Description
You are given a tree (an acyclic undirected connected graph) with n nodes. The tree nodes are numbered from 1 to n.
Each node has a color, white or black. All the nodes are black initially.
We will ask you to perfrom some instructions of the following form:
- 0 u : ask for how many nodes are connected to u, two nodes are connected iff all the node on the path from u to v (inclusive u and v) have a same color.
- 1 u : toggle the color of u(that is, from black to white, or from white to black).
Input
The first line contains a number n denoted how many nodes in the tree(1 ≤ n ≤ 105). The next n - 1 lines, each line has two numbers (u, v) describe a edge of the tree(1 ≤ u, v ≤ n). The next line contains a number m denoted how many operations we are going to process(1 ≤ m ≤ 105). The next m lines, each line describe a operation (t, u) as we mentioned above(0 ≤ t ≤ 1, 1 ≤ u ≤ n).
Output
For each query operation, output the corresponding result.
Sample Input
5
1 2
1 3
1 4
1 5
3
0 1
1 1
0 1
Sample Output
5
1
题解：
先将树定个根，然后建立黑白两棵lct
对于黑树，时刻满足如果有边u-v，则v一定是黑点（u不一定是黑点）
且对于黑树上的某个点，记录一下通过虚边和它相连的黑点有多少，lct上维护这条链的总和，白树同理
对于单点改色，假如它是黑点，要在黑树上把它和它父亲节点断开，然后再在白树上把它和它父亲节点连起来，白点同理
对于单点查询，假如它是黑点，把它access上去后要判断这条链最上面的那个点是不是黑点，不是就要输出最上面那个点的儿子节点，白点同理
code：
1 #include<cstdio>
2 #include<iostream>
3 #include<cmath>
4 #include<cstring>
5 #include<algorithm>
6 using namespace std;
7 char ch;
8 bool ok;
9 void read(int &x){
10 for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch==‘-‘) ok=1;
11 for (x=0;isdigit(ch);x=x*10+ch-‘0‘,ch=getchar());
12 if (ok) x=-x;
13 }
14 const int maxn=100005;
15 const int maxm=200005;
16 int n,q,op,a,b,tot,now[maxn],son[maxm],pre[maxm],col[maxn],fa[maxn];
17 bool flag[maxn];
18 struct LCT{
19 int id,fa[maxn],son[maxn][2],tot[maxn],sum[maxn];
20 void init(int x,int op){tot[x]+=op,sum[x]+=op;}
21 int which(int x){return son[fa[x]][1]==x;}
22 bool isroot(int x){return son[fa[x]][0]!=x&&son[fa[x]][1]!=x;}
23 void update(int x){
24 sum[x]=tot[x];
25 if (son[x][0]) sum[x]+=sum[son[x][0]];
26 if (son[x][1]) sum[x]+=sum[son[x][1]];
27 }
28 void rotate(int x){
29 int y=fa[x],z=fa[y],d=which(x),dd=which(y);
30 fa[son[x][d^1]]=y,son[y][d]=son[x][d^1],fa[x]=fa[y];
31 if (!isroot(y)) son[z][dd]=x;
32 fa[y]=x,son[x][d^1]=y,update(y),update(x);
33 }
34 void splay(int x){
35 while (!isroot(x)){
36 if (isroot(fa[x])) rotate(x);
37 else if (which(fa[x])==which(x)) rotate(fa[x]),rotate(x);
38 else rotate(x),rotate(x);
39 }
40 }
41 void access(int x){
42 for (int p=0;x;x=fa[x]){
43 splay(x);
44 if (son[x][1]) tot[x]+=sum[son[x][1]];
45 if (p) tot[x]-=sum[p];
46 son[x][1]=p,update(p=x);
47 }
48 }
49 void cut(int x,int y){
50 if (!y) return;
51 access(x),splay(x),fa[son[x][0]]=0,son[x][0]=0,update(x);
52 }
53 void link(int x,int y){
54 if (!y) return;
55 access(y),splay(y),splay(x),fa[x]=y,son[y][1]=x,update(y);
56 }
57 int find_left(int x){for (access(x),splay(x);son[x][0];x=son[x][0]); return x;}
58 void query(int x){
59 int t=find_left(x); splay(t);
60 printf("%d\n",col[t]==id?sum[t]:sum[son[t][1]]);
61 }
62 }T[2];
63 void put(int a,int b){pre[++tot]=now[a],now[a]=tot,son[tot]=b;}
64 void add(int a,int b){put(a,b),put(b,a);}
65 void dfs(int u){
66 for (int p=now[u],v=son[p];p;p=pre[p],v=son[p])
67 if (v!=fa[u]) fa[v]=u,T[1].link(v,u),dfs(v);
68 }
69 int main(){
70 read(n),T[0].id=0,T[1].id=1;
71 for (int i=1;i<n;i++) read(a),read(b),add(a,b);
72 for (int i=1;i<=n;i++) col[i]=1;
73 for (int i=1;i<=n;i++) T[1].init(i,1);
74 dfs(1);
75 for (read(q);q;q--){
76 read(op),read(a);
77 if (op) T[col[a]].cut(a,fa[a]),T[col[a]].init(a,-1),col[a]^=1,T[col[a]].init(a,1),T[col[a]].link(a,fa[a]);
78 else T[col[a]].query(a);
79 }
80 return 0;
81 }