Misha and Changing Handles

傷城~ 2021-11-05 16:32 167阅读 0赞

Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.

Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.

Input

The first line contains integer *q* (1 ≤ *q* ≤ 1000), the number of handle change requests.

Next *q* lines contain the descriptions of the requests, one per line.

Each query consists of two non-empty strings *old* and *new*, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings *old* and *new* are distinct. The lengths of the strings do not exceed 20.

The requests are given chronologically. In other words, by the moment of a query there is a single person with handle *old*, and handle *new* is not used and has not been used by anyone.

Output

In the first line output the integer *n* — the number of users that changed their handles at least once.

In the next *n* lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, *old* and *new*, separated by a space, meaning that before the user had handle *old*, and after all the requests are completed, his handle is *new*. You may output lines in any order.

Each user who changes the handle must occur exactly once in this description.

Examples

Input

5
    Misha ILoveCodeforces
    Vasya Petrov
    Petrov VasyaPetrov123
    ILoveCodeforces MikeMirzayanov
    Petya Ivanov

Output

3
    Petya Ivanov
    Misha MikeMirzayanov
    Vasya VasyaPetrov123
    
    题意：给了n组数据，表示每个用户的旧密码与新密码，现在问每个用户更改密码是否成功。输出用户修改成功的旧密码与新密码。
    这里用map来解决问题，存入数据的时候，需要逆向思维，以新密码为key，旧密码为value，这样便于检查某个密码是否已经存在。

//以新密码为key，旧密码为value存入map，这样便于检查某个密码是否已经存在
    #include<cstdio>
    #include<iostream>
    #include<map>
    using namespace std;
    int main()
    {
        int n,m,cnt=0;
        map<string,string>mp;
        map<string,string>::iterator it;
        scanf("%d",&n);
        string a,b;
        for(int i=0;i<n;i++){
            cin>>a>>b;
            if(!mp.count(a)){
                mp[b]=a;
            }else{
          //如果该旧密码在map中已经存在了，则更新密码并删除旧密码
                mp[b]=mp[a];
                mp.erase(a);
            }
        }
        cout<<mp.size()<<endl;
        for(it=mp.begin();it!=mp.end();it++){
            cout<<it->second<<" "<<it->first<<endl;
        }
        return 0;
    }

转载于:https://www.cnblogs.com/LJHAHA/p/11179969.html