题解:一眼ODT (这题数据出的好狠阿 ODT不让过直接T到倒数第三组 机智的我写了sb线段树 不会2倍空间的线段树 就只能开bool防止炸内存了 感觉就是一眼题 就是数据很强
#include <bits/stdc++.h>
const int MAXN=4e5+10;
using namespace std;
vector<int>vec[MAXN];
int a[MAXN];int p[MAXN],num[MAXN],cnt,fp[MAXN];
int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}
void dfs(int v,int fa){
p[v]=++cnt;fp[p[v]]=v;num[v]=1;
for(int i=0;i<vec[v].size();i++){
if(vec[v][i]!=fa)dfs(vec[v][i],v),num[v]+=num[vec[v][i]];
}
}
bool d[MAXN<<2][61];int flag[MAXN<<2];
void push(int x){
if(flag[x]){
for(int i=1;i<=60;i++)d[x<<1][i]=d[x<<1|1][i]=0;
d[x<<1][flag[x]]=d[x<<1|1][flag[x]]=1;
flag[x<<1]=flag[x];flag[x<<1|1]=flag[x];
flag[x]=0;
}
}
void up(int x){
for(int i=1;i<=60;i++)d[x][i]=0;
for(int i=1;i<=60;i++)d[x][i]=(d[x<<1][i]|d[x<<1|1][i]);
}
void built(int rt,int l,int r){
if(l==r){d[rt][a[fp[l]]]=1;return ;}
int mid=(l+r)>>1;
built(rt<<1,l,mid);
built(rt<<1|1,mid+1,r);
up(rt);
}
void update(int rt,int l,int r,int ql,int qr,int vul){
if(ql<=l&&r<=qr){
flag[rt]=vul;
for(int i=1;i<=60;i++)d[rt][i]=0;d[rt][vul]=1;return ;
}
int mid=(l+r)>>1;
push(rt);
if(ql<=mid)update(rt<<1,l,mid,ql,qr,vul);
if(qr>mid)update(rt<<1|1,mid+1,r,ql,qr,vul);
up(rt);
}
int ans[61];
void querty(int rt,int l,int r,int ql,int qr){
if(ql<=l&&r<=qr){
for(int i=1;i<=60;i++)ans[i]=(ans[i]|d[rt][i]);return ;
}
int mid=(l+r)>>1;
push(rt);
if(ql<=mid)querty(rt<<1,l,mid,ql,qr);
if(qr>mid)querty(rt<<1|1,mid+1,r,ql,qr);
up(rt);
}
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);
dfs(1,0);
int op,vul;int lx,rx;
built(1,1,n);
for(int i=1;i<=m;i++){
scanf("%d",&op);
if(op==1){
scanf("%d%d",&u,&vul);
lx=p[u];rx=p[u]+num[u]-1;
update(1,1,n,lx,rx,vul);
}
else{
scanf("%d",&u);
lx=p[u];rx=p[u]+num[u]-1;
for(int j=1;j<=60;j++)ans[j]=0;
querty(1,1,n,lx,rx);
int cnt=0;
for(int j=1;j<=60;j++)if(ans[j])cnt++;
printf("%d
",cnt);
}
}
return 0;
}
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.
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.
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.
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
2
3
4
5
1
2
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
6
1
3
3
2
1
2
3
5
5
1
2
2
1
1
1
2
3