bzoj3637: Query on a tree VI

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：

 #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 ;
 }