bzoj1803: Spoj1487 Query on a tree III

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\n",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;
}