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#include <bits/stdc++.h>
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using namespace std;
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const int N = 500010, M = N << 1;
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#define int long long
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// 链式前向星
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int e[M], h[N], idx, w[M], ne[M];
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void add(int a, int b, int c = 0) {
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e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
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}
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int n, m, k;
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int sz[N]; // 以u为根节点是否有人
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int g[N]; // 从u出发把u子树上的人都送回家再回到u所需要的时间
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int mx1[N]; // u的子树中最长链的长度
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int mx2[N]; // u的子树中次长链
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int up[N]; // 不在u的子树内，距离u最远的那个人的家到u的距离
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int id[N]; // 最长链条是哪条链
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int ans[N]; // 从u出发把所有点都送回家再回到u的结果
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// ans[i] -max(up[i],mx1[i]);
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void dfs1(int u, int fa) {
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (v == fa) continue;
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dfs1(v, u);
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if (sz[v] == 0) continue;
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g[u] += g[v] + 2 * w[i];
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int now = mx1[v] + w[i];
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if (now >= mx1[u]) {
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mx2[u] = mx1[u];
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mx1[u] = now;
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id[u] = v;
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} else if (now >= mx2[u])
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mx2[u] = now;
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sz[u] += sz[v];
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}
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}
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void dfs2(int u, int fa) {
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for (int i = h[u]; ~i; i = ne[i]) {
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int v = e[i];
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if (v == fa) continue;
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if (sz[v] == k) {
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ans[v] = g[v];
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up[v] = 0;
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} else if (sz[v] == 0) {
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ans[v] = ans[u] + 2 * w[i];
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up[v] = max(up[u], mx1[u]) + w[i];
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} else if (sz[v] && sz[v] != k) {
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ans[v] = ans[u];
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if (id[u] == v)
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up[v] = max(mx2[u], up[u]) + w[i];
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else
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up[v] = max(up[u], mx1[u]) + w[i];
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}
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dfs2(v, u);
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}
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}
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signed main() {
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// 初始化链式前向星
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memset(h, -1, sizeof h);
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cin >> n >> k; // n个节点,要接k个人
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for (int i = 1; i < n; i++) { // n-1条边
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int a, b, c;
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cin >> a >> b >> c;
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add(a, b, c), add(b, a, c);
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}
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for (int i = 1; i <= k; i++) {
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int x;
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cin >> x;
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sz[x] = 1; // x节点有一个人
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}
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// 第一次dfs
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dfs1(1, 0);
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ans[1] = g[1];
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// 第二次dfs
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dfs2(1, 0);
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for (int u = 1; u <= n; u++) cout << ans[u] - max(up[u], mx1[u]) << endl;
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} |
