codeforces CF620E New Year Tree DFS序 线段树 状压

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E. New Year Tree

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time limit per test: 3 seconds memory limit per test: 256 megabytes input: standard input output: standard output

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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:

1.Change the colours of all vertices in the subtree of the vertex $ v $ to the colour $ c $.
2.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 ≤ 4·10^5) $
— the number of vertices in the tree and the number of the queries.

The second line contains $ n $ integers $ c_i (1 ≤ ci ≤ 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

input1

 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

output1

 2
 3
 4
 5
 1
 2

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input2

 23 30
 1 2 2 6 5 3 2 1 1 1 2 4 5 3 4 4 3 3 3 3 3 4 6
 1 2
 1 3
 1 4
 2 5
 2 6
 3 7
 3 8
 4 9
 4 10
 4 11
 6 12
 6 13
 7 14
 7 15
 7 16
 8 17
 8 18
 10 19
 10 20
 10 21
 11 22
 11 23
 2 1
 2 5
 2 6
 2 7
 2 8
 2 9
 2 10
 2 11
 2 4
 1 12 1
 1 13 1
 1 14 1
 1 15 1
 1 16 1
 1 17 1
 1 18 1
 1 19 1
 1 20 1
 1 21 1
 1 22 1
 1 23 1
 2 1
 2 5
 2 6
 2 7
 2 8
 2 9
 2 10
 2 11
 2 4

output2

 6
 1
 3
 3
 2
 1
 2
 3
 5
 5
 1
 2
 2
 1
 1
 1
 2
 3

题目大意

• 给你一棵树，每个结点刚开始的时候都有一个颜色，

• 现有操作

1. $ u \quad col $ ：给这个结点及其子树染上 $ col $ 这种颜色
2. $ u $ ：查询以 $ u $ 为根节点的子树的所有颜色种类

• 颜色数 $ \le 60 \quad n \le 10^5 $

思路

• 我们看到这道题的颜色最多只有 $ 60 $ 种，所以理所应当的想到突破口就是颜色，
  $ 60 $ 的范围可以考虑状压，其实我们在乎的只是有或者没有该颜色，所以我们可以用或运算来合并两个块

• 只有修改子树或者查询子树的操作，所以直接 $ dfs $ 序就可以

• 对 $ dfs $ 序维护一个线段树，支持区间修改区间或，统计区间或之后得到的数在二进制下 $ 1 $ 的个数，即为答案

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define int long long  
#define N 400005
inline int read() {
    register char ch;
    while(!isdigit(ch=getchar()));
    register int x=ch^'0';
    while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
    return x;
}
vector<int>e[N];
int n,m,c[N],id[N],in[N],out[N],tim,sum[N<<2],lzy[N<<2];
void dfs(int u,int fa){
    id[in[u]=++tim]=u;
    for(int i=0;i<e[u].size();++i){
        int v=e[u][i];
        if(v!=fa) dfs(v,u);
    }
    out[u]=tim;
}
void build(int o,int l,int r){
    if(l==r){
        sum[o]=c[id[l]];
        return;
    }
    int mid=l+r>>1;
    build(o<<1,l,mid); build(o<<1|1,mid+1,r);
    sum[o]=sum[o<<1]|sum[o<<1|1]; 
}
inline void pushdown(int o){
    sum[o<<1]=sum[o<<1|1]=lzy[o<<1]=lzy[o<<1|1]=lzy[o];
    lzy[o]=0;
}
void updata(int o,int l,int r,int L,int R,int val){
    if(L<=l&&r<=R){
        sum[o]=val;
        lzy[o]=val;
        return;
    }
    if(lzy[o]) pushdown(o);
    int mid=l+r>>1;
    if(L>mid) updata(o<<1|1,mid+1,r,L,R,val);
    else if(R<=mid) updata(o<<1,l,mid,L,R,val);
    else {
        updata(o<<1,l,mid,L,R,val);
        updata(o<<1|1,mid+1,r,L,R,val);
    }
    sum[o]=sum[o<<1]|sum[o<<1|1];
}
int query(int o,int l,int r,int L,int R){
    if(L<=l&&r<=R) return sum[o];
    if(lzy[o]) pushdown(o);
    int mid=l+r>>1;
    if(L>mid) return query(o<<1|1,mid+1,r,L,R);
    else if(R<=mid) return query(o<<1,l,mid,L,R);
    else return query(o<<1,l,mid,L,R)|query(o<<1|1,mid+1,r,L,R);
    sum[o]=sum[o<<1]|sum[o<<1|1];
}
inline int calc(int x){
    int res=0; 
    while(x){ ++res; x-=x&(-x); }
    return res;
}
signed main(){
    n=read(); m=read();
    for(int i=1;i<=n;++i) c[i]=1ll<<(read()-1);
    for(int i=1;i<n;++i){
        int u=read(),v=read();
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs(1,0);
    build(1,1,n);
    while(m--){
        int opt=read(),u=read();
        if(opt==1) updata(1,1,n,in[u],out[u],1ll<<(read()-1));
        else printf("%lld\n",calc(query(1,1,n,in[u],out[u])));
    }
    return 0;
}