传送门

题意：

实现如下操作：

思路：

fhq-treap板子辣，不要被他的编号误导了，我们还是需要按照排名来分裂，从上到下标号$1-n$即可，注意在按照排名分裂的时候$findrank$函数与按照权值分裂是不同的，因为他不满足二叉树的性质，那么怎么得到排名呢？可以发现其中序遍历在他前面遍历的个数$+1$就是他的排名，所以我们往上暴跳记录一下有多少在他前面就好了，复杂度$O(logn)$。
其他的都是一些基本操作了，代码很好写，基本没bug。

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,m,x,y,z,h,k,root,tot;
int p[N],has[N];
struct Node {int l,r;int rank,val,id,size;int fa;
}tr[N<<2];void pushup(int u) {tr[u].size=tr[tr[u].l].size+tr[tr[u].r].size+1;tr[tr[u].l].fa=u; tr[tr[u].r].fa=u;
}int newnode(int id,int val) {int u=++tot;tr[u]={0,0,rand(),val,id,1};return u;
}void split(int u,int k,int &x,int &y) {if(!u) { x=y=0; return; }if(k<=tr[tr[u].l].size) y=u,split(tr[u].l,k,x,tr[u].l);else x=u,split(tr[u].r,k-tr[tr[u].l].size-1,tr[u].r,y);pushup(u);
}int merge(int u,int v) {if(!u||!v) return u+v;if(tr[u].rank<tr[v].rank) {tr[u].r=merge(tr[u].r,v);pushup(u);return u;}else {tr[v].l=merge(u,tr[v].l);pushup(v);return v;}
}int findrank(int v) {int ans=tr[v].size-tr[tr[v].r].size;while(v!=root) {if(v==tr[tr[v].fa].r) ans+=tr[tr[v].fa].size-tr[v].size;v=tr[v].fa;}return ans;
}int findsize(int k) {int u=root;while(1) {if(tr[tr[u].l].size>=k) u=tr[u].l;else if(tr[tr[u].l].size+1==k) return tr[u].val;else k-=tr[tr[u].l].size+1,u=tr[u].r;}
}int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);scanf("%d%d",&n,&m);for(int i=1;i<=n;i++) scanf("%d",&p[i]),root=merge(root,newnode(i,p[i])),has[p[i]]=i;while(m--) {char op[10]; int xx;scanf("%s%d",op,&xx);if(op[0]=='T') {int id=has[xx];int rank=findrank(id);split(root,rank,x,y);split(x,rank-1,x,z);root=merge(merge(z,x),y);}else if(op[0]=='B') {int id=has[xx];int rank=findrank(id);split(root,rank,x,y);split(x,rank-1,x,z);root=merge(merge(x,y),z);}else if(op[0]=='I') {int yy; scanf("%d",&yy);if(!yy) continue;int id=has[xx];int rank=findrank(id);if(yy>0) {split(root,rank,x,y); split(x,rank-1,x,z);split(y,1,y,h); root=merge(merge(merge(x,y),z),h);}else {split(root,rank,x,y); split(x,rank-1,x,z);split(x,rank-2,x,h); root=merge(merge(merge(x,z),h),y);}}else if(op[0]=='A') {int id=has[xx];int rank=findrank(id);printf("%d\n",rank-1);} else {printf("%d\n",findsize(xx));}}return 0;
}
/**/