BZOJ3639 Query on a tree VII

原题链接：https://www.lydsy.com/JudgeOnline/problem.php?id=3639

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 Wi, of each node(|Wi| ≤ 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

题目大意

询问同色联通块内的最大值，支持修改颜色，修改权值。

题解

前置知识：维护联通块。

Qtree\mathcal{Qtree}Qtree系列，板子题一家。

维护联通块的方法见上面链接，剩下的就是维护一个子树的最大值了。此时，如果仅仅维护左右儿子的最大值，显然是不行的，Splay\mathcal{Splay}Splay时也没法继续维护。所以，我们需要给每个节点开一个multiset\mathcal{multiset}multiset，将所有虚子树的最大值都丢进去，就可以做到O(log2n)O(log_2n)O(log2n)更新了。

其他维护大同小异。

借此学习一波STL\mathcal{STL}STL姿势。

代码

#include<bits/stdc++.h>
#define ls son[v][0]
#define rs son[v][1]
#define inf INT_MAX
#define C lct[col[a]]
using namespace std;
const int M=1e5+5,N=M<<1;
int col[M],fa[M],val[M],n,m;
vector<int>mmp[M];
struct LCT{int son[N][2],dad[N],mx[N];multiset<int>s[N];LCT(){mx[0]=-inf;}bool notroot(int v){return son[dad[v]][0]==v||son[dad[v]][1]==v;}void up(int v){mx[v]=max(val[v],max(mx[ls],mx[rs]));if(!s[v].empty())mx[v]=max(mx[v],*s[v].rbegin());}void spin(int v){int f=dad[v],ff=dad[f],k=son[f][1]==v,w=son[v][!k];if(notroot(f))son[ff][son[ff][1]==f]=v;son[v][!k]=f;son[f][k]=w;if(w)dad[w]=f;dad[f]=v;dad[v]=ff;up(f);}void splay(int v){int f,ff;while(notroot(v)){f=dad[v];ff=dad[f];if(notroot(f))spin((son[f][0]==v)^(son[ff][0]==f)?v:f);spin(v);}up(v);}void access(int v){for(int f=0;v;v=dad[f=v]){splay(v);if(rs)s[v].insert(mx[rs]);if(rs=f)s[v].erase(s[v].find(mx[f]));up(v);}}int root(int v){access(v);splay(v);while(ls)v=ls;splay(v);return v;}void link(int v){splay(v);int f=dad[v]=fa[v];access(f);splay(f);son[f][1]=v;up(f);}void cut(int v){access(v);splay(v);ls=dad[ls]=0;up(v);}
}lct[2];
void dfs(int v,int f)
{fa[v]=f;int to;for(int i=mmp[v].size()-1;i>=0;--i){to=mmp[v][i];if(to==f)continue;dfs(to,v);}lct[col[v]].link(v);
}
void in()
{int a,b;scanf("%d",&n);for(int i=1;i<n;++i)scanf("%d%d",&a,&b),mmp[a].push_back(b),mmp[b].push_back(a);for(int i=1;i<=n;++i)scanf("%d",&col[i]);for(int i=1;i<=n;++i)scanf("%d",&val[i]);
}
void ac()
{int op,a;dfs(1,n+1);scanf("%d",&m);for(int i=1;i<=m;++i){scanf("%d%d",&op,&a);switch(op){case 0:printf("%d\n",C.mx[C.son[C.root(a)][1]]);break;case 1:C.cut(a);col[a]^=1;C.link(a);break;case 2:C.access(a);C.splay(a);scanf("%d",&val[a]);C.up(a);}}
}
int main(){in();ac();}