「题解」Codeforces 825G Tree Queries

点权转边权，把边权设为两个端点的 \(\min\)，然后发现询问 \(x\) 的答案，就是询问 \(x\) 与所有黑点的虚树，边权的 \(\min\) 是多少。假设要判定答案是否 \(\geq k\)，那么就是询问 \(x\) 只经过 \(\geq k\) 是否能到达所有黑点，于是想到建立 Kruskal 重构树，那么 \(x\) 与所有黑点的 LCA 的权值即为答案。时间复杂度为 \(\mathcal{O}(n+q\log n)\)．

严格比 CF1628E 弱，没救了。。。

#include<cstdio>
#include<vector>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<ctime>
#include<random>
#include<assert.h>
#define pb emplace_back
#define mp make_pair
#define fi first
#define se second
#define dbg(x) cerr<<"In Line "<< __LINE__<<" the "<<#x<<" = "<<x<<'\n'
#define dpi(x,y) cerr<<"In Line "<<__LINE__<<" the "<<#x<<" = "<<x<<" ; "<<"the "<<#y<<" = "<<y<<'\n'
#define DE(fmt,...) fprintf(stderr, "Line %d : " fmt "\n",__LINE__,##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>pii;
typedef pair<ll,int>pli;
typedef pair<ll,ll>pll;
typedef pair<int,ll>pil;
typedef vector<int>vi;
typedef vector<ll>vll;
typedef vector<pii>vpii;
typedef vector<pll>vpll;
template<typename T>T cmax(T &x, T y){return x=x>y?x:y;}
template<typename T>T cmin(T &x, T y){return x=x<y?x:y;}
template<typename T>
T &read(T &r){
	r=0;bool w=0;char ch=getchar();
	while(ch<'0'||ch>'9')w=ch=='-'?1:0,ch=getchar();
	while(ch>='0'&&ch<='9')r=r*10+(ch^48),ch=getchar();
	return r=w?-r:r;
}
template<typename T1,typename... T2>
void read(T1 &x,T2& ...y){read(x);read(y...);}
const int N=2000010;
int n,m;
struct Edge{
	int x,y,w;
}e[N];
vi eg[N];
int tot,val[N],fa[N];
int getfa(int x){return fa[x]==x?x:fa[x]=getfa(fa[x]);}
int dep[N],top[N],pa[N],son[N],siz[N];
void dfs1(int x){
	siz[x]=1;
	for(auto v:eg[x]){
		dep[v]=dep[x]+1;pa[v]=x;
		dfs1(v);siz[x]+=siz[v];
		if(siz[v]>siz[son[x]])son[x]=v;
	}
}
void dfs2(int x,int t){
	top[x]=t;
	if(son[x])dfs2(son[x],t);
	for(auto v:eg[x])if(v!=son[x])dfs2(v,v);
}
int LCA(int x,int y){
	while(top[x]!=top[y]){
		if(dep[top[x]]<dep[top[y]])swap(x,y);
		x=pa[top[x]];
	}
	return dep[x]<dep[y]?x:y;
}
signed main(){
	#ifdef do_while_true
//		assert(freopen("data.in","r",stdin));
//		assert(freopen("data.out","w",stdout));
	#endif
	read(n,m);
	for(int i=1;i<n;i++)read(e[i].x,e[i].y),e[i].w=min(e[i].x,e[i].y);
	sort(e+1,e+n,[&](const Edge &x,const Edge &y){return x.w>y.w;});
	tot=n;
	for(int i=1;i<=n;i++)fa[i]=val[i]=i;
	for(int i=1;i<n;i++){
		int x=e[i].x,y=e[i].y,w=e[i].w;
		x=getfa(x);y=getfa(y);
		++tot;val[tot]=w;
		eg[tot].pb(x);eg[tot].pb(y);
		fa[x]=fa[y]=fa[tot]=tot;
	}
	dfs1(tot);
	dfs2(tot,tot);
	int las=0,t=0;
	while(m--){
		int op,x;read(op,x);
		x=(las+x)%n+1;
		if(op==1){
			if(!t)t=x;
			else t=LCA(t,x);
		}
		else{
			las=val[LCA(t,x)];
			cout<<las<<'\n';
		}
	}
    #ifdef do_while_true
//		cerr<<'\n'<<"Time:"<<1.0*clock()/CLOCKS_PER_SEC*1000<<" ms"<<'\n';
	#endif
	return 0;
}