题意
一棵树,每个点初始有个点权和颜色
(0 u) :询问所有(u,v) 路径上的最大点权,要满足(u,v) 路径上所有点的颜色都相同
$1 u (:反转)u$ 的颜色
(2 u w) :把(u) 的点权改成(w)
(color_i∈[0,1],w_i∈[−10^9,10^9],n,m≤10^5)
Sol
(LCT)
和(QTREE6)一样，黑白两棵(LCT)
不过这次我们用数据结构维护虚子树内的最大权的同色点
可以用(multiset)，但我还是习惯可删除的(priority\_queue)
然后每个点维护一下所有子树的最大权的同色点
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 5);
const int INF(2e9);
typedef int Arr[_];
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
Arr w;
struct Heap{
priority_queue <int> A, B;
IL void Push(RG int x){
A.push(x);
}
IL void Del(RG int x){
B.push(x);
}
IL int Top(){
while(!B.empty() && B.top() == A.top()) A.pop(), B.pop();
return A.empty() ? -INF : A.top();
}
};
struct LCT{
Arr fa, ch[2], mxv;
Heap mx[_];
IL int Son(RG int x){
return ch[1][fa[x]] == x;
}
IL int Isroot(RG int x){
return ch[0][fa[x]] != x && ch[1][fa[x]] != x;
}
IL void Update(RG int x){
mxv[x] = max(max(mxv[ch[0][x]], mxv[ch[1][x]]), max(w[x], mx[x].Top()));
}
IL void Rotate(RG int x){
RG int y = fa[x], z = fa[y], c = Son(x);
if(!Isroot(y)) ch[Son(y)][z] = x; fa[x] = z;
ch[c][y] = ch[!c][x], fa[ch[c][y]] = y;
ch[!c][x] = y, fa[y] = x, Update(y);
}
IL void Splay(RG int x){
for(RG int y = fa[x]; !Isroot(x); Rotate(x), y = fa[x])
if(!Isroot(y)) Son(x) ^ Son(y) ? Rotate(x) : Rotate(y);
Update(x);
}
IL void Access(RG int x){
for(RG int y = 0; x; y = x, x = fa[x]){
Splay(x);
mx[x].Push(mxv[ch[1][x]]), mx[x].Del(mxv[y]);
ch[1][x] = y, Update(x);
}
}
IL int Findroot(RG int x){
Access(x), Splay(x);
while(ch[0][x]) x = ch[0][x];
Splay(x);
return x;
}
IL void Link(RG int x, RG int y){
if(!y) return;
Access(y), Splay(x), Splay(y);
fa[x] = y, ch[1][y] = x, Update(y);
}
IL void Cut(RG int x, RG int y){
if(!y) return;
Access(x), Splay(x);
ch[0][x] = fa[ch[0][x]] = 0, Update(x);
}
} T[2];
Arr fa, col;
int n, m;
vector <int> G[_];
IL void Dfs(RG int u, RG int ff){
for(RG int i = 0, l = G[u].size(); i < l; ++i){
RG int v = G[u][i];
if(v == ff) continue;
T[col[v]].Link(v, u), fa[v] = u;
Dfs(v, u);
}
}
int main(RG int argc, RG char *argv[]){
n = Input();
for(RG int i = 1; i < n; ++i){
RG int u = Input(), v = Input();
G[u].push_back(v), G[v].push_back(u);
}
for(RG int i = 1; i <= n; ++i) col[i] = Input();
for(RG int i = 1; i <= n; ++i) w[i] = Input();
T[0].mxv[0] = T[1].mxv[0] = -INF;
Dfs(1, 0), m = Input();
for(RG int i = 1; i <= m; ++i){
RG int op = Input(), x = Input(), ff, v, &c = col[x];
if(op == 1) T[c].Cut(x, fa[x]), c ^= 1, T[c].Link(x, fa[x]);
else if(op == 2){
v = Input(), T[c].Access(x), T[c].Splay(x);
w[x] = v, T[c].Update(x);
}
else{
T[c].Access(x), ff = T[c].Findroot(x);
if(col[ff] == c) printf("%d
", T[c].mxv[ff]);
else printf("%d
", T[c].mxv[T[c].ch[1][ff]]);
}
}
return 0;
}