Educational Codeforces Round 6 E. New Year Tree dfs+线段树

题目链接：http://codeforces.com/contest/620/problem/E

E. New Year Tree

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ≤ 4·105) — the number of vertices in the tree and the number of the queries.

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

Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ 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 tk (1 ≤ tk ≤ 2) — the type of the k-th query. For the queries of the first type then follows two integers vk, ck (1 ≤ vk ≤ n, 1 ≤ ck ≤ 60) — the number of the vertex whose subtree will be recoloured with the colour ck. For the queries of the second type then follows integer vk (1 ≤ vk ≤ 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

2
3
4
5
1
2

input

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

output

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

题意：给你一颗树，根为1，每个点有一个颜色，颜色<=60;

　　　q个询问；

　　　1：表示区间修改根为v的子树，颜色改成c；

　　  2：表示求根为v的子树区间的不同颜色个数；

思路：dfs序处理子树问题，颜色小于60，直接状态压缩存；

　　　线段树维护，区间或，延迟标记；

　　　还有一个细节处理，dfs序有点的顺序会改变，利用一个flag数组处理即可；详见代码；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=5e5+,M=1e6+,inf=1e9+;
const ll INF=1e18+,mod=;
ll tree[N<<],lazy[N<<];
int in[N],out[N],tot,a[N],flag[N];
void pushup(int pos)
{
    tree[pos]=(tree[pos<<]|tree[pos<<|]);
}
void pushdown(int pos)
{
    if(lazy[pos])
    {
        tree[pos<<]=lazy[pos];
        tree[pos<<|]=lazy[pos];
        lazy[pos<<]=lazy[pos];
        lazy[pos<<|]=lazy[pos];
        lazy[pos]=0LL;
    }
}
void build(int l,int r,int pos)
{
    lazy[pos]=0LL;
    if(l==r)
    {
        tree[pos]=1LL<<(a[flag[l]]-);
        return;
    }
    int mid=(l+r)>>;
    build(l,mid,pos<<);
    build(mid+,r,pos<<|);
    pushup(pos);
}
void update(int L,int R,int l,int r,int pos,ll c)
{
    if(L<=l&&r<=R)
    {
        tree[pos]=c;
        lazy[pos]=c;
        return;
    }
    pushdown(pos);
    int mid=(l+r)>>;
    if(L<=mid)
    update(L,R,l,mid,pos<<,c);
    if(R>mid)
    update(L,R,mid+,r,pos<<|,c);
    pushup(pos);
}
ll query(int L,int R,int l,int r,int pos)
{
    if(L<=l&&r<=R)
    {
        return tree[pos];
    }
    pushdown(pos);
    int mid=(l+r)>>;
    ll ans=0LL;
    if(L<=mid)
        ans=(ans|query(L,R,l,mid,pos<<));
    if(R>mid)
        ans=(ans|query(L,R,mid+,r,pos<<|));
    return ans;
}
struct is
{
    int v,nex;
}edge[N<<];
int head[N<<],edg;
int n,p;
void init()
{
    memset(head,-,sizeof(head));
    edg=;
    tot=;
}
void add(int u,int v)
{
    edg++;
    edge[edg].v=v;
    edge[edg].nex=head[u];
    head[u]=edg;
}
void dfs(int u,int fa)
{
    in[u]=++tot;
    for(int i=head[u];i!=-;i=edge[i].nex)
    {
        int v=edge[i].v;
        if(v==fa)continue;
        dfs(v,u);
    }
    out[u]=tot;
}
char ch[];
int main()
{
    int n,q;
    while(~scanf("%d%d",&n,&q))
    {
        init();
        for(int i=;i<=n;i++)
        scanf("%d",&a[i]);
        for(int i=;i<n;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        dfs(,-);
        for(int i=;i<=n;i++)
        flag[in[i]]=i;
        build(,n,);
        while(q--)
        {
            int flag,p;
            scanf("%d%d",&flag,&p);
            if(flag==)
            {
                int c;
                scanf("%d",&c);
                update(in[p],out[p],,n,,1LL<<(c-));
            }
            else
            {
                ll temp=query(in[p],out[p],,n,);
                int ans=;
                while(temp)
                {
                    ans+=temp%;
                    temp>>=;
                }
                printf("%d\n",ans);

            }
        }
    }
    return ;
}