每日一题——QQandIP

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 ]  

题目大意是一个IP可以登陆好几个QQ，一个QQ可以在好几个IP地址登陆。现在要找出某个IP登陆了多少QQ，和一个QQ在哪些IP地址登陆了。

这题就是用map，没什么难的。直接贴代码了。

#include<iostream>
#include <vector>
#include <string>
#include <map>
#include<set>
#include <algorithm>

using namespace std;
int main()
{
	int n;
	string _ip;
	string qq;
	set<string> IP;
	set<string> QQ;
	map<string, vector<string>> _mapQQ;
	map<string, vector<string>> _mapIP;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> qq >> _ip;
		_mapQQ[qq].push_back(_ip);
		_mapIP[_ip].push_back(qq);
	}
	cin >> _ip >> qq;
	if (!_mapIP[_ip].empty())
	{
		for (int i = 0; i < _mapIP[_ip].size(); i++)
			QQ.insert(_mapIP[_ip][i]);
	}
	if (!_mapQQ[qq].empty())
	{
		for (int i = 0; i < _mapQQ[qq].size(); i++)
			IP.insert(_mapQQ[qq][i]);
	}

	string sip_qq;
	string sqq_ip;
	sip_qq += _ip;
	sip_qq += " ==> [ ";
	set<string>::iterator it; //定义前向迭代器  
	for (it = QQ.begin(); it != QQ.end(); it++)
	{
		sip_qq += *it;
		sip_qq += " ";
	}
	sip_qq += "]";
	sqq_ip += qq;
	sqq_ip += " ==> [ ";
	for (it = IP.begin(); it != IP.end(); it++)
	{
		sqq_ip += *it;
		sqq_ip += " ";
	}
	sqq_ip += "]";
	if (QQ.empty())
		cout << "no qq"<<endl;
	else {
		cout << sip_qq << endl;
	}
	if (IP.empty())
		cout << "no ip" << endl;
	else
	{
		cout << sqq_ip << endl;
	}
	return 0;
}