[SPOJ-PT07J] Query on tree III (主席树)

题意翻译

你被给定一棵带点权的n个点的有根数，点从1到n编号。

定义查询 query(x,k): 寻找以x为根的k大点的编号（从小到大排序第k个点）

假设没有两个相同的点权。

输入格式： 第一行为整数n，第二行为点权，接下来n-1行为树边，接下来一行为整数m，下面m行为两个整数x,k，代表query(x,k)

输出格式： m行，输出每次查询的结果。

Translated by @vegacx

题目描述

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.

输入输出格式

输入格式：

The first line contains one integer n (1 <= n <= 10 ^{5}5 ). The next line contains n integers _l {i}i​ (0 <= _l {i}i​ <= 10 ^{9}9 ) 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}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)

输出格式：

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.

输入输出样例

输入样例#1：

5

1 3 5 2 7

1 2

2 3

1 4

3 5

4

2 3

4 1

3 2

3 2

输出样例#1：

5

4

5

5

Solution

这道题,直接维护一遍树上的DFS序,然后记录每个节点 siz,再用主席树维护即可。

然后查询的时候查询 ( b[i] , b[i]+siz[i]-1 ) ,需要注意的是输出的是编号。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn=100005;
struct sj{int to,next;}line[maxn*2];
int size,n,m,num,q,siz[maxn];
int head[maxn],id[maxn],tot,idx[maxn];
int a[maxn],b[maxn],c[maxn],T[maxn];
int L[maxn*20],R[maxn*20],sum[maxn*20];
void add(int x,int y)
{
    line[++size].to=y;
    line[size].next=head[x];
    head[x]=size;
}
int old[maxn];
void dfs(int x)
{
    siz[x]=1;
    a[++num]=c[x];
    id[x]=num; b[num]=a[num];
    old[num]=x;
    for(int i=head[x];i;i=line[i].next)
    {
        int tt=line[i].to;
        if(!siz[tt])
        {
        	dfs(tt);
        	siz[x]+=siz[tt];
		}
    }
}

int build(int l,int r)
{
	int rt=++tot;
	if(l!=r)
	{
		int mid=(l+r)/2;
		L[rt]=build(l,mid);
		R[rt]=build(mid+1,r);
	}
	return rt;
}

int update(int pre,int l,int r,int x)
{
    int rt=++tot;
    L[rt]=L[pre];
    R[rt]=R[pre];
    sum[rt]=sum[pre]+1;
    int mid=(l+r)/2;
    if(l!=r)
    {
        if (x<=mid) L[rt]=update(L[pre],l,mid,x);
        else R[rt]=update(R[pre],mid+1,r,x);
    }
    return rt;
}

int query(int x,int y,int l,int r,int k)
{
	if(l==r)return l;
	int now=sum[L[y]]-sum[L[x]];
	int mid=(l+r)/2;
	if(now>=k) return query(L[x],L[y],l,mid,k);
	else return query(R[x],R[y],mid+1,r,k-now);
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%d",&c[i]);
	for(int i=1;i<n;i++)
	{
		int x,y; scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	dfs(1);
	sort(b+1,b+1+n);
    m=unique(b+1,b+1+n)-b-1;
    T[0]=build(1,m);
    for(int i=1;i<=n;i++)
    {
	a[i]=lower_bound(b+1,b+1+m,a[i])-b,
        T[i]=update(T[i-1],1,m,a[i]);
			idx[a[i]]=old[i];
	}
	scanf("%d",&q);
    while(q--)
    {
    	int x,k;
        scanf("%d%d",&x,&k);
        int p=query(T[id[x]-1],T[id[x]+siz[x]-1],1,m,k);
        printf("%d\n",idx[p]);
    }
}
/*
5
1 3 5 2 7
1 2
2 3
1 4
3 5
4
2 3
4 1
3 2
3 2
*/