RT，楼主每次提交都WA on test 5 ...

简单说一下思路：dfs得到每个节点的子树节点数，然后算出这条边的weight，即$subtree[u] \cdot (N-subtree[u])$. 然后分类讨论一下$M$ 和$N-1$的大小，贪心地选取权值。

long long该开的应该也都开了...请大佬们帮帮萌新吧，谢谢啦！

代码如下，码风还行：

const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;
const int maxm = 6e4 + 5;

int subtree[maxn], p[maxm], N, M, T, cnt;
ll weight[maxn];

vector<int> G[maxn];

void dfs(int u, int k) {
	subtree[u] = 1;
	for (int i = 0; i < (int) G[u].size(); i++) {
		int v = G[u][i];
		if (v == k) continue;
		dfs(v, u);
		subtree[u] += subtree[v];
	}
	if (u != 1) weight[++cnt] = (ll) subtree[u] * (ll) (N - subtree[u]) % mod;
}

void solve() {
	ll ans = 0;
	cin >> N;
	rep(i, 1, N) G[i].clear();
	memset(weight, 0, sizeof(weight));
	cnt = 0;
	rep(i, 1, N - 1) {
		int u, v;
		cin >> u >> v;
		G[u].pb(v), G[v].pb(u);
	}
	dfs(1, -1);
	cin >> M;
	rep(i, 1, M) cin >> p[i];
	sort(p + 1, p + M + 1, greater<int>());
	sort(weight + 1, weight + N, greater<int>());
	if (M <= N - 1) {
		for (int i = 1; i <= M; i++) ans = ((ll) ans + (ll) weight[i] * p[i]) % mod;
		for (int i = M + 1; i <= N - 1; i++) ans = ((ll) ans + (ll) weight[i]) % mod;
	} else {
		ll mul = 1;
		rep(i, 1, M - N + 2) mul = (ll) mul * (ll) p[i] % mod;
		ans = (ll) weight[1] * (ll) mul % mod;
		rep(i, 2, N - 1) ans = ((ll) ans + (ll) weight[i] * p[M - N + 1 + i]) % mod; 
	}
	cout << ans << endl;
}