[洛谷P2634][国家集训队]聪聪可可

题目大意：给你一棵树，随机选两个点，求它们之间路径长度是$3$的倍数的概率

题解：点分治，求出当前状态的重心，然后求出经过重心的答案，接着分治每棵子树。注意考虑重复计算的情况

卡点：无

C++ Code：

#include <cstdio>
#include <algorithm>
#define maxn 20010
const int inf = 0x3f3f3f3f;
inline int min(int a, int b) {return a < b ? a : b;}
inline int max(int a, int b) {return a > b ? a : b;}

int head[maxn], cnt;
struct Edge {
	int to, nxt, w;
} e[maxn << 1];
inline void add(int a, int b, int c) {
	e[++cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
	e[++cnt] = (Edge) {a, head[b], c}; head[b] = cnt;
}

int n, ans;
bool vis[maxn];

namespace Center_of_Gravity {
	#define n nodenum
	int root, MIN, n;
	int sz[maxn];
	void __getroot(int u, int fa) {
		sz[u] = 1;
		int MAX = 0;
		for (int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			if (v != fa && !vis[v]) {
				__getroot(v, u);
				sz[u] += sz[v];
				MAX = max(sz[v], MAX);
			}
		}
		MAX = max(n - sz[u], MAX);
		if (MAX < MIN) MIN = MAX, root = u;
	}
	int getroot(int rt, int nodenum) {
		Center_of_Gravity::n = nodenum;
		MIN = inf;
		__getroot(rt, 0);
		return root;
	}
	#undef n
}
using Center_of_Gravity::getroot;

int V[3];
void getlist(int u, int fa, int val) {
	V[val]++;
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa && !vis[v]) getlist(v, u, (val + e[i].w) % 3);
	}
}
int calc(int u, int val = 0) {
	V[0] = V[1] = V[2] = 0;
	getlist(u, 0, val);
	return (V[1] * V[2] << 1) + V[0] * V[0];
}
void dfs(int u) {
	ans += calc(u);
	vis[u] = true;
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (!vis[v]) {
			ans -= calc(v, e[i].w);
			dfs(getroot(v, Center_of_Gravity::sz[v]));
		}
	}
}

int main() {
	scanf("%d", &n);
	for (int i = 1, a, b, c; i < n; i++) {
		scanf("%d%d%d", &a, &b, &c);
		add(a, b, c % 3);
	}
	dfs(getroot(1, n));
	int tmp = std::__gcd(ans, n * n);
	printf("%d/%d\n", ans / tmp, n * n / tmp);
	return 0;
}