【题目链接】

对于u这个子树，显然如果一条边(u, v)减去之后，v这个子树就变成子问题了。

TreeDP一下就行了。

/* Footprints In The Blood Soaked Snow */
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long LL;

const int maxn = 1000005;

int n, head[maxn], cnt, size[maxn];
LL ans;

struct _edge {
	int v, w, next;
} g[maxn << 1];

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline int add(int u, int v, int w) {
	g[cnt] = (_edge){v, w, head[u]};
	head[u] = cnt++;
}

inline int intabs(int x) {
	return x > 0 ? x : -x;
}

inline void dfs(int x, int f) {
	size[x] = 1;
	for(int i = head[x]; ~i; i = g[i].next) if(g[i].v ^ f) {
		dfs(g[i].v, x);
		size[x] += size[g[i].v];
		ans += (LL)intabs((n - size[g[i].v]) - size[g[i].v]) * g[i].w;
	}
}

int main() {
	n = iread();
	for(int i = 1; i <= n; i++) head[i] = -1; cnt = 0;

	for(int i = 1; i < n; i++) {
		int u = iread(), v = iread(), w = iread();
		add(u, v, w); add(v, u, w);
	}

	dfs(1, 0);
	printf("%lld\n", ans);
	return 0;
}