像我这种写法的，$\operatorname{get\ rt}$ 函数当传入的 $u$ 不等于最终的 $rt$ 时，则 $sz$ 数组是以 $u$ 为根而非以 $rt$ 为根，那么在 $\operatorname{divide}$ 函数中再次 $\operatorname{get \ rt}$ 前，若枚举到的 $v$ 恰为 $u$ 的父亲，则 $sum$ 的初始值就不是当前子树的大小了。可这样仍然可以 $AC$，这是为什么？

#include <iostream>
using namespace std;
typedef long long ll;
const ll N = 1e4 + 10, M = 1e2 + 10, INF = 1e7 + 10;
ll n, m, u, v, w, cnt, tot, Min, sum, rt, head[N], sz[N], dis[N], d[N], q[INF], ask[M];
bool del[N], ans[M], judge[INF];
ll rd() {
	ll sum = 0; bool f = 0; char ch = getchar();
	while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
	while (ch >= '0' && ch <= '9') sum = (sum << 1) + (sum << 3) + (ch ^ 48), ch = getchar();
	return f ? -sum : sum;
}
struct Edge { ll nxt, to, w; } e[N << 1];
void add(ll u, ll v, ll w) { e[++ cnt] = {head[u], v, w}; head[u] = cnt; }
void get_rt(ll u, ll pre) {
	sz[u] = 1; ll mx = 0;
	for (ll i = head[u], v; i; i = e[i].nxt) {
		v = e[i].to;
		if (v == pre || del[v]) continue;
		get_rt(v, u);
		sz[u] += sz[v], mx = max(mx, sz[v]);
	}
	mx = max(mx, sum - sz[u]);
	if (mx < Min) Min = mx, rt = u;
}
void get_dis(ll u, ll pre) {
	dis[++ tot] = d[u];
	for (ll i = head[u], v; i; i = e[i].nxt) {
		v = e[i].to;
		if (v == pre || del[v]) continue;
		d[v] = d[u] + e[i].w;
		get_dis(v, u);
	}
}
void calc(ll u) {
	del[u] = judge[0] = 1; ll top = 0;
	for (ll i = head[u], v; i; i = e[i].nxt) {
		v = e[i].to;
		if (del[v]) continue;
		tot = 0, d[v] = e[i].w;
		get_dis(v, u);
		for (ll j = 1; j <= tot; ++ j)
			for (ll k = 1; k <= m; ++ k)
				if (ask[k] >= dis[j])
					ans[k] |= judge[ask[k] - dis[j]];
		for (ll j = 1; j <= tot; ++ j)
			if (dis[j] < INF)
				q[++ top] = dis[j], judge[dis[j]] = 1;
	}
	for (ll i = 1; i <= top; ++ i) judge[q[i]] = 0;
}
void divide(ll u) {
	calc(u);
	for (ll i = head[u], v; i; i = e[i].nxt) {
		v = e[i].to;
		if (del[v]) continue;
		Min = sum = sz[v];
		get_rt(v, 0);
		divide(rt);
	}
}
signed main() {
	n = rd(), m = rd();
	for (ll i = 1; i < n; ++ i) u = rd(), v = rd(), w = rd(), add(u, v, w), add(v, u, w);
	for (ll i = 1; i <= m; ++ i) ask[i] = rd();
	Min = sum = n, get_rt(1, 0);
	divide(rt);
	for (ll i = 1; i <= m; ++ i) puts(ans[i] ? "AYE" : "NAY");
	return 0;
}