bzoj3639: Query on a tree VII

Description

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：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<set>
 7 using namespace std;
 8 char ch;
 9 bool ok;
10 void read(int &x){
11     for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1;
12     for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());
13     if (ok) x=-x;
14 }
15 const int maxn=100005;
16 const int maxm=200005;
17 int n,q,op,a,b,v,tot,now[maxn],son[maxm],pre[maxm],fa[maxn],col[maxn];
18 struct lct{
19     int id,fa[maxn],son[maxn][2],val[maxn],maxv[maxn];
20     multiset<int> rest[maxn];
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         maxv[x]=val[x];
25         if (!rest[x].empty()) maxv[x]=max(maxv[x],*rest[x].rbegin());
26         if (son[x][0]) maxv[x]=max(maxv[x],maxv[son[x][0]]);
27         if (son[x][1]) maxv[x]=max(maxv[x],maxv[son[x][1]]);
28     }
29     void rotate(int x){
30         int y=fa[x],z=fa[y],d=which(x),dd=which(y);
31         fa[son[x][d^1]]=y,son[y][d]=son[x][d^1],fa[x]=fa[y];
32         if (!isroot(y)) son[z][dd]=x;
33         fa[y]=x,son[x][d^1]=y,update(y),update(x);
34     }
35     void splay(int x){
36         while (!isroot(x)){
37             if (isroot(fa[x])) rotate(x);
38             else if (which(x)==which(fa[x])) rotate(fa[x]),rotate(x);
39             else rotate(x),rotate(x);
40         }
41     }
42     void access(int x){
43         for (int p=0;x;x=fa[x]){
44             splay(x);
45             if (son[x][1]) rest[x].insert(maxv[son[x][1]]);
46             if (p) rest[x].erase(rest[x].find(maxv[p]));
47             son[x][1]=p,update(p=x);
48         }
49     }
50     void link(int x,int y){
51         if (!y) return;
52         access(y),splay(y),splay(x),fa[x]=y,son[y][1]=x,update(y);
53     }
54     void cut(int x,int y){
55         if (!y) return;
56         access(x),splay(x),fa[son[x][0]]=0,son[x][0]=0,update(x);
57     }
58     int find_root(int x){for (access(x),splay(x);son[x][0];x=son[x][0]); return x;}
59     void query(int x){
60         int t=find_root(x); splay(t);
61         printf("%d
",col[t]==id?maxv[t]:maxv[son[t][1]]);
62     }
63     void modify(int x,int v){access(x),splay(x),val[x]=v,update(x);}
64 }T[2];
65 void put(int a,int b){pre[++tot]=now[a],now[a]=tot,son[tot]=b;}
66 void add(int a,int b){put(a,b),put(b,a);}
67 void dfs(int u){
68     for (int p=now[u],v=son[p];p;p=pre[p],v=son[p])
69         if (v!=fa[u]) fa[v]=u,T[col[v]].link(v,u),dfs(v);
70 }
71 int main(){
72     read(n),T[0].id=0,T[1].id=1;
73     for (int i=1;i<n;i++) read(a),read(b),add(a,b);
74     for (int i=1;i<=n;i++) read(col[i]);
75     for (int i=1;i<=n;i++) read(T[0].val[i]),T[0].maxv[i]=T[0].val[i];
76     for (int i=1;i<=n;i++) T[1].maxv[i]=T[1].val[i]=T[0].val[i];
77     dfs(1);
78     for (read(q);q;q--){
79         read(op),read(a);
80         if (op==0) T[col[a]].query(a);
81         else if (op==1) T[col[a]].cut(a,fa[a]),col[a]^=1,T[col[a]].link(a,fa[a]);
82         else read(v),T[0].modify(a,v),T[1].modify(a,v);
83     }
84     return 0;
85 }