【LOJ #6066】「2017 山东一轮集训 Day3」第二题（二分答案 / 树哈希 / 括号序列）

传送门

首先显然二分答案

其实我第一眼想得长链剖分维护树哈希

实际上由于这个子树有先后顺序
于是可以看做括号序列

每个点的$k$子树就是若干区间
哈希判一下即可

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool 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 ull unsigned long long
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=1e9+7;
ull bas=137;
inline int add(int a,int b){return (a+=b)>=mod?(a-mod):a;}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return (r>=mod)?(r%mod):r;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=(r>=mod)?(r%mod):r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
inline int fix(int x){return (x<0)?x+mod:x;}
cs int N=100005;
ull pw[N*2];
vector<int> e[N];
int fa[N][17],n,mxdep[N];
int in[N],id[N*2],tot,out[N];
ull has[N*2];
void dfs1(int u){
	for(int i=1;i<=16;i++)fa[u][i]=fa[fa[u][i-1]][i-1];
	in[u]=++tot,id[tot]=1;
	mxdep[u]=0;
	for(int &v:e[u]){
		fa[v][0]=u,dfs1(v);
		chemx(mxdep[u],mxdep[v]+1);
	}
	out[u]=++tot,id[tot]=mod-1;
}
map<ull,int>vis;
vector<int> p[N];
inline int jump(int u,int k){
	for(int i=16;~i;i--)if(k&(1<<i))u=fa[u][i];
	return u;
}
inline bool comp(int a,int b){return in[a]<in[b];}
inline ull calc(int l,int r){return has[r]-has[l-1]*pw[r-l+1];}
inline bool check(int k){
	for(int i=1;i<=n;i++)
	p[jump(i,k+1)].pb(i);
	bool fg=0;
	for(int i=1;i<=n;i++)if(mxdep[i]>=k){
		sort(p[i].bg(),p[i].end(),comp);
		ull res=0,l=in[i];
		for(int &v:p[i]){
			if(l<in[v])
			res=res*pw[in[v]-l]+calc(l,in[v]-1);
			l=out[v]+1;
		}
		res=res*pw[out[i]-l+1]+calc(l,out[i]);
		if(vis.count(res)){fg=1;break;}
		else vis[res]=1;
	}
	for(int i=1;i<=n;i++)p[i].clear();
	vis.clear();
	return fg;
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read();
	pw[0]=1;
	for(int i=1;i<=2*n;i++)pw[i]=pw[i-1]*bas;
	for(int i=1;i<=n;i++){
		int x=read();
		while(x--)e[i].pb(read());
	}
	dfs1(1);
	for(int i=1;i<=tot;i++)has[i]=has[i-1]*bas+id[i];
	int l=1,r=n,res=0;
	while(l<=r){
		int mid=(l+r)>>1;
		if(check(mid))l=mid+1,res=mid;
		else r=mid-1;
	}
	cout<<res<<'
';
	return 0;
}