這道題要求區間反轉。。。好東西。。

對於Splay：把l-1旋到根，把r+1旋到根的右兒子，這樣r+1的左兒子就是整個區間了，然後對這個區間打個tg

註意要插-Inf和Inf到樹裏面，防止越界，坐標要+1

#include<cstdio>
#include<iostream>
#define R register int
using namespace std;
const int N=100010,Inf=0x3f3f3f3f;
inline int g() {
    R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch==‘
 
-‘?-1:fix;
    do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
}
int n,m,tot,rt;
struct node{
    int fa,ch[2],sz,tg,vl;
    #define fa(x) t[x].fa
    #define ch(x,i) t[x].ch[i]
    #define sz(x) t[x].sz
    #define tg(x) t[x].tg
    #define vl(x) t[x].vl
    #define ls ch(x,0)
    #define
 
 rs ch(x,1)
}t[N];
inline void upd(int x) {sz(x)=sz(ls)+sz(rs)+1;}
inline void spread(int x) {if(tg(x)) tg(ls)^=1,tg(rs)^=1,tg(x)=0,swap(ls,rs);}
inline void rot(int x) {
    R y=fa(x),d=ch(y,1)==x;
    if(fa(y)) ch(fa(y),ch(fa(y),1)==y)=x;
    fa(x)=fa(y); fa(ch(y,d)=ch(x,d^1))=y;
    fa(ch(x,d
 
^1)=y)=x; upd(y);
}
inline void Splay(int x,int f) {
    while(fa(x)!=f) {
        R y=fa(x); if(fa(y)!=f) 
            rot((ch(y,1)==x)==(ch(fa(y),1)==y)?y:x);
        rot(x);
    } upd(x); if(!f) rt=x;
}
inline int build(int f,int l,int r) {
    if(l>r) return 0; R md=l+r>>1,x=++tot;
    fa(x)=f,++sz(x),vl(x)=md-1;
    ls=build(x,l,md-1);rs=build(x,md+1,r);
    upd(x); return x;
}
inline int get(int pos) { R x=rt;
    while(1) {spread(x); R s=sz(ls); 
        if(pos==s+1) return x;
        if(pos<=s) x=ls;
        else x=rs,pos-=(s+1);
    }
}
inline void reverse(int l,int r) {
    l=get(l),r=get(r+2); 
    Splay(l,0),Splay(r,l); tg(ch(r,0))^=1;
} int s; 
inline void print(int x) { spread(x);
    if(!x) return ; print(ls); 
    if(vl(x)>=1&&vl(x)<=n) printf("%d ",vl(x)); print(rs);
} 
signed main() {
    n=g(),m=g(); rt=build(0,1,n+2);
    for(R i=1;i<=m;++i) {R l=g(),r=g(); reverse(l,r);}
    print(rt); putchar(‘\n‘);
}

對於FHQ Treap，先把[1,l-1]和[l,n]split出來，再把[l,r]和[r+1,n]split出來，在代表區間的子樹的根節點打tg，註意這道題的的split是按rank的

不過split完忘了upd。。。我真實沙雕qwq。。。

#include<cstdio>
#include<iostream>
#include<cstdlib>
#define R register int
#define ls(x) ch[x][0]
#define rs(x) ch[x][1]
using namespace std;
const int N=100010;
inline int g() {
    R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch==‘-‘?-1:fix;
    do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
}
int n,m,tot,rt;
int ch[N][2],sz[N],vl[N],dat[N];
bool tg[N];
inline void upd(int x) {sz[x]=sz[ls(x)]+sz[rs(x)]+1;}
inline int cre(int v) {R x=++tot; vl[x]=v,sz[x]=1,dat[x]=rand(); upd(x); return x;}
inline void spread(int x) {if(tg[x]) swap(ls(x),rs(x)),tg[ls(x)]^=1,tg[rs(x)]^=1,tg[x]=0;}
inline void split(int o,int rk,int& x,int& y) {
    if(!o) {x=y=0; return ;} spread(o);
    if(sz[ls(o)]<rk) {x=o; split(rs(o),rk-sz[ls(o)]-1,rs(o),y); upd(x);}
    else {y=o; split(ls(o),rk,x,ls(o)); upd(y);}
}
inline int merge(int x,int y) {
    if(!x||!y) return x+y;
    if(dat[x]<dat[y]) {spread(x); rs(x)=merge(rs(x),y); upd(x); return x;}
    else {spread(y); ls(y)=merge(x,ls(y)); upd(y); return y;}
}
inline void print(int x) {if(!x) return ; spread(x); print(ls(x)); printf("%d ",vl[x]); print(rs(x));}
signed main() { srand(100023323);
    n=g(),m=g(); for(R i=1;i<=n;++i) rt=merge(rt,cre(i)); R x=0,y=0,z=0;
    for(R i=1;i<=m;++i) { R l=g(),r=g();
        split(rt,l-1,x,y); split(y,r-l+1,y,z); tg[y]^=1; merge(x,merge(y,z));
    } print(rt); while(1);
}

2019.05.06

Luogu P3391 文藝平衡樹（Splay or FHQ Treap）