算典05_例题_11_UVA-814

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The Letter Carrier’s Rounds

题意

本题的任务为模拟发送邮件时MTA（邮件传输代理）之间的交互。所谓MTA，就是email地址格式user@mtaname的“后面部分”。当某人从user1@mta1发送给另一个人user2@mta2时，这两个MTA将会通信。如果两个收件人属于同一个MTA，发送者的MTA只需与这个MTA通信一次就可以把邮件发送给这两个人。

题解

map
set
看代码

#include <algorithm>#include <iostream>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <cstring>#include <cstdio>#include <cmath>#define met(a, b) memset(a, b, sizeof(a));#define IN freopen("in.txt", "r", stdin);using namespace std;typedef long long LL;const int maxn = 1e6 + 5;const int INF = (1 << 31) - 1;string s, t, user1, mta1, user2, mta2;int k;set<string> addr;void solve(const string& s, string& user, string& mta) {    int k = s.find('@');    user = s.substr(0,k);    mta = s.substr(k+1);}int main() {    #ifdef _LOCAL    IN;    #endif // _LOCAL    //insert    while(cin >> s && s != "*") {        cin >> s >> k;        while(k--) {cin >> t; addr.insert(t + "@" + s); }    }    while(cin >> s && s != "*") {        solve(s, user1, mta1);        vector<string> mta;        map<string, vector<string> > dest;        set<string> vis;        while(cin >> t && t != "*") {            solve(t, user2, mta2);            if(vis.count(t)) continue;            vis.insert(t);            if(!dest.count(mta2)) {                mta.push_back(mta2);                dest[mta2] = vector<string>();            }            dest[mta2].push_back(t);        }        getline(cin, t);        string data;        while(getline(cin, t) && t[0] != '*') data += "     " + t + "\n";        for(int i = 0; i < mta.size(); i++) {            string mta2 = mta[i];            vector<string> users = dest[mta2];            cout << "Connection between " << mta1 << " and " << mta2 <<endl;            cout << "     HELO " << mta1 << "\n";            cout << "     250\n";            cout << "     MAIL FROM:<" << s << ">\n";            cout << "     250\n";            bool ok = false;            for(int i = 0; i < users.size(); i++) {                cout << "     RCPT TO:<" << users[i] << ">\n";                if(addr.count(users[i])) {                    ok = true;                    cout << "     250\n";                } else cout << "     550\n";            }            if(ok) {                cout << "     DATA\n";                cout << "     354\n";                cout << data;                cout << "     .\n";                cout << "     250\n";            }            cout << "     QUIT\n";            cout << "     221\n";        }    }    return 0;}

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