CF893F 题解

CF893F 题解

$\text{Description}$

给定一棵点带权有根树,多次询问以某个节点为根的子树内距离该点不超过 $k$ 的节点中的最小点权。本题强制在线

$\text{Solution}$

问题实际上是询问一个子树内一个深度区间内的点权最小值,容易想到对每个点开权值线段树,下标为深度,维护区间的点权最小值。从而可以利用线段树合并解决问题。为了做到在线,线段树合并时必须额外开空间存储。

$\text{Code}$
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#include<bits/stdc++.h>
#define REG register
#define MAXN 100005
#define FORE(i,now) for(REG int i=head[now];i;i=ed[i].nxt)
using namespace std;
inline int read(){
REG int x(0);
REG char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) x=(x*10)+(c^48),c=getchar();
return x;
}
inline int Min(int a,int b){return a<b?a:b;}

int n,q,a[MAXN],dep[MAXN];

struct Edge{
int v,nxt;
}ed[MAXN<<1];
int head[MAXN],cnt;
inline void adde(int u,int v){ed[++cnt]=(Edge){v,head[u]},head[u]=cnt;}
inline void add(int u,int v){adde(u,v),adde(v,u);}

int rt[MAXN],tot;
#define Ls(now) (t[now].ls)
#define Rs(now) (t[now].rs)
#define mid ((l+r)>>1)
struct Tree{
int val,ls,rs;
}t[MAXN<<8];
inline void Push_up(int now){t[now].val=Min(t[Ls(now)].val,t[Rs(now)].val);}
void Update(int& now,int l,int r,int pos,int status){
if(!now) t[now=++tot].val=1e9+7;
t[now].val=Min(t[now].val,status);
if(l==r) return;
if(mid>=pos) Update(Ls(now),l,mid,pos,status);
else Update(Rs(now),mid+1,r,pos,status);
Push_up(now);
}
int Askmin(int now,int l,int r,int x,int y){
if(!now) return 1e9+7;
if(l>=x&&r<=y) return t[now].val;
int res=1e9+7;
if(mid>=x) res=Min(res,Askmin(Ls(now),l,mid,x,y));
if(mid<y) res=Min(res,Askmin(Rs(now),mid+1,r,x,y));
return res;
}
int Merge(int a,int b){
if(!a||!b) return a|b;
int now=++tot;
t[now].val=Min(t[a].val,t[b].val);
Ls(now)=Merge(Ls(a),Ls(b));
Rs(now)=Merge(Rs(a),Rs(b));
return now;
}

void dfs(int now,int fa){
dep[now]=dep[fa]+1;
Update(rt[now],1,n,dep[now],a[now]);
FORE(i,now){
int v=ed[i].v;
if(v==fa) continue;
dfs(v,now);
rt[now]=Merge(rt[now],rt[v]);
}
}

void Work(){
n=read(),rt[0]=read();
for(REG int i=1;i<=n;++i) a[i]=read();
for(REG int i=1;i<n;++i) add(read(),read());
t[0].val=1e9+7;
dfs(rt[0],0);
int lastans=0;
q=read();
while(q--){
int x=read(),k=read();
x=(x+lastans)%n+1;
k=(k+lastans)%n;
printf("%d\n",lastans=Askmin(rt[x],1,n,dep[x],Min(dep[x]+k,n)));
}
}

int main(){
Work();
}
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