题目链接：这里
题意：按照时间顺序给了n对新旧字符串，其中如果满足a->b&&b->c那么最后只会存在a->c，然后让输出最后字符串对。
解法：水题，按照题意模拟就可以了，用map

//CF 501B

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1010;
map <string, string> mp;
int n;
string s1, s2;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> s1 >> s2;
        if(i == 0) mp[s1] = s2;
        else{
            int flag = 0;
            for(map <string, string> :: iterator it = mp.begin(); it != mp.end(); it++){
                if(it->second == s1){
                    mp[it->first] = s2;
                    flag = 1;
                }
            }
            if(flag == 0) mp[s1] = s2;
        }
    }
    cout << mp.size() << endl;
    for(map <string, string> :: iterator it = mp.begin(); it != mp.end(); it++){
        cout << it -> first << " " << it -> second << endl;
    }
    return 0;
}