[luogu3369/bzoj3224]普通平衡树(splay模板、平衡树初探)
解题关键:splay模板题整理。
如何不加入极大极小值?(待思考)
#include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cstdlib> using namespace std; typedef long long ll; const int N = 100005; int ch[N][2],par[N],val[N],cnt[N],size[N],ncnt,root; bool chk(int x){ return ch[par[x]][1]==x; } void pushup(int x){ size[x]=size[ch[x][0]]+size[ch[x][1]]+cnt[x];//和线段树不同的是需要+上自身的cnt } void rotate(int x){ int y=par[x],z=par[y],k=chk(x),w=ch[x][k^1]; ch[y][k]=w;par[w]=y; ch[z][chk(y)]=x;par[x]=z; ch[x][k^1]=y;par[y]=x; pushup(y);pushup(x); } void splay(int x,int goal=0){ while(par[x]!=goal){ int y=par[x],z=par[y]; if(z!=goal){ if(chk(x)==chk(y)) rotate(y); else rotate(x); } rotate(x); } if(!goal) root=x; } void insert(int x){ int cur=root,p=0; while(cur&&val[cur]!=x){ p=cur; cur=ch[cur][x>val[cur]]; } if(cur){ cnt[cur]++; }else{ cur=++ncnt; if(p) ch[p][x>val[p]]=cur; ch[cur][0]=ch[cur][1]=0; par[cur]=p;val[cur]=x; cnt[cur]=size[cur]=1; } splay(cur); } void find(int x){ int cur=root; if(!cur) return; while(ch[cur][x>val[cur]]&&x!=val[cur]){ cur=ch[cur][x>val[cur]]; } splay(cur); } //从0开始 int kth(int k){ k++; int cur=root; while(1){ if(ch[cur][0]&&k<=size[ch[cur][0]]){ cur=ch[cur][0]; }else if(k>size[ch[cur][0]]+cnt[cur]){ k-=size[ch[cur][0]]+cnt[cur]; cur=ch[cur][1]; }else{ return cur; } } } int rnk(int x){ find(x); if(val[root]>=x) return size[ch[root][0]]; else return size[ch[root][0]]+cnt[root]; } int pre(int x){ find(x); if(val[root]<x) return root; int cur=ch[root][0]; while(ch[cur][1]) cur=ch[cur][1]; return cur; } int succ(int x) { find(x); if(val[root]>x) return root; int cur=ch[root][1]; while(ch[cur][0]) cur=ch[cur][0]; return cur; } void remove(int x) { int last=pre(x),nxt=succ(x); splay(last);splay(nxt,last); int del=ch[nxt][0]; if(cnt[del]>1){ cnt[del]--; splay(del); } else ch[nxt][0]=0; } void init(){ insert(-2e9); insert(2e9); } int n,op,x; int main() { scanf("%d",&n); init(); while(n--){ scanf("%d%d",&op,&x); switch(op){ case 1: insert(x); break; case 2: remove(x); break; case 3: printf("%d\n",rnk(x)); break; case 4: printf("%d\n",val[kth(x)]); break; case 5: printf("%d\n",val[pre(x)]); break; case 6: printf("%d\n",val[succ(x)]); break; } } }