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\n", T[c].mxv[ff]);
            else printf("%d\n", T[c].mxv[T[c].ch[1][ff]]);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/cjoieryl/p/8653038.html