Description
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.comaddresses, 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, addressesACM.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: addressespolycarp+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.
Input
The first line of the input contains a positive integer n(1 ≤ n ≤ 2·104) — the number of email addresses in Polycarp's address book.
The following n lines contain the email addresses, one per line. It is guaranteed that all of them are correct. All the given lines are distinct. The lengths of the addresses are from 3 to 100, inclusive.
Output
Print the number of groups k and then in k lines print the description of every group.
In the i-th line print the number of addresses in the group and all addresses that belong to the i-th group, separated by a space. It is allowed to print the groups and addresses in each group in any order.
Print the email addresses exactly as they were given in the input. Each address should go to exactly one group.
Sample Input
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：于所有的邮箱，都是由login@domain这样的形式构成，而且字符都是不区分大小写的。 我们有一种特殊类型的邮箱——@bmail.com，这种邮箱除了不区分大小写外—— 1，'@'之前的'.'，有等同于无 2，'@'之前的第一个'+'之后的字符可以忽略不计 然后其他字符相同的被认定为邮箱相同。 现在给你n（2e4）个邮箱，让你输出每个邮箱出现的次数与所有这个邮箱的原始串。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
const double eps = 1e-10;
const int inf =0x7f7f7f7f;
const double pi=acos(-1);
const int maxn=20000+100;
char s[maxn][105];
char s2[maxn][105];
map<string,int> mp;
vector<int> vec[maxn];
int num=0;
void init(int k)
{
for(int i=1;s[k][i]!=' ';i++)
if(s[k][i]>='A'&&s[k][i]<='Z') s2[k][i]=tolower(s[k][i]);
else s2[k][i]=s[k][i];
int p=1;for(;s[k][p]!='@';p++);
if(strcmp(s2[k]+p+1,"bmail.com")==0)
{
int flag=0,cnt=0;
for(int i=1;s[k][i]!=' ';i++)
if((s[k][i]=='.'||flag)&&i<p) continue;
else if(s[k][i]=='+') flag=1;
else s2[k][++cnt]=tolower(s[k][i]);
s2[k][cnt+1]=' ';
}
if(!mp[s2[k]+1])
{num++;
mp[s2[k]+1]=num;
vec[num].push_back(k);}
else {int kk=mp[s2[k]+1];vec[kk].push_back(k);}
}
int main()
{
int n;
while(~scanf("%d",&n))
{
MM(s,' ');MM(s2,' ');num=0;mp.clear();
for(int i=1;i<=n;i++)
{
scanf("%s",s[i]+1);
init(i);
}
printf("%d
",num);
for(int i=1;i<=num;i++)
{
printf("%d ",vec[i].size());
for(int j=0;j<vec[i].size();j++)
{
int k=vec[i][j];
printf("%s ",s[k]+1);
}
printf("
");
}
for(int i=1;i<=num;i++) vec[i].clear();
}
return 0;
}
分析：比赛时确实做出来了，不过用的哈希，而且并没有充分运用好字符串函数，导致写的时间久，
改进：
1.统计字符串个数不一定用哈希，还可以用map<string,int>；
2.比较字符串可以直接调用strcmp()函数；
3，字母小写转大写可以用toupper()，大写转小写可以用tolower();