bzoj2152 聪聪可可 (树形dp)

大意: 给定树, 随机选两点, 求两点距离是3的倍数的概率.

树形dp入门水题, 枚举每个点作为lca时的答案即可.

#include <iostream>
#include <queue>
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

const int N = 1e6+10;
int n, dp[N][3];
struct _ {int to,w;};
vector<_> g[N];
ll ans;

void dfs(int x, int fa, int d) {
	for (int i=0; i<g[x].size(); ++i) {
		int y = g[x][i].to;
		if (y==fa) continue;
		dfs(y,x,(d+g[x][i].w)%3);
		ans += 2ll*dp[y][d];
		REP(i,0,2) ans += 2ll*dp[x][i]*dp[y][(2*d-i+3)%3];
		REP(i,0,2) dp[x][i] += dp[y][i];
	}
	++dp[x][d],++ans;
}

int main() {
	scanf("%d", &n);
	REP(i,2,n) {
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		w %= 3;
		g[u].push_back({v,w});
		g[v].push_back({u,w});
	}
	dfs(1,0,0);
	ll x = ans, y = (ll)n*n, z = gcd(x,y);
	x /= z, y /= z;
	printf("%lld/%lld
", x,y);
}