ll T;

while(~scanf("%d",&T)){

while(T--) {

= = ...

思路：

用秩合并，看了题解才发现 if(fx == fy)要输出当前集合的秩而不是0。。。

#include <cstdio>

#include <iostream>

#include <algorithm>

#include <string.h>

#include <vector>

#include <map>

#include <queue>

#include <math.h>

#include <string>

#include <set>

using namespace std;

#define mod 1000000007

#define ll int

#define N 100100

ll r[N], f[N];

ll find(ll x){return x==f[x]?x:f[x] = find(f[x]);}

ll Union(ll x, ll y){

	ll fx = find(x), fy = find(y);

	if(fx==fy)return r[fx];

	if(r[fx]>r[fy])

		swap(fx,fy);

	f[fx] = fy;

	r[fy] += r[fx];

	return r[fy];

}

ll n;

void init(){

	for(ll i = 0; i < N; i++) r[i] = 1;

	for(ll i = 0; i < N; i++) f[i] = i;

}

#define Word_Len 50500

#define Sigma_size 95

struct Trie{

	ll ch[Word_Len][Sigma_size];	 //Word_Len是字典树的节点数 若都是小写字母Sigma_size=26

	ll Have_word[Word_Len];		 //这个节点下有几个单词

	ll val[Word_Len];				 // 这个节点附带的信息。初始化为0表示这个节点不存在单词，所以节点带的信息必须!=0

	ll sz ;						 //当前节点数

	ll pre[Word_Len];

	char he[Word_Len];

	ll Newnode(){memset(ch[sz], 0, sizeof(ch[sz])); val[sz]=Have_word[sz]=0; return sz++;}

	void init()							 //初始化字典树

	{ sz = 0; Newnode();}//初始化

	ll idx(char c){return c-32;}	 //字符串编号

	ll insert(char *s){	 //把v数字加给 s单词最后一个字母

		ll u = 0, len = strlen(s);

		for(ll i = 0; i < len; i++){

			ll c = idx(s[i]);

			if(!ch[u][c])		     //节点不存在就新建后附加

			{

				he[sz] = s[i];

				val[sz] = val[u]+1;

				pre[sz] = u;

				ch[u][c] = Newnode();

			}

			u = ch[u][c];

			Have_word[u]++;

		}			//如今的u就是这个单词的最后一个位置

		return u;

	}

	ll find_word(char *s){

		ll u = 0, len = strlen(s);

		for(ll i = 0; i < len; i++){

			ll c = idx(s[i]);

			if(!ch[u][c])return 0;		//节点不存在

			u = ch[u][c];

		}

		return Have_word[u];

	}

	void Have_name(char *s, ll now){

		ll len = val[now];

		s[len--] = 0;

		ll cc = now;

		while(cc)

		{

			s[len--] = he[cc];

			cc = pre[cc];

		}

	}

};

Trie ac;

char a[1005], b[1005];

int main(){

	ll T;

	while(~scanf("%d",&T)){

		while(T--) {

			cin>>n;

			init();

			ac.init();

			while(n--)

			{

				cin>>a>>b;

				ll u, v;

				u = ac.insert(a);

				v = ac.insert(b);

				cout<<Union(u,v)<<endl;

			}

		}

	}

	return 0;

}