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
1 2
1 3
1 4
1 5
3
0 1
1 1
0 1
Sample Output
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：
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
char ch;
bool ok;
void read(int &x){
for (ok=,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=;
for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
if (ok) x=-x;
}
const int maxn=;
const int maxm=;
int n,q,op,a,b,tot,now[maxn],son[maxm],pre[maxm],col[maxn],fa[maxn];
bool flag[maxn];
struct LCT{
int id,fa[maxn],son[maxn][],tot[maxn],sum[maxn];
void init(int x,int op){tot[x]+=op,sum[x]+=op;}
int which(int x){return son[fa[x]][]==x;}
bool isroot(int x){return son[fa[x]][]!=x&&son[fa[x]][]!=x;}
void update(int x){
sum[x]=tot[x];
if (son[x][]) sum[x]+=sum[son[x][]];
if (son[x][]) sum[x]+=sum[son[x][]];
}
void rotate(int x){
int y=fa[x],z=fa[y],d=which(x),dd=which(y);
fa[son[x][d^]]=y,son[y][d]=son[x][d^],fa[x]=fa[y];
if (!isroot(y)) son[z][dd]=x;
fa[y]=x,son[x][d^]=y,update(y),update(x);
}
void splay(int x){
while (!isroot(x)){
if (isroot(fa[x])) rotate(x);
else if (which(fa[x])==which(x)) rotate(fa[x]),rotate(x);
else rotate(x),rotate(x);
}
}
void access(int x){
for (int p=;x;x=fa[x]){
splay(x);
if (son[x][]) tot[x]+=sum[son[x][]];
if (p) tot[x]-=sum[p];
son[x][]=p,update(p=x);
}
}
void cut(int x,int y){
if (!y) return;
access(x),splay(x),fa[son[x][]]=,son[x][]=,update(x);
}
void link(int x,int y){
if (!y) return;
access(y),splay(y),splay(x),fa[x]=y,son[y][]=x,update(y);
}
int find_left(int x){for (access(x),splay(x);son[x][];x=son[x][]); return x;}
void query(int x){
int t=find_left(x); splay(t);
printf("%d\n",col[t]==id?sum[t]:sum[son[t][]]);
}
}T[];
void put(int a,int b){pre[++tot]=now[a],now[a]=tot,son[tot]=b;}
void add(int a,int b){put(a,b),put(b,a);}
void dfs(int u){
for (int p=now[u],v=son[p];p;p=pre[p],v=son[p])
if (v!=fa[u]) fa[v]=u,T[].link(v,u),dfs(v);
}
int main(){
read(n),T[].id=,T[].id=;
for (int i=;i<n;i++) read(a),read(b),add(a,b);
for (int i=;i<=n;i++) col[i]=;
for (int i=;i<=n;i++) T[].init(i,);
dfs();
for (read(q);q;q--){
read(op),read(a);
if (op) T[col[a]].cut(a,fa[a]),T[col[a]].init(a,-),col[a]^=,T[col[a]].init(a,),T[col[a]].link(a,fa[a]);
else T[col[a]].query(a);
}
return ;
}