\(\mathcal{Description}\)

  OurTeam.

  給定一棵 \(n\) 個點的樹形隨機的帶邊權樹，求所有含奇數條邊的路徑中位數之和。樹形生成方式為隨機取不連通兩點連邊直到全部連通。

  \(n\le32000\)。

\(\mathcal{Solution}\)

  考慮用中位數的標準姿勢統計每條邊的貢獻——小於它的設為 \(-1\)，大於它的設為 \(+1\)，邊權相等按編欽定大小關係。那麼這條邊的貢獻就是路徑兩端權值加和為 \(0\) 的路徑對數（顯然每對路徑連起來都是奇數路徑）。

  令 \(f(u,i)\) 表示 \(u\) 子樹內到 \(u\) 路徑權值和為 \(i\)

  複雜度 \(\mathcal O(n^2)\)，不過可以算出帶一個 \(\frac{1}6\) 的常數，所以可過。

\(\mathcal{Code}\)

#include <queue>
#include <cstdio>
#include <cstring>

typedef long long LL;

const int MAXN = 32000, MAXSQRT = 180, MAXV = 1e6;
int n, ecnt, head[MAXN + 5], fa[MAXN + 5], dep[MAXN + 5], faw[MAXN + 5];
int mempool[MAXN * MAXSQRT * 2], *f[MAXN + 5], *frepos = mempool, *g;
int pre[MAXN + 5], buc[MAXV + 5], ori[MAXN + 5], down[MAXN + 5];

struct Edge { int to, cst, nxt; } graph[MAXN * 2 + 5];
struct EdgeSet { int u, v, w; } eset[MAXN + 5];

inline int rint () {
	int x = 0; char s = getchar ();
	for ( ; s < '0' || '9' < s; s = getchar () );
	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
	return x;
}

inline int min_ ( const int a, const int b ) { return a < b ? a : b; }
inline int max_ ( const int a, const int b ) { return a < b ? b : a; }

inline void link ( const int s, const int t, const int c ) {
	graph[++ ecnt] = { t, c, head[s] };
	head[s] = ecnt;
}

inline void BFS ( const int s ) {
	static std::queue<int> que;
	que.push ( s ); dep[s] = 1;
	while ( ! que.empty () ) {
		int u = que.front (); que.pop ();
		for ( int i = head[u], v; i; i = graph[i].nxt ) {
			if ( ! dep[v = graph[i].to] ) {
				pre[v] = u, dep[v] = dep[u] + 1;
				que.push ( v );
			}
		}
	}
}

inline void init ( const int u ) {
	dep[u] = 0;
	for ( int i = head[u], v; i; i = graph[i].nxt ) {
		if ( ( v = graph[i].to ) ^ fa[u] ) {
			fa[v] = u, faw[v] = 1, down[graph[i].cst] = v, init ( v );
			if ( dep[v] + 1 > dep[u] ) dep[u] = dep[v] + 1;
		}
	}
	f[u] = frepos += dep[u] + 1, frepos += dep[u] + 1;
}

inline void DP ( const int u ) {
	*f[u] = 1;
	for ( int i = head[u], v; i; i = graph[i].nxt ) {
		if ( ( v = graph[i].to ) ^ fa[u] ) {
			DP ( v );
			for ( int j = -dep[v]; j <= dep[v]; ++ j ) f[u][j + 1] += f[v][j];
		}
	}
}

inline LL update ( const int s ) {
	int l = -dep[s], r = dep[s]; faw[s] = -1;
	for ( int u = fa[s], v = s, d1 = 0, d2 = 0; u; u = fa[v = u] ) {
		d1 += v ^ s ? faw[v] : 1, d2 += faw[v];
		for ( int i = -dep[s]; i <= dep[s]; ++ i ) {
			f[u][i + d1] -= f[s][i];
			f[u][i + d2] += f[s][i];
		}
		for ( int i = max_ ( -dep[u], l ), t = min_ ( dep[u], r ), d = d2 + 1; i <= t; ++ i ) {
			g[i + d] += f[u][i] - f[v][i - faw[v]];
		}
		l -= faw[u], r -= faw[u];
	}
	LL ret = 0;
	for ( int i = -dep[s]; i <= dep[s]; ++ i ) {
		ret += 1ll * f[s][i] * g[-i];
		g[-i] = 0;
	}
	return ret;
}

int main () {
	n = rint (); int mxw = 0;
	for ( int i = 1, u, v, w; i < n; ++ i ) {
		u = rint (), v = rint (), w = rint ();
		eset[i] = { u, v, w }, ++ buc[w];
		if ( w > mxw ) mxw = w;
	}
	for ( int i = 2; i <= mxw; ++ i ) buc[i] += buc[i - 1];
	for ( int i = 1; i < n; ++ i ) {
		eset[i].w = buc[ori[buc[eset[i].w]] = eset[i].w] --;
		link ( eset[i].u, eset[i].v, eset[i].w );
		link ( eset[i].v, eset[i].u, eset[i].w );
	}
	BFS ( 1 );
	int s = 0, t = 0;
	for ( int i = 1; i <= n; ++ i ) {
		if ( dep[i] > dep[s] ) s = i;
		dep[i] = pre[i] = 0;
	}
	BFS ( s );
	for ( int i = 1; i <= n; ++ i ) if ( dep[i] > dep[t] ) t = i;
	for ( int i = dep[s] + dep[t] - 2 >> 1; i --; t = pre[t] );
	init ( t ), DP ( t );
	g = frepos += dep[t] + 1, frepos += dep[t] + 1;
	LL ans = 0;
	for ( int i = 1; i < n; ++ i ) ans += ori[i] * update ( down[i] );
	printf ( "%lld\n", ans );
	return 0;
}