懶得復制，戳我戳我

Solution:

• 還是一個\(Splay\)，我們只用多存一個值\(rad\)來維護二叉樹，然後用數組存下每個書對應的值是多少
• \(Top\)操作，我是把\(s\)旋轉到根節點，然後刪除，將\(s\)對應的\(rad\)值調至最小，然後插入就可以
• \(Bottom\)操作，就是和\(Top\)相反，刪除後把\(rad\)調整至最大，然後插入
• \(Insert\)操作，\(s==0\)不用操作，\(-1\)時，就只用把\(s\)旋轉到根節點，把左子樹最大值旋轉至左子樹根節點，然後交換兩個節點信息，並且交換數組儲存的每個書對應的\(rad\)，\(1\)就是相反的旋轉右子樹最小值就可以
• \(Ask\)操作，就只用把\(s\)旋轉到根節點，輸出左子樹大小
• \(Query\)操作，就是找第\(s\)大的\(rad\)值對應的書籍編號就可以了
• 一開始想怎麽找編號好久，發現編號是\(1-n\)就好操作了

Code:

//It is coded by Ning_Mew on 4.12
#include<bits/stdc++.h>
#define ls(x) node[x].ch[0]
#define rs(x) node[x].ch[1]
#define fa(x) node[x].fa
#define root node[0].ch[1]
using namespace std;

const
 
 int maxn=8e4+10,INF=1e9+7;

int n,m,tot,vall[maxn],low=0,up=0;
struct Node{
  int fa,ch[2],val,num,size;//num-> bian hao   val->rand()
}node[maxn];

void update(int x){node[x].size=node[ls(x)].size+node[rs(x)].size+1;}
void connect(int x,int fa,int how){node[x].fa=fa;node[fa].ch[how]=x;}
int ident(int
 
 x){return x==node[fa(x)].ch[0]?0:1;}
void rorate(int x){
  int Y=fa(x),R=fa(Y);int Yson=ident(x),Rson=ident(Y);
  connect(node[x].ch[Yson^1],Y,Yson);
  connect(Y,x,Yson^1);
  connect(x,R,Rson);
  update(Y);update(x);
}
void Splay(int x,int goal){
  int to=fa(goal);
  while(fa(x)!=to){
    if(fa(fa(x))==to)rorate(x);
    else if(ident(x)==ident(fa(x)))rorate(fa(x)),rorate(x);
    else rorate(x),rorate(x);
  }
}
int newnode(int x,int rad,int fa){node[++tot].val=rad;node[tot].num=x;node[tot].size=1;node[tot].fa=fa;return tot;}
void ins(int x,int rad){
  //cout<<x<<‘ ‘<<rad<<endl;
  int now=root;
  if(!root){newnode(x,rad,0);root=tot;return;}
  else{
    while(1){
      node[now].size++;
      int nxt=rad<node[now].val?0:1;
      if(!node[now].ch[nxt]){
    int p=newnode(x,rad,now);
    node[now].ch[nxt]=p;Splay(p,root);return;
      }
      now=node[now].ch[nxt];
    }
  }return;
}
int find(int x,int rad){
  int now=root;
  while(1){
    if(!now)return 0;
    if(node[now].val==rad){Splay(now,root);return now;}
    int nxt=rad<node[now].val?0:1;
    now=node[now].ch[nxt];
  }
}
void del(int x,int rad,int opt,int sum){//1->del 2->swap
  int pos=find(x,rad);
  if(!pos)return;
  if(opt==1){
    if(!ls(pos)&&!rs(pos)){root=0;return;}
    if(!ls(pos)){root=rs(pos);fa(root)=0;return;}
    else{
      int left=ls(pos);
      while(rs(left))left=rs(left);
      Splay(left,ls(pos));
      connect(rs(pos),left,1);
      connect(left,0,1);update(left);
    }
  }else{
    if(sum==0)return;
    if(sum==-1&&!ls(pos))return;
    if(sum==1&&!rs(pos))return;
    if(sum==-1){
      int left=ls(pos);
      while(rs(left))left=rs(left);
      Splay(left,ls(pos));
      swap(vall[ node[pos].num ],vall[ node[left].num ]);
      swap(node[pos].num,node[left].num);
    }else{
      int right=rs(pos);
      while(ls(right))right=ls(right);
      Splay(right,rs(pos));
      swap(vall[ node[pos].num ],vall[ node[right].num ]);
      swap(node[pos].num,node[right].num);
    }
  }
}
int rak(int x,int rad){int pos=find(x,rad);return node[ls(pos)].size;}
int kth(int x,int rad){
  int now=root;
  while(1){
    if(x-node[ls(now)].size==1){
      Splay(now,root);return node[now].num;}
    int nxt=x<=node[ls(now)].size?0:1;
    if(nxt){x=x-(node[ls(now)].size+1);}
    now=node[now].ch[nxt];
  }
}
string s;
int main(){
  scanf("%d%d",&n,&m);
  int x,t;
  for(int i=1;i<=n;i++){
    scanf("%d",&x);
    ins(x,++up);vall[x]=up;
  }
  //cout<<"ins finished"<<endl;
  for(int i=1;i<=m;i++){
    cin>>s;scanf("%d",&x);
    if(s=="Top"){
      del(x,vall[x],1,0);vall[x]=--low;
      ins(x,vall[x]);
    }
    if(s=="Bottom"){
      del(x,vall[x],1,0);vall[x]=++up;
      ins(x,vall[x]);
    }
    if(s=="Insert"){
      scanf("%d",&t); del(x,vall[x],2,t);
    }
    if(s=="Ask"){printf("%d\n",rak(x,vall[x]));}
    if(s=="Query"){printf("%d\n",kth(x,vall[x]));}
  }
  return 0;
}

【題解】 [ZJOI2006]書架 （Splay）