HDU 3172 Virtual Friends 并与正确集中检查 -秩

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;
}