luoguP2495 [SDOI2011]消耗战

https://www.luogu.org/problemnew/show/P2495

Dp 方程很显然

设 Dp[u] 表示——使 u 不与其子树中任意一个关键点联通的最小代价

设 w[a, b] 表示 a 与 b 之间的边的权值。

• 若 son[i] 不是关键点，Dp[u] = Dp[u] + min
• 若 son[i] 是关键点，Dp[u] = Dp[u] + w[u][son[i]]

但这样复杂度很显然是不对的，所以我们考虑虚树

什么，你还不会虚树？那就去跟 zzq 学吧 https://www.cnblogs.com/zzqsblog/p/5560645.html

我们发现 k 的总和与 n 同级，所以用虚树优化这个 Dp，建出虚树，在虚树上 Dp 即可

#include <bits/stdc++.h>
#define X first
#define Y second
#define mp make_pair
using namespace std;

typedef long long ll;
const int N = 250000 + 5, LG2 = 18;

vector < pair <int, int> > G[N], G2[N];
int pre[N][LG2 + 1], dep[N], mx[N][LG2 + 1], id[N], dfn;
int n, m, k, h[N], sta[N], len, MX;
ll f[N];
bool book[N];

void init(int u, int fa) {
	pre[u][0] = fa; dep[u] = dep[fa] + 1; id[u] = ++dfn;
	for(int i = 1; i <= LG2; i++) {
		pre[u][i] = pre[pre[u][i - 1]][i - 1];
		mx[u][i] = min(mx[u][i - 1], mx[pre[u][i - 1]][i - 1]);
	}
	for(vector < pair <int, int> > :: iterator it = G[u].begin(); it != G[u].end(); it++) {
		int v = it -> X; if(v != fa) mx[v][0] = it -> Y, init(v, u);
	}
}

int LCA(int x, int y) {
	MX = INT_MAX;
	if(dep[x] > dep[y]) swap(x, y);
	for(int i = LG2; i >= 0; i--)
		if(dep[pre[y][i]] >= dep[x])
			MX = min(MX, mx[y][i]), y = pre[y][i];
	if(x == y) return x;
	for(int i = LG2; i >= 0; i--)
		if(pre[x][i] != pre[y][i]) {
			MX = min(MX, min(mx[x][i], mx[y][i]));
			x = pre[x][i], y = pre[y][i];
		}
	return pre[x][0];
}

bool cmp(int x, int y) {return id[x] < id[y];}

void DP(int u) {
	f[u] = 0;
	for(vector < pair <int, int> > :: iterator it = G2[u].begin(); it != G2[u].end(); it++) {
		int v = it -> X; DP(v);
		if(book[v]) f[u] += it -> Y;
		else f[u] += min(f[v], (ll)it -> Y);
	}
}

int main() {
	cin >> n;
	for(int i = 1; i < n; i++) {
		int a, b, c;
		scanf("%d %d %d", &a, &b, &c);
		G[a].push_back( mp(b, c) );
		G[b].push_back( mp(a, c) ); 
	}
	init(1, 0); cin >> m;
	while(m--) {
		scanf("%d", &k);
		for(int i = 1; i <= k; i++) {
			scanf("%d", &h[i]);
			book[h[i]] = 1;
		}
		sort(h + 1, h + k + 1, cmp);
		sta[len = 1] = 1; G2[1].clear();
		for(int i = 1; i <= k; i++) {
			if(h[i] == 1) continue;
			int lca = LCA(h[i], sta[len]);
			if(lca != sta[len]) {
				while(id[lca] < id[sta[len - 1]]) {
					LCA(sta[len - 1], sta[len]);
					G2[sta[len - 1]].push_back( mp(sta[len], MX) );
					len--;
				}
				if(id[lca] > id[sta[len - 1]]) {
					G2[lca].clear();
					LCA(lca, sta[len]);
					G2[lca].push_back( mp(sta[len], MX) );
					sta[len] = lca;
				} else LCA(lca, sta[len]), G2[lca].push_back( mp(sta[len], MX) ), len--;
			}
			G2[h[i]].clear(); sta[++len] = h[i];
		}
		for(int i = 1; i < len; i++) LCA(sta[i], sta[i + 1]), G2[sta[i]].push_back( mp(sta[i + 1], MX) );
		DP(1); printf("%lld\n", f[1]);
		for(int i = 1; i <= k; i++) book[h[i]] = 0;
	}
	return 0;
}