数组开2e4就RE， 下载数据本地能过， 把数组开2e5直接CE，？？？？？？

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 2e4 + 10;

template < typename T > inline void read(T &x) {
	x = 0; T ff = 1, ch = getchar();
	while (!isdigit(ch)) {
		if (ch == '-') ff = -1;
		ch = getchar(); 
	}
	while (isdigit(ch)) {
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar(); 
	}
	x *= ff; 
}

int n, tot = 0, root, node, m[MAXN], lin[MAXN], siz[MAXN];
int ans1 = 0, ans2, dis[MAXN], vis[MAXN], hav[5], cnt[5], a[5] = {0, 2, 1};
struct edge {
	int y, v, next;
}e[MAXN << 1];

inline void add(int xx, int yy, int vv) {
	e[++tot].y = yy;
	e[tot].v = vv;
	e[tot].next = lin[xx];
	lin[xx] = tot;
}

inline int Gcd(int x, int y) {
	if (x % y == 0) return y;
	return Gcd(y, x % y);
}

inline void getroot(int x, int fa) {
	siz[x] = 1, m[x] = 0;
	for (int i = lin[x]; i; i = e[i].next) {
		int y = e[i].y;
		if (y == fa || vis[y]) continue;
		getroot(y, x);
		siz[x] += siz[y];
		m[x] = max(m[x], siz[y]); 
	}
	m[x] = max(m[x], node - siz[x]);
	if (m[x] < m[root]) root = x; 
}

inline void getdis(int x, int fa) {
	++cnt[dis[x] % 3];
	for (int i = lin[x], y; i; i = e[i].next) {
		if (vis[y = e[i].y] || y == fa) continue;
		dis[y] = dis[x] + e[i].v;
		getdis(y, x);
	} 
}

inline void calc(int x) {
	hav[0] = 1;
	for (int i = lin[x], y; i; i = e[i].next) {
		if (vis[y = e[i].y]) continue;
		cnt[0] = cnt[1] = cnt[2] = 0;
		dis[y] = e[i].v;
		getdis(y, x);
		for (int j = 0; j <= 2; ++j) ans1 += cnt[j] * hav[a[j]];
		for (int j = 0; j <= 2; ++j) hav[j] += cnt[j];
	} 
	hav[0] = hav[1] = hav[2] = 0;
}

inline void dfz(int x) {
	calc(x);
	vis[x] = true;
	for (int i = lin[x], y; i; i = e[i].next) {
		if (vis[y = e[i].y]) continue;
		root = 0;
		m[0] = node = siz[y];
		getroot(y, 0); getroot(root, 0);
		dfz(root);
	}
}

int main() {
	read(n);
	for (int i = 1; i < n; ++i) {
		int x, y, v;
		read(x), read(y), read(v);
		v %= 3;
		add(x, y, v);
		add(y, x, v);
	}
	m[0] = node = n;
	getroot(1, 0); getroot(root, 0);
	dfz(root);
	ans1 = ans1 * 2 + n; 
	ans2 = n * n;
	int g = Gcd(ans1, ans2);
	printf("%d/%d", ans1 / g, ans2 / g);
	return 0;
}