We can answer the queries offline (meaning that we process them in an order that is convenient for us, not that which is given by the input). Like the silver version, run a DFS from farm $1$. Suppose that we're currently processing the farm $x.$ Store a stack $ord$ containing all the nodes on the path from $1$ to $x,$ and also a stack $stor[t]$ for each type $t$ containing the pairs $(y,depth[y])$ for all farms $y$ with type $t$ on the path from $1$ to $x.$

Suppose that we want to update the answer for a query $i$ having an endpoint $A_i=x$ (the reasoning for $B_i=x$ is similar). We need to check whether the last farm $y$ in the stack corresponding to $C_i$ actually lies on the path from $A_i$ to $B_i$. Let $L$ be the least common ancestor of $A_i$ and $B_i.$ First, we can check whether $y$ is an ancestor of $B_i$ in $O(1)$ using the preorder and postorder traversals of the tree. If not, then $y$ lies on the path between $A_i$ and $L_i,$ so the answer to this query must be 1. If yes, then it's still possible for $y$ to lie on the path from $A_i$ to $B_i$ if $y=L_i$.

One option is to actually find $L_i$ and compare it to $y,$ but this is unnecessary. Instead, if $y\neq A_i$ then consider the farm $Y=ord[depth[y]+1].$ If $Y$ is an ancestor of $B_i,$ then $y$ is clearly not the LCA. Otherwise, $A_i$ and $B_i$ lie in different child subtrees of $y,$ implying that $y$ is the LCA.

Thus, this problem is solvable in $O(N+M)$ time.

#include <bits/stdc++.h>
using namespace std;
 
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vpi;
 
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)
 
#define mp make_pair
#define pb push_back
#define f first
#define s second
 
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rsz resize
 
const int MX = 100005;

void setIO(string name) {
    ios_base::sync_with_stdio(0); cin.tie(0);
    freopen((name+".in").c_str(),"r",stdin);
    freopen((name+".out").c_str(),"w",stdout);
}

int N,M,T[MX],C[MX];
bool ok[MX];
vi adj[MX];
array<int,2> dat[MX];
vi todo[MX];
pi range[MX];
 
int co = 0;
void dfs(int x, int y) {
	range[x].f = co++;
	trav(t,adj[x]) if (t != y) dfs(t,x);
	range[x].s = co-1;
}
 
vpi stor[MX];
vi ord;
 
bool anc(int a, int b) {
	return range[a].f <= range[b].f && range[b].s <= range[a].s;
}
 
void dfs2(int x, int y) {
	stor[T[x]].pb({x,sz(ord)}); ord.pb(x);
	trav(t,todo[x]) if (sz(stor[C[t]])) {
		pi y = stor[C[t]].back();
		if (y.f == x) ok[t] = 1;
		else {
			int Y = ord[y.s+1];
			// x is one of endpoints for query t
			assert(dat[t][0] == x || dat[t][1] == x); 
			if (!anc(Y,dat[t][0]+dat[t][1]-x)) ok[t] = 1;
		}
	}
	trav(t,adj[x]) if (t != y) dfs2(t,x);
	stor[T[x]].pop_back(); ord.pop_back();
}
 
int main() {
	setIO("milkvisits"); 
	cin >> N >> M;
	FOR(i,1,N+1) cin >> T[i];
	F0R(i,N-1) {
		int A,B; cin >> A >> B;
		adj[A].pb(B), adj[B].pb(A);
	}
	dfs(1,0);
	F0R(i,M) {
		cin >> dat[i][0] >> dat[i][1] >> C[i];
		F0R(j,2) todo[dat[i][j]].pb(i);
	}
	dfs2(1,0);
	F0R(i,M) {
		if (ok[i]) cout << 1;
		else cout << 0;
	}
	cout << "\n";
}

Bonus: solve the problem online in $O((N+M)\log N).$