分析：

人名字符串用map映射一下就可以了，这个是一个存储int的并查集，有优化。

代码：

#include <iostream>
#include <vector>
#include <map>
using namespace std;

class UnionFindSets
{
public:
	UnionFindSets();
	UnionFindSets(int n);
	void link(int x, int y);
	bool isLink(int x, int y);
private:
	vector<int>fa;
};
UnionFindSets::UnionFindSets() {
	fa.clear();
}
UnionFindSets::UnionFindSets(int n) {
	fa.clear();
	for (int i = 0; i <= n; i++) {
		fa.push_back(i);
	}
}
void UnionFindSets::link(int x, int y) {
	while (x != fa[x]) {
		fa[x] = fa[fa[x]];
		x = fa[x];
	}
	while (y != fa[y]) {
		fa[y] = fa[fa[y]];
		y = fa[y];
	}
	fa[x] = y;
}
bool UnionFindSets::isLink(int x, int y) {
	while (x != fa[x]) {
		fa[x] = fa[fa[x]];
		x = fa[x];
	}
	while (y != fa[y]) {
		fa[y] = fa[fa[y]];
		y = fa[y];
	}
	return x == y;
}
map<string, int>mp;
int main()
{
	int T, n = 0;
	cin >> T;
	UnionFindSets ans = UnionFindSets(2e5 + 10);
	while (T--) {
		int op;
		string a, b;
		cin >> op >> a >> b;
		if (op == 0) {
			if (mp.count(a) == 0) {
				mp[a] = n++;
			}
			if (mp.count(b) == 0) {
				mp[b] = n++;
			}
			ans.link(mp[a], mp[b]);
		}
		else {
			if (mp.count(a) == 0) {
				mp[a] = n++;
			}
			if (mp.count(b) == 0) {
				mp[b] = n++;
			}
			if (ans.isLink(mp[a], mp[b]) == true) {
				cout << "yes" << endl;
			}
			else {
				cout << "no" << endl;
			}
		}
	}
}