codeforces 589A Email Aliases(map)

Polycarp has quite recently learned about email aliases. Of course, he used to suspect that the case of the letters doesn‘t matter in email addresses. He also learned that a popular mail server in Berland bmail.com ignores dots (characters ‘.‘) and all the part of an address from the first character "plus" (‘+‘) to character "at" (‘@‘) in a login part of email addresses.

Formally, any email address in this problem will look like "login@domain", where:

• a "login" is a non-empty sequence of lowercase and uppercase letters, dots (‘.‘) and pluses (‘+‘), which starts from a letter;
• a "domain" is a non-empty sequence of lowercase and uppercase letters and dots, at that the dots split the sequences into non-empty words, consisting only from letters (that is, the "domain" starts from a letter, ends with a letter and doesn‘t contain two or more consecutive dots).

When you compare the addresses, the case of the characters isn‘t taken into consideration. Besides, when comparing the bmail.com addresses, servers ignore the dots in the login and all characters from the first character "plus" (‘+‘) to character "at" (‘@‘) in login part of an email address.

For example, addresses saratov@example.com and SaratoV@Example.Com correspond to the same account. Similarly, addresses ACM.ICPC.@bmail.com and A.cmIcpc@Bmail.Com also correspond to the same account (the important thing here is that the domains of these addresses are bmail.com). The next example illustrates the use of character ‘+‘ in email address aliases: addresses polycarp+contest@BMAIL.COM, Polycarp@bmail.com and polycarp++acm+icpc@Bmail.Com also correspond to the same account on the server bmail.com. However, addresses a@bmail.com.ru and a+b@bmail.com.ru are not equivalent, because ‘+‘ is a special character only for bmail.com addresses.

Polycarp has thousands of records in his address book. Until today, he sincerely thought that that‘s exactly the number of people around the world that he is communicating to. Now he understands that not always distinct records in the address book represent distinct people.

Help Polycarp bring his notes in order by merging equivalent addresses into groups.

6
ICPC.@bmail.com
p+con+test@BMAIL.COM
P@bmail.com
a@bmail.com.ru
I.cpc@Bmail.Com
a+b@bmail.com.ru

4
2 ICPC.@bmail.com I.cpc@Bmail.Com 
2 p+con+test@BMAIL.COM P@bmail.com 
1 a@bmail.com.ru 
1 a+b@bmail.com.ru 
题意：有两种邮件第一种：@bmail 第二种L：@bmain.xxx。对于第一种邮件格式我们要求在@前的 . 忽略，对于@前第一个出现的+后面的内容直到@全部删除，给你n个邮件地址，让你把它们分组，并把每组所包含的
邮件数 和 邮件都打印出来。
分析： map搞。

 1 /*************************************************************************
 2     > File Name: 365A.cpp
 3     > Author: 
 4     > Mail: 
 5     > Created Time: 2016年07月29日 星期五 20时54分32秒
 6  ************************************************************************/
 7 
 8 #include<iostream>
 9 #include<bits/stdc++.h>
10 using namespace std;
11 const int maxn = 2e4 + 7;
12 map<string,int> mp;
13 map<string,int> :: iterator it;
14 int num[maxn];
15 vector<string> a[maxn];
16 char cnt[] ={"bmail.com"};
17 char str[105],str1[105];
18 
19 int main()
20 {
21     int n;
22     cin >> n;
23     int id = 0;
24     mp.clear();
25     while(n--)
26     {
27         cin >> str;
28         strcpy(str1,str);
29         int st;
30         for(int i = 0; str1[i]; i++)
31         {
32             str1[i] = tolower(str1[i]);
33             if(str1[i] == ‘@‘) st = i; 
34         }
35         if(strcmp(str1 + st + 1,cnt) == 0)
36         {
37             int t = 0;
38             for(int i = 0; i< st; i++)
39             {
40                 if(str1[i] == ‘.‘) continue;
41                 if(str1[i] == ‘+‘) break;
42                 str1[t++] = str1[i];
43             }
44             for(int i = st; str1[i];i++)
45             {
46                 str1[t++] = str1[i]; 
47             }
48             str1[t] = 0;
49         }
50         if(mp.find(str1) == mp.end()) mp[str1] = ++id;
51         int ss = mp[str1];
52         ++num[ss];
53         a[ss].push_back(str);
54     }
55     cout << id << endl;
56     for(int i = 1; i<= id; i++)
57     {
58         cout << num[i] << endl;
59         for(int j = 0; j< a[i].size(); j++) cout << ‘ ‘ << a[i][j];
60         cout << endl;
61     }
62     return 0;
63 }