/*51 nod 1515 明辨是非(并查集合并)problem:两种操作：x y 1: 如果第x,第y个数可以相同,则输出YES,并令他们相同. 否则输出NOx y 0: 如果第x,第y个数可以不相同 ......solve:相同可以用并查集来维护.  但是不同则不行, 如果a,b不同, b,c不同.但是a,c可以相同. 开始脑子抽了都用并查集 卒...先set记录一下每个数与其不同的数有哪些. 然后判断两个数是否不相等时直接进行查找.并要判断他们各自所在的并查集合并的时候再把set处理一下就好. printf一直TL,换成putsAC.hhh-2016/09/04-17:18:40*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <vector>#include <math.h>#include <queue>#include <set>#include <map>#define lson  i<<1#define rson  i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define scanfi(a) scanf("%d",&a)#define scanfs(a) scanf("%s",a)#define scanfl(a) scanf("%I64d",&a)#define scanfd(a) scanf("%lf",&a)#define key_val ch[ch[root][1]][0]#define eps 1e-7#define inf 0x3f3f3f3f3f3f3f3fusing namespace std;const ll mod = 1000000007;const int maxn = 200010;const double PI = acos(-1.0);struct node{    int x,y,kind;} qry[maxn];int id[maxn];set<int>q[maxn];map<int,int> mp;set<int>::iterator it;int par[maxn];int fin(int x){    if(par[x] == x) return x;    return par[x] = fin(par[x]);}void unio(int x,int y){    if(x == y)        return ;    if(q[x].size() > q[y].size())    {        int t = x;        x = y;         y = t ;    }    par[x] = y;    for(it = q[x].begin(); it != q[x].end(); it++)    {        q[y].insert(*it);    }}int main(){//    freopen("in.txt","r",stdin);//    freopen("out.txt","w",stdout);    int n;    scanfi(n);    int cnt = 0;    for(int i = 1; i <= n; i++)    {        par[i] = i;        scanfi(qry[i].x),scanfi(qry[i].y);        scanfi(qry[i].kind);        id[cnt++] = qry[i].x,id[cnt++] = qry[i].y;    }    sort(id,id+cnt);    int total = unique(id,id+cnt)-id;    for(int i = 0; i < total; i++)        mp[id[i]] = i;//        cout << total <<endl;//        for(int i = 0;i <= total;i++)//            cout <<id[i] <<" " ;//        cout <<endl;    for(int i =1 ; i <= n; i++)    {//            cout << qry[i].x <<" " <<qry[i].y << " "<<qry[i].kind <<endl;        int x = mp[qry[i].x];        int y =  mp[qry[i].y];        int tx = fin(x),ty = fin(y),flag =0;        if(qry[i].kind == 1)        {//                cout << x <<" " <<y <<endl;            if(q[tx].size() > q[ty].size())            {                it = q[ty].find(tx);                if(it != q[ty].end())                    flag = 1;                for(it = q[ty].begin(); !flag && it != q[ty].end(); it++)                {                    if(fin(*it) == tx)                    {                        flag = 1;                        break;                    }                }            }            else            {                it = q[tx].find(ty);                if(it != q[tx].end())                    flag = 1;                for(it = q[tx].begin(); !flag &&it != q[tx].end(); it++)                {                    if(fin(*it) == ty)                    {                        flag = 1;                        break;                    }                }            }            if(flag)                puts("NO");            else            {                puts("YES");                unio(tx,ty);            }        }        else        {            if(tx == ty)                puts("NO");            else            {                puts("YES");                q[tx].insert(ty);                q[ty].insert(tx);            }        }    }    return 0;}