CodeForces 620E:New Year Tree(dfs序+线段树)

E. New Year Tree
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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:

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 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

 因为只有60种颜色，所以用二进制数表示颜色的种类。

#include <cstdio>
#include <string.h>
using namespace std;
const int MAXN=400005;
typedef long long LL;
struct Edge{
    int to,net;
}es[MAXN+MAXN];
int head[MAXN],tot;
int n,m;
int val[MAXN],Hash[MAXN];
void addedge(int u,int v)
{
    es[tot].to=v;
    es[tot].net=head[u];
    head[u]=tot++;
}
int lch[MAXN],rch[MAXN],key;
void dfs(int u,int fa)
{
    lch[u]=++key;
    Hash[lch[u]]=val[u];
    for(int i=head[u];i!=-1;i=es[i].net)
    {
        if(es[i].to!=fa)
        {
            dfs(es[i].to,u);
        }
    }
    rch[u]=key;
}
struct Node{
    int l,r;
    LL color,lazy;
}a[MAXN*3];
void build(int rt,int l,int r)
{
    a[rt].l=l;
    a[rt].r=r;
    a[rt].lazy=0;
    if(l==r)
    {
        a[rt].color=1LL<<Hash[l];
        return ;
    }
    int mid=(l+r)>>1;
    build(rt<<1,l,mid);
    build((rt<<1)|1,mid+1,r);
    a[rt].color=(a[rt<<1].color | a[(rt<<1)|1].color);
}
void pushDown(int rt)
{
    a[rt<<1].color=a[rt].color;
    a[(rt<<1)|1].color=a[rt].color;
    a[rt<<1].lazy=a[rt].lazy;
    a[(rt<<1)|1].lazy=a[rt].lazy;
    a[rt].lazy=0;
}
void update(int rt,int l,int r,int v)
{
    if(a[rt].l==l&&a[rt].r==r)
    {
        a[rt].color=1LL<<v;
        a[rt].lazy=1LL<<v;
        return ;
    }
    if(a[rt].lazy!=0)
    {
        pushDown(rt);
    }
    int mid=(a[rt].l+a[rt].r)>>1;
    if(r<=mid)
    {
        update(rt<<1,l,r,v);
    }
    else if(mid<l)
    {
        update((rt<<1)|1,l,r,v);
    }
    else
    {
        update(rt<<1,l,mid,v);
        update((rt<<1)|1,mid+1,r,v);
    }
    a[rt].color=(a[rt<<1].color | a[(rt<<1)|1].color);
}
LL res;
void query(int rt,int l,int r)
{
    if(a[rt].l==l&&a[rt].r==r)
    {
        res|=a[rt].color;
        return ;
    }
    if(a[rt].lazy!=0)
    {
        pushDown(rt);
    }
    int mid=(a[rt].l+a[rt].r)>>1;
    if(r<=mid)
    {
        query(rt<<1,l,r);
    }
    else if(mid<l)
    {
        query((rt<<1)|1,l,r);
    }
    else
    {
        query(rt<<1,l,mid);
        query((rt<<1)|1,mid+1,r);
    }
    a[rt].color=(a[rt<<1].color | a[(rt<<1)|1].color);
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(head,-1,sizeof(head));
        tot=0;
        key=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&val[i]);
        }
        for(int i=0;i<n-1;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        dfs(1,-1);
        build(1,1,n);
        for(int i=0;i<m;i++)
        {
            int type;    
            scanf("%d",&type);
            if(type==1)
            {
                int x,y;
                scanf("%d%d",&x,&y);
                update(1,lch[x],rch[x],y);
            }
            else
            {
                int x;
                scanf("%d",&x);
                res=0;
                query(1,lch[x],rch[x]);
                int sum=0;
                while(res>0)
                {
                    if(res&1)    sum++;
                    res>>=1;
                }
                printf("%d\n",sum);
            }
        }
    }
    return 0;
}