New Year Tree
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
题意：给出一个无向树和每个顶点的颜色，然后进行q次操作，1.修改区间颜色，2.求顶点x为根所在子树的颜色数
思路:dfs序(+线段树(常规染色操作)
#include<bits/stdc++.h> #define ll long long #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define pb push_back #define P pair<int,int> using namespace std; const int maxn = 4*1e5+7; ll sum[maxn<<2]; int head[maxn],Next[maxn*20],To[maxn*20],cnt; int in[maxn],out[maxn],pos[maxn],tot,c[maxn],lazy[maxn<<2]; void add(int u,int v) { Next[++cnt]=head[u]; head[u]=cnt; To[cnt]=v; } void dfs(int u,int fa) { in[u]=++tot; pos[tot]=u;//记录下当前的节点,后面建树对应c[pos[l]] for(int i=head[u]; i!=-1; i=Next[i]) { if(To[i]==fa) continue; dfs(To[i],u); } out[u]=tot; } void push_up(int rt) { sum[rt]=sum[rt<<1]|sum[rt<<1|1]; } void push_down(int rt) { if(lazy[rt]) { lazy[rt<<1]=lazy[rt<<1|1]=lazy[rt]; sum[rt<<1]=sum[rt<<1|1]=(1LL<<lazy[rt]); lazy[rt]=0; } } void build(int l,int r,int rt) { if(l==r) { sum[rt]=1LL<<c[pos[l]]; return; } int m=(l+r)>>1; build(lson); build(rson); push_up(rt); } void updata(int l,int r,int rt,int L,int R,int val) { if(L<=l&&r<=R) { sum[rt]=1LL<<val; lazy[rt]=val; return; } push_down(rt); int m=(l+r)>>1; if(L<=m) updata(lson,L,R,val); if(R>m) updata(rson,L,R,val); push_up(rt); } ll query(int l,int r,int rt,int L,int R) { if(L<=l&&r<=R) return sum[rt]; push_down(rt); int m=(l+r)>>1; ll ans=0; if(L<=m) ans|=query(lson,L,R); if(R>m) ans|=query(rson,L,R); return ans; } int main() { int n,m,u,v; scanf("%d %d",&n,&m); for(int i=1; i<=n; i++) scanf("%d",&c[i]); memset(head,-1,sizeof(head)); for(int i=1; i<n; i++) { scanf("%d %d",&u,&v); add(u,v); add(v,u); } dfs(1,-1); build(1,n,1); while(m--) { int op,val,x; scanf("%d",&op); if(op==1) { scanf("%d %d",&x,&val); updata(1,n,1,in[x],out[x],val); } else { scanf("%d",&x); ll ans=query(1,n,1,in[x],out[x]); int ct=0; while(ans) { ct+=(ans%2); ans/=2; } printf("%d\n",ct); } } }
PS:摸鱼怪的博客分享，欢迎感谢各路大牛的指点~