题面
A. 树
思路
先说结论, 每条边的贡献次数等于其左右两侧子树大小的最小值.
证明
#include "iostream"
#include "vector"using namespace std;typedef pair<int, int> pii;
typedef long long ll;constexpr int N = 2e5 + 10;int n, sz[N];
ll ans = 0;
vector<pii> e[N];void dfs(int u, int fa) {sz[u] = 1;for (auto [v, w] : e[u]) {if (v == fa) continue;dfs(v, u);ans += 2ll * min(sz[v], n - sz[v]) * w;sz[u] += sz[v];}
}void init() {scanf("%d", &n);for (int i = 1, u, v, w; i < n; ++i) {scanf("%d %d %d", &u, &v, &w);e[u].emplace_back(v, w);e[v].emplace_back(u, w);}
}void calculate() {dfs(1, 0);printf("%lld", ans);
}void solve() {init();calculate();
}int main() {solve();return 0;
}
