题目意思就是，road命令连接第A坐标和第B坐标的点。而line则是查看纵坐标C拉出的扫描

线过几个联通块，而联通块中的数量和是多少。

（具体的可以查看原题）

题解：其实是线段树一个简单的模拟，即并查集找联通块和联通块中点的数量，然后每次更

新删除A点和B点的信息，重新建立A点和B点连接后的信息就行了。

#include<iostream>
#include<cstring>
#include<sstream>
#include<string>
#include<cstdio>
#include<cctype>
#include<vector>
#include<queue>
#include<cmath>
#include<stack>
#include<list>
#include<set>
#include<map>
#include<algorithm>
#define fi first
#define se second
#define MP make_pair
#define P pair<int,int>
#define PLL pair<ll,ll>
#define Sca(x) scanf("%d",&x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x)
#define Scl2(x,y) scanf("%lld%lld",&x,&y)
#define Scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define Pri(x) printf("%d\n",x)
#define Prl(x) printf("%lld\n",x)
#define For(i,x,y) for(int i=x;i<=y;i++)
#define _For(i,x,y) for(int i=x;i>=y;i--)
#define FAST_IO std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define ll long long
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;
const double Pi = acos(-1.0);
using namespace std;
template <class T>void tomax(T&a,T b){ a=max(a,b); } 
template <class T>void tomin(T&a,T b){ a=min(a,b); }
const int N=1e5+5;
struct Segt{
    #define lc (p<<1)
    #define rc (p<<1|1)
    #define MID (tree[p].l+tree[p].r)>>1
    struct Tree{
        int l,r;
        int uck,val,lz[2];
        inline void init(int left,int right){ l=left; r=right; uck=0; val=0; lz[0]=0; lz[1]=0; }
        void updata(int uckv,int valv){ uck+=uckv; val+=valv; lz[0]+=uckv; lz[1]+=valv; }
    }tree[N<<2|1];
    void pushdown(int p){  tree[p].uck=tree[lc].uck+tree[rc].uck;  tree[p].val=tree[lc].val+tree[rc].val; }
    void pushup(int p){
        int one=tree[p].lz[0],two=tree[p].lz[1];
        tree[lc].updata(one,two);
        tree[rc].updata(one,two);
        tree[p].lz[0]=0; tree[p].lz[1]=0;
    }
    void build(int l,int r,int p){
        tree[p]=Tree{l,r,0,0,0,0};
        if(l==r) return ;
        int mid=MID;
        build(l,mid,lc);
        build(mid+1,r,rc);
    }
    void updata(int l,int r,int p,int uckv,int valv){
        if(l>r) return ;
        if(tree[p].l>=l&&tree[p].r<=r){ tree[p].updata(uckv,valv); return ; }
        if(tree[p].lz[0] || tree[p].lz[1]) pushup(p);
        int mid=MID;
        if(l<=mid) updata(l,r,lc,uckv,valv);
        if(r>mid) updata(l,r,rc,uckv,valv);
        pushdown(p);
    }
    P getans(int pos,int p){
        if(tree[p].l==pos && tree[p].r==pos) return MP(tree[p].uck,tree[p].val);
        if(tree[p].lz[0] || tree[p].lz[1]) pushup(p);
        int mid=MID;
        if(pos<=mid) return getans(pos, lc);
        if(pos>mid) return getans(pos, rc);
    }
}t;
struct Up{ int minn,maxx; } up[N];
int num[N],tmp[N],uck[N],ans[N];
int len=0;
int find(int x){ return uck[x]==x ? x : uck[x]=find(uck[x]); }
void merge(int x,int y){
    int t1(find(x)),t2(find(y));
    if(t1!=t2){
        t.updata(up[t1].minn+1,up[t1].maxx,1,-1,-ans[t1]);
        t.updata(up[t2].minn+1,up[t2].maxx,1,-1,-ans[t2]);
        uck[t2]=t1;
        ans[t1]+=ans[t2];
        tomax(up[t1].maxx,up[t2].maxx); tomin(up[t1].minn,up[t2].minn);
        t.updata(up[t1].minn+1,up[t1].maxx,1,1,ans[t1]);
    }
}
inline void dis(int n){
    sort(tmp+1,tmp+1+n);
    len=unique(tmp+1,tmp+1+n) - tmp - 1;
    For(i,1,n) num[i]=lower_bound(tmp+1,tmp+1+len,num[i]) - tmp;
}
inline void init(int n){
    dis(n);
    For(i,1,n){ uck[i]=i; ans[i]=1; up[i]=Up{num[i], num[i]}; }
}
int main(){
    int T; Sca(T);
    while(T--){
        int n; Sca(n);
        For(i,1,n){ int x; Sca(x); Sca(num[i]); tmp[i]=num[i]; }
        init(n);
        t.build(1,len,1);
        int m; Sca(m);
        while(m--){
            char cmd[10]; scanf("%s",cmd);
            if(cmd[0]=='r'){ int x,y; Sca2(x,y); x++; y++; merge(x,y); }
            else if(cmd[0]=='l'){
                double x; scanf("%lf",&x);
                x+=0.5;
                x = lower_bound(tmp+1,tmp+1+len,x) - tmp;
                if(x==len+1) puts("0 0");
                else{
                    P out = t.getans(x,1);
                    printf("%d %d\n",out.fi,out.se);
                }
            }
        }
    }
}