Description
You are given a node-labeled rooted tree with n nodes. Define the query (x, k): Find the node whose label is k-th largest in the subtree of the node x. Assume no two nodes have the same labels.
Input
The first line contains one integer n (1 <= n <= 10^5). The next line contains n integers li (0 <= li <= 109) which denotes the label of the i-th node. Each line of the following n - 1 lines contains two integers u, v. They denote there is an edge between node u and node v. Node 1 is the root of the tree. The next line contains one integer m (1 <= m <= 10^4) which denotes the number of the queries. Each line of the next m contains two integers x, k. (k <= the total node number in the subtree of x)
Output
For each query (x, k), output the index of the node whose label is the k-th largest in the subtree of the node x.
按dfs序建可持久化线段树,子树对应连续的区间,可以直接查询kth
#include<cstdio> inline int _int(){ int x=0,c=getchar(); while(c>57||c<48)c=getchar(); while(c>47&&c<58)x=x*10+c-48,c=getchar(); return x; } const int N=100010; int n; int v[N],rt[N],id[N],idr[N],idp=0; int es[N*2],enx[N*2],e0[N],ep=2; int ch[N*33][2],sz[N*33],ids[N*33],p=0; int ins(int w,int x,int id){ int u=++p,u0=u; for(int i=30;~i;i--){ int d=x>>i&1; ch[u][d^1]=ch[w][d^1]; sz[u=ch[u][d]=++p]=sz[w=ch[w][d]]+1; } ids[u]=id; return u0; } void dfs(int w,int pa){ id[w]=++idp; rt[idp]=ins(rt[idp-1],v[w],w); for(int i=e0[w];i;i=enx[i]){ int u=es[i]; if(u!=pa)dfs(u,w); } idr[w]=idp; } void kth(int w1,int w2,int k){ for(int i=30;~i;i--){ int s=sz[ch[w1][0]]-sz[ch[w2][0]]; if(s<k){ k-=s; w1=ch[w1][1]; w2=ch[w2][1]; }else{ w1=ch[w1][0]; w2=ch[w2][0]; } } printf("%d ",ids[w1]); } int main(){ n=_int(); for(int i=1;i<=n;i++)v[i]=_int(); for(int i=1;i<n;i++){ int a=_int(),b=_int(); es[ep]=b;enx[ep]=e0[a];e0[a]=ep++; es[ep]=a;enx[ep]=e0[b];e0[b]=ep++; } dfs(1,0); for(int q=_int();q;q--){ int x=_int(); kth(rt[idr[x]],rt[id[x]-1],_int()); } return 0; }