[洛谷P2495][SDOI2011]消耗战

题目大意：有一棵$n(nleqslant2.5 imes10^5)$个节点的带边权的树，$m$个询问，每次询问给出$k(sumlimits_{i=1}^mk_ileqslant5 imes10^5)$个点，要求用最小的代价砍断一些边，使得$1$号点与这$k$个点都不连通，输出最小代价

题解：先考虑一个一次$O(n)$的$DP$，可以把所有子树内没有特殊点的点先删去，令$f(i)$表示把他于他子树内所有的特殊点割断所需要的代价，即为$sumlimits_{vin son_u}min{w,f_v}$（$w$为这条边边权，若$v$为关键点，$f_v=inf$）。这样的复杂度是$O(n^2)$的，不可以通过。

可以建虚树，一般建一棵虚树的复杂度是$O(klog_2n)$的，它可以使得建出来的树中只包含最多$2k$个节点。于是就可以通过本题

卡点：最多$2k$个点，$kleqslant n$，没有开两倍的空间

C++ Code：

#include <algorithm>
#include <cctype>
#include <cstdio>
#include <cstring>
namespace std {
	struct istream {
#define M (1 << 25 | 3)
		char buf[M], *ch = buf - 1;
		inline istream() {
#ifndef ONLINE_JUDGE
			freopen("input.txt", "r", stdin);
#endif
			fread(buf, 1, M, stdin);
		}
		inline istream& operator >> (int &x) {
			while (isspace(*++ch));
			for (x = *ch & 15; isdigit(*++ch); ) x = x * 10 + (*ch & 15);
			return *this;
		}
#undef M
	} cin;
	struct ostream {
#define M (1 << 25 | 3)
		char buf[M], *ch = buf - 1;
		inline ostream& operator << (int x) {
			if (!x) {*++ch = '0'; return *this;}
			static int S[11], *top; top = S;
			while (x) {*++top = x % 10 ^ 48; x /= 10;}
			for (; top != S; --top) *++ch = *top;
			return *this;
		}
		inline ostream& operator << (long long x) {
			if (!x) {*++ch = '0'; return *this;}
			static int S[20], *top; top = S;
			while (x) {*++top = x % 10 ^ 48; x /= 10;}
			for (; top != S; --top) *++ch = *top;
			return *this;
		}
		inline ostream& operator << (const char x) {*++ch = x; return *this;}
		inline ~ostream() {
#ifndef ONLINE_JUDGE
			freopen("output.txt", "w", stdout);
#endif
			fwrite(buf, 1, ch - buf + 1, stdout);
		}
#undef M
	} cout;
}

#define maxn 250010

int head[maxn], cnt;
struct Edge {
	int to, nxt, w;
} e[maxn << 1];
inline void addedge(int a, int b, int c) {
	e[++cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
}

int in[maxn], out[maxn], idx, dep[maxn];
namespace Tree {
#define M 17
	int fa[maxn][M + 1], Min[maxn][M + 1];
	void dfs(int u) {
		in[u] = ++idx;
		for (int i = 1; i <= M; i++) {
			fa[u][i] = fa[fa[u][i - 1]][i - 1];
			Min[u][i] = std::min(Min[u][i - 1], Min[fa[u][i - 1]][i - 1]);
		}
		for (int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			if (v != *fa[u]) {
				*fa[v] = u;
				*Min[v] = e[i].w;
				dep[v] = dep[u] + 1;
				dfs(v);
			}
		}
		out[u] = idx;
	}
	inline int LCA(int x, int y) {
		if (dep[x] < dep[y]) std::swap(x, y);
		for (int i = dep[x] - dep[y]; i; i &= i - 1) x = fa[x][__builtin_ctz(i)];
		if (x == y) return x;
		for (int i = M; ~i; i--) if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
		return *fa[x];
	}
	const int inf = 0x3f3f3f3f;
	inline int query(int x, int k) {
		int res = inf;
		for (int i = k; i; i &= i - 1) {
			res = std::min(res, Min[x][__builtin_ctz(i)]);
			x = fa[x][__builtin_ctz(i)];
		}
		return res;
	}
#undef M
}

int n, m;
namespace Work {
	const long long inf = 0x3f3f3f3f3f3f3f3f;
	inline bool cmp(int a, int b) {return in[a] < in[b];}

	int k;
	int list[maxn << 1], S[maxn], top;
	long long f[maxn];
	bool imp[maxn];

	void dfs(int u) {
		f[u] = 0;
		for (int i = head[u]; i; i = e[i].nxt) {
			int v = e[i].to;
			dfs(v);
			f[u] += std::min(f[v], static_cast<long long> (e[i].w));
		}
		if (imp[u]) f[u] = inf, imp[u] = false;
		head[u] = 0;
	}

	void solve() {
		cnt = 0;
		std::cin >> k;
		for (int i = 0; i < k; i++) {
			std::cin >> list[i];
			imp[list[i]] = true;
		}

		std::sort(list, list + k, cmp);
		int tot = k;
		for (int i = 0; i < k - 1; i++) list[tot++] = Tree::LCA(list[i], list[i + 1]);
		list[tot++] = 1;
		tot = (std::sort(list, list + tot, cmp), std::unique(list, list + tot) - list);
		top = 0;
		for (int I = 0, i = *list; I < tot; I++, i = list[I]) {
			while (top && out[S[top]] < in[i]) top--;
			if (top) addedge(S[top], i, Tree::query(i, dep[i] - dep[S[top]]));
			S[++top] = i;
		}

		dfs(1);
		std::cout << f[1] << '
';
	}
}

int main() {
	std::cin >> n;
	for (int i = 1, u, v, w; i < n; i++) {
		std::cin >> u >> v >> w;
		addedge(u, v, w);
		addedge(v, u, w);
	}
	Tree::dfs(1);
	std::cin >> m;
	memset(head, 0, sizeof head);
	while (m --> 0) Work::solve();
	return 0;
}