BZOJ 5005 & JZOI 3959 鸡腿の乒乓

SOL： 我们发现两个有交且不相互覆盖的区间可以互相到达，我们用并查集把这样的区间缩成一个大区间，用线段树维护即可。

#pragma GCC optimize("-O2")
#include<bits/stdc++.h>
#define pb push_back
#define Mid (l+r>>1)
#define ls now<<1,l,Mid
#define rs now<<1|1,Mid+1,r
#define N 300015
using namespace std;
int A[N],l[N],r[N],L[N],R[N],f[N],oz,n,op[N];
struct qr{vector<int> c;}tr[N*2];
int tot=0;
#define sight(x) ('0'<=x&&x<='9')
inline void read(int &x){
    static char c; static int b;
    for (b=1,c=getchar();!sight(c);c=getchar())if (c=='-') b=-1;
    for (x=0;sight(c);c=getchar())x=x*10+c-48; x*=b;
}
void write(int x){if (x<10) {putchar('0'+x); return;} write(x/10); putchar('0'+x%10);}
inline void writeln(int x){ if (x<0) putchar('-'),x*=-1; write(x); putchar('\n'); }
inline void writel(int x){ if (x<0) putchar('-'),x*=-1; write(x); putchar(' '); }
void Li() {
    sort(A+1,A+A[0]+1);
    for (int i=1;i<=n;i++){
        if (op[i]==2) continue;
     l[i]=upper_bound(A+1,A+A[0]+1,l[i])-A,
     r[i]=upper_bound(A+1,A+A[0]+1,r[i])-A; }
}
int gf(int x){
    return x^f[x]?f[x]=gf(f[x]):x;
}
void Del(int now,int l,int r,int x){
    if (tr[now].c.size()) {
        int id;
        for (int i=0;i<tr[now].c.size();i++) {
            id=tr[now].c[i];
            f[gf(id)]=oz;
            L[oz]=min(L[oz],L[id]); 
            R[oz]=max(R[oz],R[id]);
        }
        tr[now].c.clear();
    }
    if (l==r) return;
    if (x<=Mid) Del(ls,x); else Del(rs,x);
}
void up(int now,int l,int r,int L,int R){
    if (L<=l&&r<=R) {
        tr[now].c.pb(oz); return;
    }
    if (L<=Mid) up(ls,L,R);
    if (R> Mid) up(rs,L,R);
}
signed main () {
    read(n);
    for (int i=1;i<=n;i++) {
        read(op[i]),read(l[i]); read(r[i]);
        if (op[i]==1) {
            A[++A[0]]=l[i],A[++A[0]]=r[i];
        }
    }
    Li(); 
    for (int t=1;t<=n;t++) {
        if (op[t]==1) {
            L[++oz]=l[t],R[oz]=r[t];
            f[oz]=oz;
            Del(1,1,A[0],l[t]); Del(1,1,A[0],r[t]);
            up(1,1,A[0],L[oz]+1,R[oz]-1);
        } else {
          int y=gf(l[t]),x=gf(r[t]);
          if (x==y||(L[x]<L[y]&&L[y]<R[x])||(L[x]<R[y]&&R[y]<R[x]))
            puts("YES");
            else puts("NO"); 
        }
    } return 0;
}