CF613D Kingdom and its Cities

题意

给定一棵树，每次询问给出 \(k\) 个点。

问最少删除多少个 节点 (不能删这 \(k\) 个点) 使得这 \(k\) 个点两两不连通。

Sol

无解的情况是 \(trivial\) 的。

判断是否有相邻的两个关键点就行了。

但是 \(dp\) 是不太 \(trivial\) 的。

设 \(f_u\) 表示使得 \(u\) 子树内的关键点两两不连通需要的最小代价。

考虑讨论当前节点是否为关键点。

发现转移时缺少一个重要信息：

如果当前点不是关键点，且儿子内只有一个儿子子树内有关键点，此时删除儿子或者当前点都是错误的。

所以我们考虑设 \(g_u = 1/0\) 表示当前点是否与关键点相连。

\[f_u = \begin{cases} vis_u = 1 & f_u = \sum g_v + f_v, g_u = 1 \cr vis_u = 0, \sum g_v \le 1 & f_u = \sum f_v, g_u = \sum g_v \cr vis_u = 0, \sum g_v > 1 & f_u = 1 + \sum f_v. g_u = 0 \end{cases} \]

简单得到 \(n ^ 2\) 算法。

优化成 \(O(n \log n)\) 只需要套一个虚树就可以了。

Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <vector>
#include <bitset>
#define int long long
#define pii pair <int, int>
using namespace std;
#ifdef ONLINE_JUDGE

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;

#endif
int read() {
	int p = 0, flg = 1;
	char c = getchar();
	while (c < '0' || c > '9') {
		if (c == '-') flg = -1;
		c = getchar();
	}
	while (c >= '0' && c <= '9') {
		p = p * 10 + c - '0';
		c = getchar();
	}
	return p * flg;
}
void write(int x) {
	if (x < 0) {
		x = -x;
		putchar('-');
	}
	if (x > 9) {
		write(x / 10);
	}
	putchar(x % 10 + '0');
}

#define fi first
#define se second


const int N = 1e5 + 5, M = 2e5 + 5;

namespace G {

array <int, N> fir;
array <int, M> nex, to;
int cnt;
void add(int x, int y) {
	cnt++;
	nex[cnt] = fir[x];
	to[cnt] = y;
	fir[x] = cnt;
}

}

namespace T {

array <int, N> fir;
array <int, M> nex, to;
int cnt;
void add(int x, int y) {
	cnt++;
	nex[cnt] = fir[x];
	to[cnt] = y;
	fir[x] = cnt;
}

}

namespace Hpt {

using G::fir; using G::nex; using G::to;

array <int, N> son, siz, fa, dep;

void dfs1(int x) {
	siz[x] = 1;
	for (int i = fir[x]; i; i = nex[i]) {
		if (to[i] == fa[x]) continue;
		fa[to[i]] = x;
		dep[to[i]] = dep[x] + 1;
		dfs1(to[i]);
		siz[x] += siz[to[i]];
		if (siz[to[i]] > siz[son[x]]) son[x] = to[i];
	}
}

array <int, N> dfn, idx, top;
int cnt;

void dfs2(int x, int Mgn) {
	cnt++;
	dfn[x] = cnt;
	idx[cnt] = x;
	top[x] = Mgn;
	if (son[x]) dfs2(son[x], Mgn);
	for (int i = fir[x]; i; i = nex[i]) {
		if (to[i] == fa[x] || to[i] == son[x]) continue;
		dfs2(to[i], to[i]);
	}
}

int lca(int x, int y) {
	while (top[x] != top[y]) {
		if (dfn[top[x]] < dfn[top[y]]) swap(x, y);
		x = fa[top[x]];
	}
	if (dfn[x] > dfn[y]) swap(x, y);
	return x;
}

}

namespace Vit {

vector <pii> isl;
array <int, N> stk;
int top;

void build() {
	sort(isl.begin(), isl.end());
	T::fir[isl[0].se] = 0; stk[top = 1] = isl[0].se;
	for (int i = 1; i < (int)isl.size(); i++) {
		pii x = isl[i]; int lcA = Hpt::lca(stk[top], x.se);
		if (lcA != stk[top]) {
			while (Hpt::dfn[stk[top - 1]] > Hpt::dfn[lcA])
				T::add(stk[top - 1], stk[top]), top--;
			if (stk[top - 1] != lcA)
				T::fir[lcA] = 0, T::add(lcA, stk[top]), stk[top] = lcA;
			else
				T::add(stk[top - 1], stk[top]), top--;
		}
		T::fir[x.se] = 0, top++, stk[top] = x.se;
	}
	for (int i = 1; i < top; i++)
		T::add(stk[i], stk[i + 1]);
}

bitset <N> vis;

pii dfs(int x) {
	pii res(0, 0);
	int tp1 = 0, tp2 = 0;
	for (int i = T::fir[x]; i; i = T::nex[i]) {
		pii now = dfs(T::to[i]);
		tp1 += now.fi, tp2 += now.se;
	}

	if (vis[x]) res.fi = tp1 + tp2, res.se = 1;
	else if (tp2 > 1) res.fi = 1 + tp1, res.se = 0;
	else res.fi = tp1, res.se = (bool)tp2;

	return res;
}

}

signed main() {
	int n = read();
	for (int i = 2; i <= n; i++) {
		int x = read(), y = read();
		G::add(x, y), G::add(y, x);
	}
	Hpt::dfs1(1), Hpt::dfs2(1, 0);
	int q = read();
	while (q--) {
		int k = read();
		Vit::isl.clear();
		bool flg = 0; T::cnt = 0;
		for (int i = 1; i <= k; i++) {
			int x = read();
			Vit::isl.push_back(make_pair(Hpt::dfn[x], x));
		}
		for (auto x : Vit::isl) Vit::vis[x.se] = 1;
		for (auto x : Vit::isl) flg |= Vit::vis[Hpt::fa[x.se]];
		Vit::build();
		if (flg) puts("-1");
		else write(Vit::dfs(Vit::stk[1]).fi), puts("");
		for (auto x : Vit::isl) Vit::vis[x.se] = 0;
	}
	return 0;
}