SPOJ 1487. Query on a tree III

SPOJ Problem Set (classical)

SPOJ 1487. Query on a tree III

Problem code: PT07J

 

 

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 <= 10⁵). The next line contains n integers l_i (0 <= l_i <= 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 <= 10⁴) which denotes the number of the queries. Each line of the nextm 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.

Example

Input:
5
1 3 5 2 7
1 2
2 3
1 4
3 5
4
2 3
4 1
3 2
3 2

Output:
5
4
5
5
---------------------------------------------------------------
题目大意：询问树上某棵子树上第K大的节点。
解题思路：用一次dfs把树搞成括号序列，记录每个节点的开始和结束时间戳，然后就变成了区间第k大值，用划分树模板即可。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#define MID(x,y) ( ( x + y ) >> 1 )
#define L(x) ( x << 1 )
#define R(x) ( x << 1 | 1 )
#define BUG puts("here!!!")
#define LL long long
using namespace std;

const int N=100005;
int n,m;
int val[40][N],sorted[N],toleft[40][N];
bool cmp(int a,int b)
{
    return a<b;
}
struct node
{
    int l,r;
    int getmid()
    {
        return (l+r)>>1;
    }
} tree[N * 4];
void build(int rt,int l,int r,int d)
{
    tree[rt].l=l;
    tree[rt].r=r;
    if(l==r) return;
    int mid=(l+r) >> 1;
    int same=mid-l+1;
    for(int i=l; i<=r; ++i)
        if (val[d][i]<sorted[mid])same--;
    int lpos=l;
    int rpos=mid+1;
    for(int i=l; i<=r; ++i)
    {
        if(i==l)toleft[d][i]=0;
        else toleft[d][i]=toleft[d][i - 1];
        if(val[d][i]<sorted[mid])
        {
            toleft[d][i]++;
            val[d+1][lpos++]=val[d][i];
        }
        else if(val[d][i]>sorted[mid])val[d+1][rpos++]=val[d][i];
        else if(same)
        {
            same--;
            toleft[d][i]++;
            val[d+1][lpos++]=val[d][i];
        }
        else val[d+1][rpos++]=val[d][i];
    }
    build(L(rt),l,mid,d+1);
    build(R(rt),mid+1,r,d+1);
}
int query(int rt, int l, int r, int d, int k)
{
    if(l==r)return val[d][l];
    int s,ss;
    if(l==tree[rt].l)
    {
        s=toleft[d][r];
        ss=0;
    }
    else
    {
        s=toleft[d][r]-toleft[d][l-1];
        ss=toleft[d][l-1];
    }
    if(k<=s)
    {
        int left=tree[rt].l+ss;
        int right=left+s-1;
        return query(L(rt),left,right,d+1,k);
    }
    else
    {
        int b=l-tree[rt].l-ss;
        int left=tree[rt].getmid()+b+1;
        int right=left+r-l-s;
        return query(R(rt),left,right,d+1,k-s);
    }
}

int lab[N],one[N],two[N],eid,num;
int head[N],ed[N<<1],nxt[N<<1];
map<int,int>ha;

void addedge(int s,int e)
{
    ed[eid]=e;
    nxt[eid]=head[s];
    head[s]=eid++;
}

void dfs(int s,int f)
{
    one[s]=num;
    sorted[num]=lab[s];
    val[0][num++]=lab[s];
    for(int i=head[s]; ~i; i=nxt[i])
        if(ed[i]!=f)dfs(ed[i],s);
    two[s]=num-1;
}

int main()
{
//    freopen("/home/axorb/in","r",stdin);
    ha.clear();
    scanf("%d",&n);
    eid=0;
    memset(head,-1,sizeof(head));
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&lab[i]);
        ha[lab[i]]=i;
    }
    for(int i=1; i<n; i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        addedge(a,b);
        addedge(b,a);
    }
    num=0;
    dfs(1,-1);
sort(sorted,sorted+n,cmp);
build(1,0,n-1,0);
    int q;
    scanf("%d",&q);
    while(q--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        printf("%d\n",ha[query(1,one[a],two[a],0,b)]);
    }
}