CodeForces 620E New Year Tree 线段树

最新推荐文章于 2019-08-18 23:06:37 发布

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本文介绍了一种利用线段树优化树形DP的方法，解决了改变子树颜色及查询子树中不同颜色数量的问题。通过DFS序记录子树范围，并采用二进制表示颜色存在状态。

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The New Year holidays are over, but Resha doesn't want to throw away the New Year tree. He invited his best friends Kerim and Gural to help him to redecorate the New Year tree.

The New Year tree is an undirected tree with n vertices and root in the vertex 1.

You should process the queries of the two types:

Change the colours of all vertices in the subtree of the vertex v to the colour c.
Find the number of different colours in the subtree of the vertex v.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 4·10⁵) — the number of vertices in the tree and the number of the queries.

The second line contains n integers c_i (1 ≤ c_i ≤ 60) — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers x_j, y_j (1 ≤ x_j, y_j ≤ n) — the vertices of the j-th edge. It is guaranteed that you are given correct undirected tree.

The last m lines contains the description of the queries. Each description starts with the integer t_k (1 ≤ t_k ≤ 2) — the type of the k-th query. For the queries of the first type then follows two integers v_k, c_k (1 ≤ v_k ≤ n, 1 ≤ c_k ≤ 60) — the number of the vertex whose subtree will be recoloured with the colour c_k. For the queries of the second type then follows integer v_k (1 ≤ v_k ≤ n) — the number of the vertex for which subtree you should find the number of different colours.

Output

For each query of the second type print the integer a — the number of different colours in the subtree of the vertex given in the query.

Each of the numbers should be printed on a separate line in order of query appearing in the input.

Examples

Input

7 10
1 1 1 1 1 1 1
1 2
1 3
1 4
3 5
3 6
3 7
1 3 2
2 1
1 4 3
2 1
1 2 5
2 1
1 6 4
2 1
2 2
2 3

Output

题意：一棵树上面的点都有颜色，最多61种，有两种操作，一种是把这个点以及它子树的所有节点都涂成那个颜色，另一个操作是问这个节点和它的子树里颜色数

思路：建立一棵线段上，用dfs序记录下子树范围，然后维护线段树的时候，用二进制来表示某个颜色是否存在，有61种颜色，也就是2^61次方就可以表示完所有情况，对于一个点维护出来的数，每一位都去对比，然后记录颜色数就可以了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define mem(a,x) memset(a,x,sizeof(a))
#define lowbit(x) (x & (-x))
#define maxn 500005
vector<int>vec[maxn];
int vis[maxn],le[maxn],ri[maxn],a[maxn],cnt;
ll ans;

struct node{
	int le,ri;
	ll num,lazy;
}tree[maxn * 4];

void dfs(int x){  // dfs序，建立各个点到其所有子树 的编号 
	vis[x] = 1;
	le[x] = ++cnt;
	for(int i = 0;i < vec[x].size();i++){
		int v = vec[x][i];
		if(vis[v])
			continue;
		dfs(v);
	}
	ri[x] = cnt;
}
void push_down(int i){
	if(!tree[i].lazy)
		return ;
	tree[i * 2].num = tree[i * 2 + 1].num = (1ll << tree[i].lazy);
	tree[i * 2].lazy = tree[i * 2 + 1].lazy = tree[i].lazy;
	tree[i].lazy = 0;
}
void push_up(int i){
	tree[i].num = tree[i * 2].num | tree[i * 2 + 1].num;
}
void build(int i,int l,int r){
	tree[i].lazy = 0;
	tree[i].le = l;
	tree[i].ri = r;
	if(l == r)
		return ;
	int mid = (l + r) >> 1;
	build(i * 2,l,mid);
	build(i * 2 + 1,mid + 1,r);
}
void get_pos(int i,int aim,int w){
	if(tree[i].le == aim && tree[i].ri == aim){
		tree[i].num = (1ll << w);
		return ;
	}
	push_down(i);
	int mid = (tree[i].le + tree[i].ri) >> 1;
	if(aim <= mid)
		get_pos(i * 2,aim,w);
	if(aim > mid)
		get_pos(i * 2 + 1,aim,w);
	push_up(i); 
}
void update(int i,int l,int r,ll w){
	if(tree[i].le == l && tree[i].ri == r){
		tree[i].lazy = w;
		tree[i].num = (1ll << w); //更新颜色 
		return ;
	}
	push_down(i);
	int mid = (tree[i].le + tree[i].ri) >> 1;
	if(r <= mid)
		update(i * 2,l,r,w);
	else if(l > mid)
		update(i * 2 + 1,l,r,w);
	else{
		update(i * 2,l,mid,w);
		update(i * 2 + 1,mid + 1,r,w);
	}
	push_up(i);
	return ;
}
void query(int i,int l,int r){
	if(tree[i].le == l && tree[i].ri == r){
		ans |= tree[i].num;
		return ;
	}
	push_down(i); //懒惰数组，查询到了再更新 
	int mid = (tree[i].le + tree[i].ri) >> 1;
	if(r <= mid){
		query(i * 2,l,r);
	}else if(l > mid){
		query(i * 2 + 1,l,r);
	}else{
		query(i * 2,l,mid);
		query(i * 2 + 1,mid + 1,r);
	}
	push_up(i);
}
int main(){
	int n,m,u,v;
	scanf("%d %d",&n,&m);
	for(int i = 1;i <= n;i++)
		scanf("%d",&a[i]);
	for(int i = 1;i < n;i++){
		scanf("%d %d",&u,&v);
		vec[u].push_back(v);
		vec[v].push_back(u);
	}
	cnt = 0;
	dfs(1);
	build(1,1,cnt);
	for(int i = 1;i <= n;i++){
		get_pos(1,le[i],a[i]); //把每个点的颜色上传，且转化为二进制保存在线段树里 
	}
	while(m--){
		int opt,x,y;
		scanf("%d",&opt);
		if(opt == 1){
			scanf("%d %d",&x,&y);
			update(1,le[x],ri[x],y);
		}else{
			ans = 0;
			scanf("%d",&x);
			query(1,le[x],ri[x]); // 查询得到一个ans 
			int tmp = 0;
			for(int i = 1;i <= 61;i++){	//去与61个位一一比较，统计颜色数 
				if(ans & (1ll << i))
					tmp++;
			}
			printf("%d\n",tmp);
		}
	}
	return 0;
}