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@ -11,24 +11,29 @@ 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 pos[N];
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int sz[N], f[N], g[N];
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int sz[N]; // 哪个点上有人
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int len[N], id[N], slen[N];
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int f[N], g[N];
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void dfs1(int u, int fa) {
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if (pos[u]) sz[u] = 1;
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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]) {
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f[u] += f[v] + 2 * w[i];
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int now = len[v] + w[i];
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if (now >= len[u])
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slen[u] = len[u], len[u] = now, id[u] = v;
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else if (now > slen[u])
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slen[u] = now;
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}
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dfs1(v, u); // 由底向上,先递归,再更新统计信息
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// 如果v这个节点,及它的子节点上有人,那么需要汇总统计信息到sz[u]上去
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// 如果v上就没有人,那就不用统计了
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if (sz[v] == 0) continue;
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// ① u->v,v->u一来一回,路径翻倍 2*w[i]
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// ② 所有子节点都对u有贡献,所以f[u]+
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// ③ 跑完v为根的子树后,v子树的贡献要累加到u子树上,所以f[u]+=f[v]+2*w[i]
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f[u] += f[v] + 2 * w[i];
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int now = len[v] + w[i];
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if (now >= len[u])
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slen[u] = len[u], len[u] = now, id[u] = v;
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else if (now > slen[u])
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slen[u] = now;
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sz[u] += sz[v];
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}
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}
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@ -64,11 +69,12 @@ signed main() {
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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 <= m; i++) {
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for (int i = 1; i <= m; i++) { // 有m个人
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int x;
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cin >> x;
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pos[x] = 1;
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sz[x] = 1; // x号点上有人
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}
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dfs1(1, 0);
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g[1] = f[1];
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