例题5-11 邮件传输代理的交互（The Letter Carrier's Rounds，ACM/ICPC World Finals 1999，UVa814）

原题链接：https://vjudge.net/problem/UVA-814
分类：STL综合
备注：字符串以及STL容器的综合运用
前言：uDebug AndyRoswellD的数据没过，但是我没看懂他的输出的逻辑，难道有多个传输者？题目没这么写啊，好歹还是AC了。个人认为是他上传的数据有误。

代码如下：

#include<iostream>
#include<sstream>
#include<string>
#include<map>
#include<set>
#include<vector>
using namespace std;
map<string, set<string> >person;
set<string>adress;
void output(string s, vector<string>message) {
	stringstream ss(s); 
	string sender, senddress;
	ss >> sender;
	int pos = sender.find('@');
	senddress = sender.substr(pos + 1);
	string tmp, dr;
	vector<string>recdress;
	vector<string>recpt;
	map<string, set<string> >match;
	while (ss >> tmp) {
		pos = tmp.find('@');
		dr = tmp.substr(pos + 1);
		if (!match.count(dr))recdress.push_back(dr);
		if (!match[dr].count(tmp)) {
			recpt.push_back(tmp);
			match[dr].insert(tmp);
		}
	}
	for (int i = 0; i < recdress.size(); i++) {
		cout << "Connection between " << senddress << " and " << recdress[i] << "\n";
		cout << "     HELO " << senddress << "\n     250\n     MAIL FROM:<" << sender << ">\n     250\n";
		int flag = 0;
		for (int j = 0; j < recpt.size(); j++) {
			if (match[recdress[i]].count(recpt[j])) {
				cout << "     RCPT TO:<" << recpt[j] << ">\n";
				pos = recpt[j].find('@');
				string tmp2 = recpt[j].substr(0, pos);
				if (person[recdress[i]].count(tmp2)) {
					cout << "     250\n";
					flag = 1;
				}
				else cout << "     550\n";
			}
		}
		if (flag) {
			cout << "     DATA\n     354\n";
			for (int k = 0; k < message.size(); k++)
				cout << "     " << message[k] << "\n";
			cout << "     .\n     250\n";
		}
		cout << "     QUIT\n     221\n";
	}
}
int main(void){

	int n;
	string s, pepo;
	while (cin >> s){
		if (s[0] == '*')break;
		cin >> s;
		adress.insert(s);
		cin >> n;
		while (n--) {
			cin >> pepo;
			person[s].insert(pepo);
		}
	}
	getchar();
	while (getline(cin,s)){
		if (s[0] == '*')break;
		string mess;
		cin >> mess;
		getchar();
		vector<string>message;
		while (getline(cin, mess)){
			if (mess[0] == '*')break;
			message.push_back(mess);
		}
		output(s, message);
	}
	return 0;
}