$虚树+树形dp$
$虚树复习$
$虚树是用来解决每次从树中选出一些点进行统计一类问题的方法，每次把这些点建出虚树，最多不超过2*k个点$
$建虚树的过程挺简单的，就是用一个栈记录当前dfs的过程，如果出现分叉就回溯。重点在于虚点的加入，就是两个点的lca，这个判一下就行了$
$树形dp的过程比较简单，就不说了,注意虚点的统计$
#include<bits/stdc++.h> using namespace std; typedef long long ll; const int N = 1e6 + 5, inf = 0x3f3f3f3f; int n, q, k, top, dfs_clock; ll ans, ans1, ans2; int dfn[N], dep[N], fa[N][21], mark[N], a[N], st[N]; ll mn[N], mx[N], sum[N], size[N]; vector<int> G[N]; vector<pair<int, int> > g[N]; bool cmp(int i, int j) { return dfn[i] < dfn[j]; } void dfs(int u, int last) { dfn[u] = ++dfs_clock; for(int i = 0; i < G[u].size(); ++i) { int v = G[u][i]; if(v == last) { continue; } fa[v][0] = u; dep[v] = dep[u] + 1; dfs(v, u); } } int lca(int u, int v) { if(dep[u] < dep[v]) { swap(u, v); } int d = dep[u] - dep[v]; for(int i = 20; i >= 0; --i) { if(d & (1 << i)) { u = fa[u][i]; } } if(u == v) { return u; } for(int i = 20; i >= 0; --i) { if(fa[u][i] != fa[v][i]) { u = fa[u][i]; v = fa[v][i]; } } return fa[u][0]; } void dp(int u) { if(mark[u]) { mn[u] = 0; } else { mn[u] = 0x3f3f3f3f; } mx[u] = 0; sum[u] = 0; size[u] = mark[u]; for(int i = 0; i < g[u].size(); ++i) { int v = g[u][i].first, w = -g[u][i].second; dp(v); if(mark[u]) { ans += size[v] * sum[u] + size[u] * (sum[v] + w * size[v]); ans1 = min(ans1, mn[v] + w); ans2 = max(ans2, mx[v] + w + mx[u]); } else { ans += size[v] * sum[u] + size[u] * (sum[v] + w * size[v]); if(i) ans1 = min(ans1, mn[u] + mn[v] + w); if(i) ans2 = max(ans2, mx[v] + w + mx[u]); } sum[u] += sum[v] + w * size[v]; mn[u] = min(mn[u], mn[v] + w); mx[u] = max(mx[u], mx[v] + w); size[u] += size[v]; } g[u].clear(); } int main() { scanf("%d", &n); for(int i = 1; i < n; ++i) { int u, v; scanf("%d%d", &u, &v); G[u].push_back(v); G[v].push_back(u); } dfs(1, 0); for(int j = 1; j <= 20; ++j) for(int i = 1; i <= n; ++i) fa[i][j] = fa[fa[i][j - 1]][j - 1]; scanf("%d", &q); while(q--) { scanf("%d", &k); for(int i = 1; i <= k; ++i) { scanf("%d", &a[i]); mark[a[i]] = 1; } sort(a + 1, a + k + 1, cmp); st[top = 1] = 1; for(int i = 1; i <= k; ++i) { int f = lca(a[i], st[top]); while(top > 1 && dfn[f] < dfn[st[top - 1]]) { g[st[top - 1]].push_back({st[top], dep[st[top - 1]] - dep[st[top]]}); --top; } if(dfn[f] < dfn[st[top]]) { g[f].push_back({st[top], dep[f] - dep[st[top]]}); --top; } if(f != st[top]) { st[++top] = f; } if(a[i] != 1) { st[++top] = a[i]; } } while(top > 1) { g[st[top - 1]].push_back({st[top], dep[st[top - 1]] - dep[st[top]]}); --top; } ans = 0; ans1 = inf; ans2 = 0; dp(1); printf("%lld %lld %lld\n", ans, ans1, ans2); for(int i = 1; i <= k; ++i) { mark[a[i]] = 0; } } return 0; }
