【Codeforces Round #240 (Div. 1) 】E—Mashmokh's Designed Problem（Spaly）
题意：给一颗树，每个节点的儿子之间有先后关系
支持询问2点距离，给某个子树换父亲，询问最后一个深度为k的节点
考虑用维护括号序
由于深度变化是连续的，就只需要维护最大最小值
第一个操作求区间最小值，第三个判断k是否在最大最小值区间内即可
注意由于有哨兵节点，所以根的深度设为了1
#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
static char ibuf[RLEN],*ib,*ob;
(ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
char ch=gc();
int res=0,f=1;
while(!isdigit(ch))f^=ch=='-',ch=gc();
while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
const int mod=998244353,g=3;
inline int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
inline void Add(int &a,int b){a=add(a,b);}
inline int dec(int a,int b){return a>=b?a-b:a-b+mod;}
inline void Dec(int &a,int b){a=dec(a,b);}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,a=mul(a,a))(b&1)?(res=mul(res,a)):0;return res;}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=400005;
int n,m;
int dfn,in[N],out[N],idx[N];
vector<int> e[N];
namespace Splay{
int rt,fa[N],son[N][2],mn[N],mx[N],val[N],tag[N];
#define lc(u) son[u][0]
#define rc(u) son[u][1]
inline bool isrc(int u){
return rc(fa[u])==u;
}
inline void addnode(int u,int v){
mx[u]=mn[u]=val[u]=v;
}
inline void pushup(int u){
mx[u]=mn[u]=val[u];
if(lc(u))chemx(mx[u],mx[lc(u)]),chemn(mn[u],mn[lc(u)]);
if(rc(u))chemx(mx[u],mx[rc(u)]),chemn(mn[u],mn[rc(u)]);
}
inline void rotate(int v){
int u=fa[v],z=fa[u];
int t=rc(u)==v;
fa[v]=z,son[z][isrc(u)]=v;
son[u][t]=son[v][t^1],fa[son[v][t^1]]=u;
son[v][t^1]=u,fa[u]=v;
pushup(u),pushup(v);
}
inline void pushnow(int u,int v){
mx[u]+=v,mn[u]+=v,val[u]+=v,tag[u]+=v;
}
inline void pushdown(int u){
if(!tag[u])return;
if(lc(u))pushnow(lc(u),tag[u]);
if(rc(u))pushnow(rc(u),tag[u]);
tag[u]=0;
}
int stk[N],top;
inline void splay(int v,int goal){
stk[top=1]=v;
for(int x=v;fa[x];x=fa[x])stk[++top]=fa[x];
for(int i=top;i;i--)pushdown(stk[i]);
while(fa[v]!=goal){
int u=fa[v];
if(fa[u]!=goal)
isrc(u)==isrc(v)?rotate(u):rotate(v);
rotate(v);
}
if(!goal)rt=v;
}
inline void build(int &r1,int l,int r){
if(l>r){r1=0;return;}
if(l==r){r1=l;return;}
r1=(l+r)>>1;
build(lc(r1),l,r1-1);
if(lc(r1))fa[lc(r1)]=r1;
build(rc(r1),r1+1,r);
if(rc(r1))fa[rc(r1)]=r1;
pushup(r1);
}
inline int pre(int u){
splay(u,0),u=lc(u);
while(rc(u))u=rc(u);
splay(u,0);return u;
}
inline int suf(int u){
splay(u,0),u=rc(u);
while(lc(u))u=lc(u);
splay(u,0);return u;
}
inline int find(int u,int va){
while(u){
pushdown(u);
if(mx[rc(u)]>=va&&va>=mn[rc(u)])u=rc(u);
else if(val[u]==va)break;
else u=lc(u);
}
return u;
}
inline int dis(int u,int v){
u=in[u],v=in[v];
splay(u,0),splay(v,u);
int t=min(val[u],val[v]);
if(lc(u)==v&&rc(v))chemn(t,mn[rc(v)]-1);
if(rc(u)==v&&lc(v))chemn(t,mn[lc(v)]-1);
return val[u]+val[v]-2*t;
}
inline void gofa(int u,int h){
splay(in[u],0),splay(1,in[u]);
int goal=idx[find(1,val[in[u]]-h)],pr=pre(in[u]),nxt=suf(out[u]);
splay(pr,0),splay(nxt,pr);
int pp=lc(nxt);lc(nxt)=fa[pp]=0;
pushup(nxt),pushup(pr);
nxt=out[goal];
pr=pre(nxt);
splay(pr,0),splay(nxt,pr);
lc(nxt)=pp,fa[pp]=nxt;
pushnow(pp,1-h);
pushup(nxt),pushup(pr);
}
}
inline void dfs(int u,int dep){
in[u]=++dfn,idx[dfn]=u,Splay::addnode(dfn,dep);
for(int i=0;i<e[u].size();i++){
int v=e[u][i];
dfs(v,dep+1);
}
out[u]=++dfn,idx[dfn]=u,Splay::addnode(dfn,dep);
}
int main(){
n=read(),m=read();
for(int i=1;i<=n;i++){
int l=read();
for(int j=1;j<=l;j++)e[i].pb(read());
}
dfn++,dfs(1,1),dfn++;
Splay::build(Splay::rt,1,dfn);
while(m--){
int op=read();
if(op==1){
int u=read(),v=read();
cout<<Splay::dis(u,v)<<'\n';
}
if(op==2){
int u=read(),h=read();
Splay::gofa(u,h);
}
if(op==3){
int k=read();
cout<<idx[Splay::find(Splay::rt,k+1)]<<'\n';
}
}
}
