Codeforces 375 D Tree and Queries

最新推荐文章于 2020-11-24 03:01:00 发布

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最新推荐文章于 2020-11-24 03:01:00 发布

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原文链接：http://www.cnblogs.com/JYYHH/p/8609136.html

本文解析了一个使用DFS和莫队算法解决的问题：在一棵带有顶点颜色的树中，对于每个查询，找出指定子树中至少包含特定数量节点的颜色种类数。文章详细介绍了算法实现过程，并强调了莫队区间端点移动的正确顺序。

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Discription

You have a rooted tree consisting of n vertices. Each vertex of the tree has some color. We will assume that the tree vertices are numbered by integers from 1 to n. Then we represent the color of vertex v as c_v. The tree root is a vertex with number 1.

In this problem you need to answer to m queries. Each query is described by two integers v_j, k_j. The answer to query v_j, k_j is the number of such colors of vertices x, that the subtree of vertex v_j contains at least k_j vertices of color x.

You can find the definition of a rooted tree by the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory).

Input

The first line contains two integers n and m (2 ≤ n ≤ 10⁵; 1 ≤ m ≤ 10⁵). The next line contains a sequence of integers c₁, c₂, ..., c_n(1 ≤ c_i ≤ 10⁵). The next n - 1 lines contain the edges of the tree. The i-th line contains the numbers a_i, b_i(1 ≤ a_i, b_i ≤ n; a_i ≠ b_i) — the vertices connected by an edge of the tree.

Next m lines contain the queries. The j-th line contains two integers v_j, k_j(1 ≤ v_j ≤ n; 1 ≤ k_j ≤ 10⁵).

Output

Print m integers — the answers to the queries in the order the queries appear in the input.

Example

Input

8 5
1 2 2 3 3 2 3 3
1 2
1 5
2 3
2 4
5 6
5 7
5 8
1 2
1 3
1 4
2 3
5 3

Output

Input

Output

Note

A subtree of vertex v in a rooted tree with root r is a set of vertices {u : dist(r, v) + dist(v, u) = dist(r, u)}. Where dist(x, y) is the length (in edges) of the shortest path between vertices x and y.

dfs+莫队，本来挺傻的一个题，结果莫队写错了2333

以前都不是很在意莫队区间端点的移动，但是今天的事情证明了，莫队端点移动要先扩张区间，然后再缩减区间，不然有些题出现了左端点比右端点大的区间会挂掉2333

#include<bits/stdc++.h>
#define ll long long
const int maxn=100005;
using namespace std;
struct ask{
	int l,r,k,bl,num;
	bool operator <(const ask &u)const{
		return bl==u.bl?((bl&1)?r<u.r:r>u.r):bl<u.bl;
	}
}q[maxn];
int f[maxn],col[maxn];
int to[maxn*2],ne[maxn*2],hd[maxn];
int dfn[maxn],n,m,a[maxn],siz[maxn];
int cnt[maxn],dc=0,sz,ans[maxn],le,ri;

void dfs(int x,int fa){
	dfn[x]=++dc,a[dc]=col[x],siz[x]=1;
	for(int i=hd[x];i;i=ne[i]) if(to[i]!=fa){
		dfs(to[i],x);
		siz[x]+=siz[to[i]];
	}
}

inline void update(int x,int y){
	for(;x<=100000;x+=x&-x) f[x]+=y;
}

inline int query(int x){
	int an=0;
	for(;x;x-=x&-x) an+=f[x];
	return an;
}

inline void add(int x){
	if(cnt[a[x]]) update(cnt[a[x]],-1);
	cnt[a[x]]++;
	update(cnt[a[x]],1);
}

inline void del(int x){
	update(cnt[a[x]],-1);
	cnt[a[x]]--;
	if(cnt[a[x]]) update(cnt[a[x]],1);
}

inline void solve(){
	sort(q+1,q+m+1),le=1,ri=0;
	for(int i=1;i<=m;i++){
		while(ri<q[i].r) ri++,add(ri);
		while(le>q[i].l) le--,add(le);
		while(ri>q[i].r) del(ri),ri--;
		while(le<q[i].l) del(le),le++;
		ans[q[i].num]=query(100000)-query(q[i].k-1);
	}
}

int main(){
	scanf("%d%d",&n,&m),sz=sqrt(n);
	for(int i=1;i<=n;i++) scanf("%d",col+i);
	int uu,vv;
    for(int i=1;i<n;i++){
        scanf("%d%d",&uu,&vv);
        to[i]=vv,ne[i]=hd[uu],hd[uu]=i;
        to[i+n]=uu,ne[i+n]=hd[vv],hd[vv]=i+n;
    }
    dfs(1,1);  
    for(int i=1;i<=m;i++){
    	scanf("%d%d",&uu,&vv);
		q[i].num=i,q[i].k=vv,q[i].l=dfn[uu],q[i].r=dfn[uu]+siz[uu]-1;
		q[i].bl=(q[i].l-1)/sz+1; 
	}

	solve();
	
	for(int i=1;i<=m;i++) printf("%d\n",ans[i]);
	return 0;
}

转载于:https://www.cnblogs.com/JYYHH/p/8609136.html