CodeForces - 620E New Year Tree 线段树 + dfs序 + bitset

题目链接：点击查看

题意：

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.

题解：dfs序 搞一下， 剩下的就写吧，刚学了bitset，挺好用的，复杂度还低

#include<bits/stdc++.h>
using namespace std;
const int N=4e5+10;
struct node{
	int l,r;
	int laz;
	bitset<60> s;
}tree[N<<2];
int n,m;
int p[N],in[N],out[N],cnt;
vector<int> v[N];
int val[N];
void dfs(int u,int fa)
{
	in[u]=++cnt;
	p[cnt]=u;
	int l=v[u].size(),to;
	for(int i=0;i<l;i++)
	{
		to=v[u][i];
		if(to==fa) continue;
		dfs(to,u);
	}
	out[u]=cnt;
}
void pushup(int cur)
{
	tree[cur].s=tree[cur<<1].s|tree[cur<<1|1].s;
}
void build(int l,int r,int cur)
{
	int mid;
	tree[cur].l=l;
	tree[cur].r=r;
	if(l==r)
	{
		tree[cur].s[val[p[l]]-1]=1;
		return;
	}
	mid=(r+l)>>1;
	build(l,mid,cur<<1);
	build(mid+1,r,cur<<1|1);
	pushup(cur);
//	cout<<l<<" "<<r<<" "<<tree[cur].s.count()<<endl;
}
void pushdown(int cur)
{
	if(tree[cur].laz)
	{
		tree[cur<<1].laz=tree[cur].laz;
		tree[cur<<1|1].laz=tree[cur].laz;
		tree[cur<<1].s=0;
		tree[cur<<1].s[tree[cur].laz-1]=1;
		tree[cur<<1|1].s=0;
		tree[cur<<1|1].s[tree[cur].laz-1]=1;
		tree[cur].laz=0;
	}
}
void update(int pl,int pr,int cur,int val)
{
//	cout<<tree[cur].l<<" "<<tree[cur].r<<endl;
	if(pl<=tree[cur].l&&tree[cur].r<=pr)
	{
		tree[cur].s=0;tree[cur].s[val-1]=1;
		tree[cur].laz=val;
		return;
	}
	pushdown(cur);
	if(pl<=tree[cur<<1].r) update(pl,pr,cur<<1,val);
	if(pr>=tree[cur<<1|1].l) update(pl,pr,cur<<1|1,val);
	pushup(cur);
}
bitset<60> ans;
void query(int pl,int pr,int cur)
{
	if(pl<=tree[cur].l&&tree[cur].r<=pr)
	{
		ans|=tree[cur].s;
		return;
	}
	pushdown(cur);
	if(pl<=tree[cur<<1].r) query(pl,pr,cur<<1);
	if(pr>=tree[cur<<1|1].l) query(pl,pr,cur<<1|1);
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%d",&val[i]);
	int x,y;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		v[x].push_back(y);
		v[y].push_back(x);
	}
	
	dfs(1,0);
	build(1,n,1);
	int op;
	while(m--)
	{
		scanf("%d",&op);
		if(op==1)
		{
			scanf("%d%d",&x,&y);
		//	cout <<in[x]<<" "<<out[x]<<endl; 
			update(in[x],out[x],1,y);
		}
		else
		{
			ans=0;
			scanf("%d",&x);
		//	cout<<in[x]<<" "<<out[x]<<endl;
			query(in[x],out[x],1);
			printf("%d\n",ans.count());
		}
	}
	return 0;
}