bzoj3639: Query on a tree VII

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, and a weight.

We will ask you to perfrom some instructions of the following form:

• 0 u : ask for the maximum weight among the nodes which 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).
• 2 u w: change the weight of u to w.

 

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 2 lines, each line contains n number, the first line is the initial color of each node(0 or 1), and the second line is the initial weight, let's say W_i, of each node(|W_i| ≤ 10⁹).

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 ≤ 2, 1 ≤ u ≤ n, |w| ≤ 10⁹).

 

Output

For each query operation, output the corresponding result.

 

Sample Input

5
1 2
1 3
1 4
1 5
0 1 1 1 1
1 2 3 4 5
3
0 1
1 1
0 1

Sample Output

1
5

 

题解：

和http://www.cnblogs.com/chenyushuo/p/5228875.html 这个相似，不过要用set维护通过虚边相连的点的最大权值，lct维护这条链上的最大值。

code：

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<set>
 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,v,tot,now[maxn],son[maxm],pre[maxm],fa[maxn],col[maxn];
 struct lct{
     int id,fa[maxn],son[maxn][],val[maxn],maxv[maxn];
     multiset<int> rest[maxn];
     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){
         maxv[x]=val[x];
         if (!rest[x].empty()) maxv[x]=max(maxv[x],*rest[x].rbegin());
         if (son[x][]) maxv[x]=max(maxv[x],maxv[son[x][]]);
         if (son[x][]) maxv[x]=max(maxv[x],maxv[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(x)==which(fa[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][]) rest[x].insert(maxv[son[x][]]);
             if (p) rest[x].erase(rest[x].find(maxv[p]));
             son[x][]=p,update(p=x);
         }
     }
     void link(int x,int y){
         if (!y) return;
         access(y),splay(y),splay(x),fa[x]=y,son[y][]=x,update(y);
     }
     void cut(int x,int y){
         if (!y) return;
         access(x),splay(x),fa[son[x][]]=,son[x][]=,update(x);
     }
     int find_root(int x){for (access(x),splay(x);son[x][];x=son[x][]); return x;}
     void query(int x){
         int t=find_root(x); splay(t);
         printf("%d\n",col[t]==id?maxv[t]:maxv[son[t][]]);
     }
     void modify(int x,int v){access(x),splay(x),val[x]=v,update(x);}
 }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[col[v]].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++) read(col[i]);
     for (int i=;i<=n;i++) read(T[].val[i]),T[].maxv[i]=T[].val[i];
     for (int i=;i<=n;i++) T[].maxv[i]=T[].val[i]=T[].val[i];
     dfs();
     for (read(q);q;q--){
         read(op),read(a);
         if (op==) T[col[a]].query(a);
         else if (op==) T[col[a]].cut(a,fa[a]),col[a]^=,T[col[a]].link(a,fa[a]);
         else read(v),T[].modify(a,v),T[].modify(a,v);
     }
     return ;
 }