Qtree1

将边权变为这条边连接的两个点中深度更深的点的点权，这样就可以变为带修改链上最大点权。直接树链剖分即可。

下面是一份C语言代码

#include<stdio.h>
#include<string.h>
#define MAXN 10001
inline int read(){int a = 0;int f = 0;char c = getchar();while(!isdigit(c)){if(c == '-')f = 1;c = getchar();}while(isdigit(c)){a = (a << 3) + (a << 1) + (c ^ '0');c = getchar();}return f ? -a : a;
}char output[20];
inline void print(int x){int dirN = 18;if(x == 0)fwrite("0" , sizeof(char) , 1 , stdout);else{if(x < 0){x = -x;fwrite("-" , sizeof(char) , 1 , stdout);}while(x){output[--dirN] = x % 10 + 48;x /= 10;}fwrite(output + dirN , 1 , strlen(output + dirN) , stdout);}fwrite("\n" , 1 , 1 , stdout);
}inline int max(int a , int b){return a > b ? a : b;
}struct node{int l , r , maxN;
}Tree[MAXN << 2];
struct Edge{int end , upEd , w;
}Ed[MAXN << 1];
int head[MAXN] , start[MAXN << 1] , size[MAXN] , son[MAXN] , fa[MAXN] , dep[MAXN];
int top[MAXN] , ind[MAXN] , rk[MAXN] , val[MAXN] , N , cntEd , ts;inline void addEd(int a , int b , int c){Ed[++cntEd].end = b;Ed[cntEd].upEd = head[a];Ed[cntEd].w = c;head[a] = cntEd;
}void dfs1(int dir , int father , int w){val[dir] = w;dep[dir] = dep[fa[dir] = father] + 1;size[dir] = 1;int i;for(i = head[dir] ; i ; i = Ed[i].upEd)if(!dep[Ed[i].end]){dfs1(Ed[i].end , dir , Ed[i].w);size[dir] += size[Ed[i].end];if(size[son[dir]] < size[Ed[i].end])son[dir] = Ed[i].end;}
}void dfs2(int dir , int t){top[dir] = t;rk[ind[dir] = ++ts] = dir;if(!son[dir])return;dfs2(son[dir] , t);int i;for(i = head[dir] ; i ; i = Ed[i].upEd)if(Ed[i].end != fa[dir] && Ed[i].end != son[dir])dfs2(Ed[i].end , Ed[i].end);
}inline void pushup(int dir){Tree[dir].maxN = max(Tree[dir << 1].maxN , Tree[dir << 1 | 1].maxN);
}void init(int dir , int l , int r){Tree[dir].l = l;Tree[dir].r = r;if(l == r)Tree[dir].maxN = val[rk[l]];else{init(dir << 1 , l , l + r >> 1);init(dir << 1 | 1 , (l + r >> 1) + 1 , r);pushup(dir);}
}void change(int dir , int tar , int val){if(Tree[dir].l == Tree[dir].r){Tree[dir].maxN = val;return;}if(Tree[dir].l + Tree[dir].r >> 1 >= tar)change(dir << 1 , tar , val);elsechange(dir << 1 | 1 , tar , val);pushup(dir);
}int findMax(int dir , int l , int r){if(Tree[dir].l >= l && Tree[dir].r <= r)return Tree[dir].maxN;int maxN = 0;if(l <= Tree[dir].l + Tree[dir].r >> 1)maxN = max(maxN , findMax(dir << 1 , l , r));if(r > Tree[dir].l + Tree[dir].r >> 1)maxN = max(maxN , findMax(dir << 1 | 1 , l , r));return maxN;
}inline void work(int x , int y){if(x == y){print(0);return;}int tx = top[x] , ty = top[y] , maxN = -0x7fffffff;while(tx != ty)if(dep[tx] >= dep[ty]){maxN = max(maxN , findMax(1 , ind[tx] , ind[x]));x = fa[tx];tx = top[x];}else{maxN = max(maxN , findMax(1 , ind[ty] , ind[y]));y = fa[ty];ty = top[y];}if(ind[x] < ind[y])maxN = max(maxN , findMax(1 , ind[x] + 1 , ind[y]));if(ind[y] < ind[x])maxN = max(maxN , findMax(1 , ind[y] + 1 , ind[x]));print(maxN);
}int cmp(int a , int b){return dep[a] < dep[b] ? ind[b] : ind[a];
}char s[10];
int main(){int T;for(T = read() ; T ; T--){N = read();memset(head , 0 , sizeof(head));memset(dep , 0 , sizeof(dep));memset(start , 0 , sizeof(start));memset(size , 0 , sizeof(size));memset(son , 0 , sizeof(son));memset(fa , 0 , sizeof(fa));memset(top , 0 , sizeof(top));memset(ind , 0 , sizeof(ind));memset(rk , 0 , sizeof(rk));memset(val , 0 , sizeof(val));cntEd = ts = 0;int i;for(i = 1 ; i < N ; i++){start[(i << 1) - 1] = read();start[i << 1] = read();int c = read();addEd(start[(i << 1) - 1] , start[i << 1] , c);addEd(start[i << 1] , start[(i << 1) - 1] , c);}dfs1(1 , 0 , 0);dfs2(1 , 1);init(1 , 1 , N);while(scanf("%s" , s) && s[0] != 'D'){int a = read() , b = read();if(s[0] == 'Q')work(a , b);elsechange(1 , cmp(start[a << 1] , start[(a << 1) - 1]) , b);}}return 0;
}

Qtree2

同样先把边权变为较深的点的点权，那么这两个操作都可以通过倍增+维护倍增经过的所有点的点权和完成。

    #include<bits/stdc++.h>//This code is written by Itstusing namespace std;inline int read(){int a = 0;bool f = 0;char c = getchar();while(c != EOF && !isdigit(c)){if(c == '-')f = 1;c = getchar();}while(c != EOF && isdigit(c)){a = (a << 3) + (a << 1) + (c ^ '0');c = getchar();}return f ? -a : a;}const int MAXN = 10010;struct Edge{int end , upEd , w;}Ed[MAXN << 1];int jump[MAXN][21][2] , head[MAXN] , dep[MAXN];int N , cntEd;inline void addEd(int a , int b , int c){Ed[++cntEd].end = b;Ed[cntEd].upEd = head[a];Ed[cntEd].w = c;head[a] = cntEd;}void dfs(int x , int f){jump[x][0][0] = f;dep[x] = dep[f] + 1;for(int i = 1 ; jump[x][i - 1][0] ; ++i){jump[x][i][0] = jump[jump[x][i - 1][0]][i - 1][0];jump[x][i][1] = jump[x][i - 1][1] + jump[jump[x][i - 1][0]][i - 1][1];}for(int i = head[x] ; i ; i = Ed[i].upEd)if(Ed[i].end != f){jump[Ed[i].end][0][1] = Ed[i].w;dfs(Ed[i].end , x);}}inline pair < int , int > LCA(int x , int y){int sum = 0;if(dep[x] < dep[y])swap(x , y);for(int i = 20 ; i >= 0 ; --i)if(dep[x] - (1 << i) >= dep[y]){sum += jump[x][i][1];x = jump[x][i][0];}if(x == y)return make_pair(x , sum);for(int i = 20 ; i >= 0 ; --i)if(jump[x][i][0] != jump[y][i][0]){sum += jump[x][i][1] + jump[y][i][1];x = jump[x][i][0];y = jump[y][i][0];}return make_pair(jump[x][0][0] , sum + jump[x][0][1] + jump[y][0][1]);}inline int Kth(int x , int y , int k){int t = LCA(x , y).first;if(dep[x] - dep[t] + 1 >= k){--k;for(int i = 16 ; i >= 0 ; --i)if(k & (1 << i))x = jump[x][i][0];return x;}else{k = dep[x] + dep[y] - (dep[t] << 1) + 1 - k;for(int i = 16 ; i >= 0 ; --i)if(k & (1 << i))y = jump[y][i][0];return y;}}inline char getc(){char c = getchar();while(!isupper(c))c = getchar();return c = getchar();}int main(){for(int T = read() ; T ; --T){memset(head , 0 , sizeof(head));memset(jump , 0 , sizeof(jump));cntEd = 0;N = read();for(int i = 1 ; i < N ; ++i){int a = read() , b = read() , c = read();addEd(a , b , c);addEd(b , a , c);}dfs(1 , 0);int a , b , c;bool f = 1;while(f)switch(getc()){case 'I':printf("%d\n" , LCA(read() , read()).second);break;case 'T':a = read();b = read();c = read();printf("%d\n" , Kth(a , b , c));break;default:f = 0;}cout << endl;}return 0;}

Qtree3

一样是树链剖分，线段树上每一个点维护这一段区间中dfn序最大的黑点的dfn序。

#include<bits/stdc++.h>
#define MAXN 100001
using namespace std;namespace IO{const int maxn((1 << 21) + 1);char ibuf[maxn], *iS, *iT, obuf[maxn], *oS = obuf, *oT = obuf + maxn - 1, c, st[55];int f, tp;char Getc() {return (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), (iS == iT ? EOF : *iS++)) : *iS++);}void Flush() {fwrite(obuf, 1, oS - obuf, stdout);oS = obuf;}void Putc(char x) {*oS++ = x;if (oS == oT) Flush();}template <class Int> void Input(Int &x) {for (f = 1, c = Getc(); c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;for (x = 0; c <= '9' && c >= '0'; c = Getc()) x = (x << 3) + (x << 1) + (c ^ 48);x *= f;}template <class Int> void Print(Int x) {if (!x) Putc('0');if (x < 0) Putc('-'), x = -x;while (x) st[++tp] = x % 10 + '0', x /= 10;while (tp) Putc(st[tp--]);}void Getstr(char *s, int &l) {for (c = Getc(); c < 'a' || c > 'z'; c = Getc());for (l = 0; c <= 'z' && c >= 'a'; c = Getc()) s[l++] = c;s[l] = 0;}void Putstr(const char *s) {for (int i = 0, n = strlen(s); i < n; ++i) Putc(s[i]);}
}
using namespace IO;struct node{int l , r , minN;
}Tree[MAXN << 2];
struct Edge{int end , upEd;
}Ed[MAXN << 1];
int son[MAXN] , size[MAXN] , fa[MAXN] , dep[MAXN] , head[MAXN];
int top[MAXN] , ind[MAXN] , rk[MAXN] , N , cntEd , ts;inline void addEd(int a , int b){Ed[++cntEd].end = b;Ed[cntEd].upEd = head[a];head[a] = cntEd;
}void dfs1(int dir , int father){size[dir] = 1;dep[dir] = dep[fa[dir] = father] + 1;for(int i = head[dir] ; i ; i = Ed[i].upEd)if(!dep[Ed[i].end]){dfs1(Ed[i].end , dir);size[dir] += size[Ed[i].end];if(size[son[dir]] < size[Ed[i].end])son[dir] = Ed[i].end;}
}void dfs2(int dir , int t){top[dir] = t;rk[ind[dir] = ++ts] = dir;if(!son[dir])return;dfs2(son[dir] , t);for(int i = head[dir] ; i ; i = Ed[i].upEd)if(Ed[i].end != son[dir] && Ed[i].end != fa[dir])dfs2(Ed[i].end , Ed[i].end);
}inline int min(int a , int b){return a < b ? a : b;
}void init(int dir , int l , int r){Tree[dir].l = l;Tree[dir].r = r;if(l == r)Tree[dir].minN = 999999;else{init(dir << 1 , l , (l + r) >> 1);init(dir << 1 | 1 , ((l + r) >> 1) + 1 , r);Tree[dir].minN = min(Tree[dir << 1].minN , Tree[dir << 1 | 1].minN);}
}void change(int dir , int tar){if(Tree[dir].l == Tree[dir].r){Tree[dir].minN = Tree[dir].minN == 999999 ? Tree[dir].l : 999999;return;}if(tar <= (Tree[dir].l + Tree[dir].r) >> 1)change(dir << 1 , tar);elsechange(dir << 1 | 1 , tar);Tree[dir].minN = min(Tree[dir << 1].minN , Tree[dir << 1 | 1].minN);
}int findMin(int dir , int l , int r){if(Tree[dir].l >= l && Tree[dir].r <= r)return Tree[dir].minN;int minN;if(l <= (Tree[dir].l + Tree[dir].r) >> 1){minN = findMin(dir << 1 , l , r);if(minN != 999999)return minN;}if(r > (Tree[dir].l + Tree[dir].r) >> 1)return findMin(dir << 1 | 1 , l , r);return 999999;
}inline int work(int tar){int minN = 999999;while(top[tar] != 1){minN = min(minN , findMin(1 , ind[top[tar]] , ind[tar]));tar = fa[top[tar]];}minN = min(minN , findMin(1 , 1 , ind[tar]));return minN == 999999 ? -1 : rk[minN];
}int main(){int N , M;Input(N);Input(M);for(int i = 1 ; i < N ; i++){int a , b;Input(a);Input(b);addEd(a , b);addEd(b , a);}dfs1(1 , 0);dfs2(1 , 1);init(1 , 1 , N);while(M--){int a;Input(a);if(a == 0){Input(a);change(1 , ind[a]);}else{Input(a);Print(work(a));Putc('\n');}}Flush();return 0;
}

Qtree4

在动态点分的学习笔记里面

Qtree5

仍然是动态点分，对于每一个点维护其所有点分树子树中的白点到当前点的最短距离。修改和查询都暴力跳点分树。

注意到从一个子树中走到分治中心再走回去一定不优，所以并不需要维护到父亲的堆。

#include<bits/stdc++.h>
#define INF (int)1e9
//This code is written by Itst
using namespace std;inline int read(){int a = 0;bool f = 0;char c = getchar();while(c != EOF && !isdigit(c)){if(c == '-')f = 1;c = getchar();}while(c != EOF && isdigit(c)){a = (a << 3) + (a << 1) + (c ^ '0');c = getchar();}return f ? -a : a;
}const int MAXN = 100010;
struct Edge{int end , upEd;
}Ed[MAXN << 1];
int head[MAXN] , dep[MAXN] , fir[MAXN] , ST[21][MAXN << 1] , logg2[MAXN << 1] , fa[MAXN][20] , size[MAXN] , dis[MAXN][20];
int N , nowSize , minSize , minInd , cntST , cntEd;
bool vis[MAXN] , col[MAXN];
struct pq{priority_queue < int , vector < int > , greater < int > > q1 , q2;inline void maintain(){while(!q1.empty() && !q2.empty() && q1.top() == q2.top()){q1.pop();q2.pop();}}inline void push(int x){q1.push(x);}inline void pop(int x){q2.push(x);}inline int top(){maintain();return q1.empty() ? INF : q1.top();}}cur[MAXN];inline void addEd(int a , int b){Ed[++cntEd].end = b;Ed[cntEd].upEd = head[a];head[a] = cntEd;
}void init_dfs(int x , int pre){dep[x] = dep[pre] + 1;fir[x] = ++cntST;ST[0][cntST] = x;for(int i = head[x] ; i ; i = Ed[i].upEd)if(Ed[i].end != pre){init_dfs(Ed[i].end , x);ST[0][++cntST] = x;}
}inline int cmp(int a , int b){return dep[a] < dep[b] ? a : b;
}void init_st(){for(int i = 2 ; i <= cntST ; ++i)logg2[i] = logg2[i >> 1] + 1;for(int i = 1 ; 1 << i <= cntST ; ++i)for(int j = 1 ; j + (1 << i) - 1 <= cntST ; ++j)ST[i][j] = cmp(ST[i - 1][j] , ST[i - 1][j + (1 << (i - 1))]);
}inline int LCA(int x , int y){x = fir[x];y = fir[y];if(x > y)swap(x , y);int t = logg2[y - x + 1];return cmp(ST[t][x] , ST[t][y - (1 << t) + 1]);
}inline int calcLen(int x , int y){return dep[x] + dep[y] - (dep[LCA(x , y)] << 1);
}void getSize(int x){vis[x] = 1;++nowSize;for(int i = head[x] ; i ; i = Ed[i].upEd)if(!vis[Ed[i].end])getSize(Ed[i].end);vis[x] = 0;
}void getRoot(int x){int maxN = 0;vis[x] = size[x] = 1;for(int i = head[x] ; i ; i = Ed[i].upEd)if(!vis[Ed[i].end]){getRoot(Ed[i].end);size[x] += size[Ed[i].end];maxN = max(maxN , size[Ed[i].end]);}maxN = max(maxN , nowSize - size[x]);if(maxN < minSize){minSize = maxN;minInd = x;}vis[x] = 0;
}void init_dfz(int x , int p){nowSize = 0;minSize = 0x7fffffff;getSize(x);getRoot(x);x = minInd;vis[x] = 1;fa[x][0] = p;for(int i = 0 ; fa[x][i] ; ++i){fa[x][i + 1] = fa[fa[x][i]][0];dis[x][i] = calcLen(x , fa[x][i]);}for(int i = head[x] ; i ; i = Ed[i].upEd)if(!vis[Ed[i].end])init_dfz(Ed[i].end , x);
}void init(){init_dfs(1 , 0);init_st();init_dfz(1 , 0);
}inline int query(int x){int minN = cur[x].top();for(int i = 0 ; fa[x][i] ; ++i)minN = min(minN , cur[fa[x][i]].top() + dis[x][i]);return minN == INF ? -1 : minN;
}inline void modify(int x){vis[x] ^= 1;vis[x] ? cur[x].pop(0) : cur[x].push(0);for(int i = 0 ; fa[x][i] ; ++i)vis[x] ? cur[fa[x][i]].pop(dis[x][i]) : cur[fa[x][i]].push(dis[x][i]);
}int main(){N = read();for(int i = 1 ; i < N ; ++i){int a = read() , b = read();addEd(a , b);addEd(b , a);}init();for(int M = read() ; M ; --M)if(read())printf("%d\n" , query(read()));elsemodify(read());return 0;
}

Qtree6

考虑LCT。维护点的颜色连通块似乎不好做，因为是否连通实际上是和边相关的一个东西。我们考虑每一个点，当这个点是黑色的时候在黑色的LCT上将它和它的父亲连边，否则在白色的LCT上连边。这样对于黑白两色的LCT，去掉每一个连通块的root后，剩余的每一个连通块就是当前颜色的一个颜色连通块。

那么我们按照这个进行Link、Cut，并在每一次询问的时候先access保证当前点在整棵树的实链上，然后findroot并将root splay至根，最后输出根的右儿子的size即可。

值得注意的是在Link和Cut中不能够进行平常需要在其中进行的makeroot操作，因为这样会改变每一个连通块的根。我们可以这样修改：

Link：每一次进行Link操作的点都一定是当前连通块的根，所以直接splay至LCT的根即可；

Cut：因为每一次要cut的边连接的两个点的父子关系确定，所以可以不进行makeroot，直接对子节点access、splay并去掉对应的边。

至于如何维护子树size，可以维护一个变量表示当前点的虚子树size和，不难发现只有在link和access的时候会改变这个量，这样就可以进行整个子树的维护。值得注意的是维护虚子树信息在link的时候要将父节点先access、splay到LCT的根，这样可以避免更新子树大小之后的一系列pushup。

#include<bits/stdc++.h>
using namespace std;int read(){int a = 0; char c = getchar(); bool f = 0;while(!isdigit(c)){f = c == '-'; c = getchar();}while(isdigit(c)){a = a * 10 + c - 48; c = getchar();}return f ? -a : a;
}const int _ = 2e5 + 3;
int fa[_] , ch[_][2] , sz[_] , vir[_]; bool rmrk[_];struct Edge{int end , upEd;}Ed[_ << 1];
int pre[_] , head[_] , col[_] , cntEd , N , M;
void addEd(int a , int b){Ed[++cntEd] = (Edge){b , head[a]}; head[a] = cntEd;}
void dfs(int x , int p){pre[x] = p; for(int i = head[x] ; i ; i = Ed[i].upEd) if(Ed[i].end != p) dfs(Ed[i].end , x);}bool nroot(int x){return ch[fa[x]][0] == x || ch[fa[x]][1] == x;}
bool son(int x){return ch[fa[x]][1] == x;}
void up(int x){sz[x] = sz[ch[x][0]] + sz[ch[x][1]] + 1 + vir[x];}void rot(int x){bool f = son(x); int y = fa[x] , z = fa[y] , w = ch[x][f ^ 1];fa[x] = z; if(nroot(y)) ch[z][son(y)] = x;fa[y] = x; ch[x][f ^ 1] = y;ch[y][f] = w; if(w) fa[w] = y;up(y);
}void splay(int x){while(nroot(x)){if(nroot(fa[x])) rot(son(x) == son(fa[x]) ? fa[x] : x); rot(x);} up(x);}
void access(int x){for(int y = 0 ; x ; y = x , x = fa[x]){splay(x); vir[x] = vir[x] - sz[y] + sz[ch[x][1]];ch[x][1] = y; up(x);}
}
int fdrt(int x){access(x); splay(x); while(ch[x][0]) x = ch[x][0]; splay(x); return x;}
void link(int x){splay(x); int y = fa[x] = pre[x]; access(y); splay(y); vir[y] += sz[x]; up(y);}
void cut(int x){access(x); splay(x); ch[x][0] = fa[ch[x][0]] = 0; up(x);}
int qry(int x){x += col[x] * N; access(x); splay(x); splay(x = fdrt(x)); return sz[ch[x][1]];}int main(){N = read();for(int i = 1 ; i < N ; ++i){int a = read() , b = read(); addEd(a , b); addEd(b , a);}dfs(1 , 0); pre[1] = ++N;for(int i = 1 ; i <= 2 * N ; ++i) sz[i] = 1;for(int i = 1 ; i < N ; ++i){link(i); pre[i + N] = pre[i] + N;}for(M = read() ; M ; --M)if(read()){int id = read(); cut(id + N * col[id]); link(id + N * (col[id] ^= 1));}else printf("%d\n" , qry(read()));return 0;
}

Qtree7

考虑在Qtree6的基础上支持子树最值。

考虑维护所有虚子树的最大值，但是注意到我们还可能进行虚实转换使得虚子树变成实子树、不对当前点的虚子树max产生贡献，那么就不能够只使用一个标记维护max。

考虑使用multiset维护一个点的虚子树的max构成的集合，那么在虚实更换的时候对这个multiset进行插入、删除即可进行更新。

复杂度\(O(qlog^2n)\)

#include<bits/stdc++.h>
using namespace std;int read(){int a = 0; char c = getchar(); bool f = 0;while(!isdigit(c)){f = c == '-'; c = getchar();}while(isdigit(c)){a = a * 10 + c - 48; c = getchar();}return f ? -a : a;
}const int _ = 2e5 + 7;
struct Edge{int end , upEd;}Ed[_ << 1];
int head[_] , N , cntEd , pre[_] , col[_];void addEd(int a , int b){Ed[++cntEd] = (Edge){b , head[a]}; head[a] = cntEd;}
void dfs(int x , int p){pre[x] = p; for(int i = head[x] ; i ; i = Ed[i].upEd) if(Ed[i].end != p) dfs(Ed[i].end , x);}int fa[_] , ch[_][2] , val[_] , mx[_]; multiset < int > vir[_];bool nroot(int x){return ch[fa[x]][0] == x || ch[fa[x]][1] == x;}
bool son(int x){return ch[fa[x]][1] == x;}
void up(int x){mx[x] = max(max(val[x] , vir[x].size() ? *--vir[x].end() : (int)-2e9) , max(mx[ch[x][0]] , mx[ch[x][1]]));}void rot(int x){bool f = son(x); int y = fa[x] , z = fa[y] , w = ch[x][f ^ 1];fa[x] = z; if(nroot(y)) ch[z][son(y)] = x;fa[y] = x; ch[x][f ^ 1] = y;ch[y][f] = w; if(w) fa[w] = y;up(y);
}void splay(int x){while(nroot(x)){if(nroot(fa[x])) rot(son(x) == son(fa[x]) ? fa[x] : x); rot(x);} up(x);}void access(int x){for(int y = 0 ; x ; y = x , x = fa[x]){splay(x); if(ch[x][1]) vir[x].insert(mx[ch[x][1]]);if(y) vir[x].erase(vir[x].find(mx[y]));ch[x][1] = y; up(x);}
}int fdrt(int x){access(x); splay(x); while(ch[x][0]) x = ch[x][0]; splay(x); return x;}
void link(int x){splay(x); int t = pre[x]; access(t); splay(t); fa[x] = t; vir[t].insert(mx[x]); up(t);}
void cut(int x){access(x); splay(x); ch[x][0] = fa[ch[x][0]] = 0; up(x);}
void mdy(int x , int w){access(x); splay(x); val[x] = w; up(x);}
int qry(int x){access(x); splay(x); x = fdrt(x); return mx[ch[x][1]];}int main(){mx[0] = -2e9; N = read();for(int i = 1 ; i < N ; ++i){int a = read() , b = read(); addEd(a , b); addEd(b , a);}dfs(1 , 0); pre[1] = ++N;for(int i = 1 ; i < N ; ++i) col[i] = read();for(int i = 1 ; i < N ; ++i) val[i] = val[i + N] = mx[i] = mx[i + N] = read();for(int i = 1 ; i < N ; ++i) pre[i + N] = pre[i] + N;for(int i = 1 ; i < N ; ++i) link(i + col[i] * N);for(int M = read() ; M ; --M){int tp = read() , u = read();if(tp == 0) printf("%d\n" , qry(u + col[u] * N));else if(tp == 1){cut(u + col[u] * N); link(u + (col[u] ^= 1) * N);}else{int w = read(); mdy(u , w); mdy(u + N , w);}}return 0;
}