题解

可以发现每次修改的是这个点往上一条连续的链，如果我要把1改成0，需要满足这一段往上的一部分都有两个1
如果我要把0改成1，需要满足这一段往上的部分有两个0
对于每个点记录1的个数，发现我们只会把一棵树的2全部改成1或者把1全部改成2，这样加标记的时候可以同时维护是否全1或者是否全2，用lct维护，修改的时候access一遍，直接在平衡树上二分即可

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 1000005
#define eps 1e-10
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10);
}
struct node {int lc,rc,fa,val,lz;bool all[2];
}tr[MAXN];
int num[MAXN * 2],N,Q,fa[MAXN * 2],d[2];
vector<int> son[MAXN];
bool isRoot(int u) {if(!tr[u].fa) return true;return tr[tr[u].fa].lc != u && tr[tr[u].fa].rc != u;
}
bool which(int u) {return tr[tr[u].fa].lc == u;
}void addlz(int u,int v) {tr[u].val += v;tr[u].lz += v;tr[u].all[0] = tr[u].all[1] = 0;if(v == -1) tr[u].all[0] = 1;if(v == 1) tr[u].all[1] = 1;
}
void pushdown(int u) {if(tr[u].lz) {if(tr[u].lc) addlz(tr[u].lc,tr[u].lz);if(tr[u].rc) addlz(tr[u].rc,tr[u].lz);tr[u].lz = 0;}
}
void update(int u) {for(int i = 0 ; i <= 1 ; ++i) {tr[u].all[i] = (tr[u].val == i + 1) & tr[tr[u].lc].all[i] & tr[tr[u].rc].all[i];}
}
void Rotate(int u) {int v = tr[u].fa,w = tr[v].fa;if(!isRoot(v)) {(v == tr[w].lc ? tr[w].lc : tr[w].rc) = u;}int b = (u == tr[v].lc ? tr[u].rc : tr[u].lc);tr[u].fa = w;tr[v].fa = u;if(b) tr[b].fa = v;if(u == tr[v].lc) {tr[u].rc = v;tr[v].lc = b;}else {tr[u].lc = v;tr[v].rc = b;}update(v);
}
void Splay(int u) {static int que[MAXN],tot;tot = 0;int x;for(x = u ; !isRoot(x) ; x = tr[x].fa) {que[++tot] = x;}que[++tot] = x;for(int i = tot ; i >= 1 ; --i) {pushdown(que[i]);}while(!isRoot(u)) {if(!isRoot(tr[u].fa)) {if(which(tr[u].fa) == which(u)) Rotate(tr[u].fa);else Rotate(u);}Rotate(u);}update(u);
}
void Access(int u) {for(int x = 0 ; u ; x = u, u = tr[u].fa) {Splay(u);tr[u].rc = x;update(u);}
}
void dfs(int u) {for(int j = 0 ; j < 3 ; ++j) {if(son[u][j] <= N) {dfs(son[u][j]);tr[u].val += (tr[son[u][j]].val >= 2);tr[son[u][j]].fa = u;}else {tr[u].val += num[son[u][j] - N];fa[son[u][j] - N] = u;}}if(tr[u].val == 1) tr[u].all[0] = 1;if(tr[u].val == 2) tr[u].all[1] = 1;
}
void Init() {read(N);int a;for(int i = 1 ; i <= N ; ++i) {for(int j = 0 ; j < 3 ; ++j) {read(a);son[i].pb(a);}}for(int i = 1 ; i <= 2 * N + 1 ; ++i) read(num[i]);dfs(1);
}
void Solve() {read(Q);int v;d[0] = 1,d[1] = -1;tr[0].all[0] = tr[0].all[1] = 1;while(Q--) {read(v);v -= N;Access(fa[v]);Splay(fa[v]);if(tr[fa[v]].all[num[v]]) {addlz(fa[v],d[num[v]]);}else {int p = fa[v],res;while(1) {res = p;pushdown(p);if(tr[p].rc && !tr[tr[p].rc].all[num[v]]) {p = tr[p].rc;continue;}if(tr[p].val != num[v] + 1) break;if(tr[p].lc && !tr[tr[p].lc].all[num[v]]) {p = tr[p].lc;continue;}break;}Splay(res);if(tr[res].rc) addlz(tr[res].rc,d[num[v]]);tr[res].val += d[num[v]];update(res);}num[v] ^= 1;Splay(1);out(tr[1].val >= 2);enter;}
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifInit();Solve();
}