>#include <set>using namespace std;int main(){    map<string,set<string>> QQ,IP;    int count;    cin>>count;    while(count--){        string q,i;        cin>>q>>i;        QQ[q].insert(i);//利用map下标访问的特性，键值不存在则插入该键        IP[i].insert(q);    }    string i,q;    cin>>i>>q;    auto iter=IP.find(i);    if(iter==IP.end())        cout<<"no qq\n";    else{        cout<<i<<" ==> [ ";        set<string> tem=(*iter).second;        for(auto a:tem)            cout<<a<<" ";        cout<<"]\n";    }    auto itor=QQ.find(q);    if(itor==QQ.end())        cout<<"no ip\n";    else{        cout<<q<<" ==> [ ";        set<string> tem=(*itor).second;        for(auto a:tem)            cout<<a<<" ";        cout<<"]\n";    }}