题目链接：
题意：给出一个有根树，1为根节点，每个节点有权值。若干询问，询问以u为根的子树中权值第K小的节点编号。
思路：DFS一次，记录每个节点在DFS序列中的开始和结束位置。那么以u为节点的子树的所有点都在两个u之间。那么询问就转化成询问区间的第K小值，可以将DFS序列建立划分树解决。
#include <iostream>
#include <cstdio>#include <string.h>#include <algorithm>#include <cmath>#include <vector>#include <queue>#include <set>#include <stack>#include <string>#include <map>#define max(x,y) ((x)>(y)?(x):(y))#define min(x,y) ((x)<(y)?(x):(y))#define abs(x) ((x)>=0?(x):-(x))#define i64 long long#define u32 unsigned int#define u64 unsigned long long#define clr(x,y) memset(x,y,sizeof(x))#define CLR(x) x.clear()#define ph(x) push(x)#define pb(x) push_back(x)#define Len(x) x.length()#define SZ(x) x.size()#define PI acos(-1.0)#define sqr(x) ((x)*(x))#define MP(x,y) make_pair(x,y)#define EPS 1e-10#define FOR0(i,x) for(i=0;i<x;i++)#define FOR1(i,x) for(i=1;i<=x;i++)#define FOR(i,a,b) for(i=a;i<=b;i++)#define FORL0(i,a) for(i=a;i>=0;i--)#define FORL1(i,a) for(i=a;i>=1;i--)#define FORL(i,a,b)for(i=a;i>=b;i--)#define rush() int CC;for(scanf("%d",&CC);CC--;)#define Rush(n) while(scanf("%d",&n)!=-1)using namespace std;void RD(int &x){scanf("%d",&x);}void RD(i64 &x){scanf("%lld",&x);}void RD(u64 &x){scanf("%I64u",&x);}void RD(u32 &x){scanf("%u",&x);}void RD(double &x){scanf("%lf",&x);}void RD(int &x,int &y){scanf("%d%d",&x,&y);}void RD(i64 &x,i64 &y){scanf("%lld%lld",&x,&y);}void RD(u32 &x,u32 &y){scanf("%u%u",&x,&y);}void RD(double &x,double &y){scanf("%lf%lf",&x,&y);}void RD(int &x,int &y,int &z){scanf("%d%d%d",&x,&y,&z);}void RD(i64 &x,i64 &y,i64 &z){scanf("%lld%lld%lld",&x,&y,&z);}void RD(u32 &x,u32 &y,u32 &z){scanf("%u%u%u",&x,&y,&z);}void RD(double &x,double &y,double &z){scanf("%lf%lf%lf",&x,&y,&z);}void RD(char &x){x=getchar();}void RD(char *s){scanf("%s",s);}void RD(string &s){cin>>s;}void PR(int x) {printf("%d\n",x);}void PR(int x,int y) {printf("%d %d\n",x,y);}void PR(i64 x) {printf("%lld\n",x);}void PR(u32 x) {printf("%u\n",x);}void PR(u64 x) {printf("%llu\n",x);}void PR(double x) {printf("%.2lf\n",x);}void PR(char x) {printf("%c\n",x);}void PR(char *x) {printf("%s\n",x);}void PR(string x) {cout<<x<<endl;}const int mod=10007;const i64 inf=((i64)1)<<60;const double dinf=1000000000000000000.0;const int INF=2147483647;const int N=500005;struct Node{ int v,next;};Node edges[N];int head[N],e;int key[N];int n,m;void Add(int u,int v){ edges[e].v=v; edges[e].next=head[u]; head[u]=e++;}struct node{ int L,R;};node a[N<<2];int s[N],t[35][N],tot[35][N];int cnt;int pos[N][2];void dfs(int u,int pre){ s[++cnt]=key[u]; pos[u][0]=cnt; int i,v; for(i=head[u];i!=-1;i=edges[i].next) { v=edges[i].v; if(v==pre) continue; dfs(v,u); } pos[u][1]=cnt;}void build(int dep,int u,int L,int R){ a[u].L=L; a[u].R=R; if(L==R) return; int mid=(L+R)>>1; int sameNum=mid-L+1,i; for(i=L;i<=R;i++) if(t[dep][i]<s[mid]) sameNum--; int LL=L,LR=mid,RL=mid+1,RR=R; int Lnum=0,Rnum=0; int x; for(i=L;i<=R;i++) { if(i==L) tot[dep][i]=0; else tot[dep][i]=tot[dep][i-1]; x=t[dep][i]; if(x<s[mid]) { tot[dep][i]++; t[dep+1][LL+Lnum]=x; Lnum++; } else if(x>s[mid]) { t[dep+1][RL+Rnum]=x; Rnum++; } else { if(sameNum>0) { sameNum--; tot[dep][i]++; t[dep+1][LL+Lnum]=x; Lnum++; } else { t[dep+1][RL+Rnum]=x; Rnum++; } } } build(dep+1,u*2,LL,LR); build(dep+1,u*2+1,RL,RR);}int query(int dep,int u,int L,int R,int K){ if(L==R) return t[dep][L]; int x,y,xx,yy,l,r; int mid=(a[u].L+a[u].R)>>1; if(L==a[u].L) x=0; else x=tot[dep][L-1]; y=tot[dep][R]-x; if(K<=y) { l=a[u].L+x; r=a[u].L+x+y-1; return query(dep+1,u*2,l,r,K); } else { xx=L-a[u].L-x; yy=R-L+1-y; l=mid+1+xx; r=mid+1+xx+yy-1; return query(dep+1,u*2+1,l,r,K-y); }}map<int,int> mp;int main(){ RD(n); int i,j; FOR1(i,n) RD(key[i]),mp[key[i]]=i; clr(head,-1); e=0; int x,y; FOR1(i,n-1) { RD(x,y); Add(x,y); Add(y,x); } dfs(1,-1); FOR1(i,cnt) t[1][i]=s[i]; sort(s+1,s+cnt+1); build(1,1,1,cnt); RD(m); int ans; while(m--) { RD(x,y); ans=query(1,1,pos[x][0],pos[x][1],y); PR(mp[ans]); }}