[luoguP2495] [SDOI2011]消耗战（DP + 虚树）

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明显虚树。

别的题解里都是这样说的。

 

先不考虑虚树，假设只有一组询问，该如何dp？

f[u]表示把子树u中所有的有资源的节点都切掉的最优解

如果节点u需要切掉了话，$f[u]=val[u]$

否则如果u的子树中有需要切除的点的话，$f[u] = min(val[u], \sum\limits_{v是u的儿子}f[v])$

val[u]表示是根到u的路径上最小的边的权值。

最后转移到虚树上即可。

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 1000000
#define LL long long

using namespace std;

int n, m, cnt, rp, top, T;
int head[N], to[N], nex[N], dfn[N], f[N][21], q[N], deep[N], s[N];
LL ans[N], dp[N], val[N];
bool flag[N];

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void add(int x, int y, int z)
{
	to[cnt] = y;
	val[cnt] = z;
	nex[cnt] = head[x];
	head[x] = cnt++;
}

inline void dfs1(int u)
{
	int i, v;
	dfn[u] = ++rp;
	deep[u] = deep[f[u][0]] + 1;
	for(i = 0; f[u][i]; i++) f[u][i + 1] = f[f[u][i]][i];
	for(i = head[u]; ~i; i = nex[i])
	{
		v = to[i];
		if(!dfn[v])
		{
			f[v][0] = u;
			dp[v] = min(dp[u], val[i]);
			dfs1(v);
		}
	}
	head[u] = -1;
}

inline int calc_lca(int x, int y)
{
	int i, j;
	if(deep[x] < deep[y]) swap(x, y);
	for(i = 20; i >= 0; i--)
		if(deep[f[x][i]] >= deep[y]) x = f[x][i];
	if(x == y) return x;
	for(i = 20; i >= 0; i--)
		if(f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
	return f[x][0];
}

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

inline void dfs2(int u)
{
	LL sum = 0;
	int i, v;
	ans[u] = dp[u];
	for(i = head[u]; ~i; i = nex[i])
	{
		v = to[i];
		dfs2(v);
		sum += ans[v];
	}
	if(sum && !flag[u]) ans[u] = min(ans[u], sum);
	head[u] = -1;
}

inline void solve()
{
	int i, lca;
	m = read();
	top = cnt = 0;
	for(i = 1; i <= m; i++) q[i] = read(), flag[q[i]] = 1;
	sort(q + 1, q + m + 1, cmp);
	for(i = 1; i <= m; i++)
	{
		if(!top)
		{
			s[++top] = q[i];
			continue;
		}
		lca = calc_lca(q[i], s[top]);
		while(dfn[lca] < dfn[s[top]])
		{
			if(dfn[lca] >= dfn[s[top - 1]])
			{
				add(lca, s[top], 0);
				if(s[--top] != lca) s[++top] = lca;
				break;
			}
			add(s[top - 1], s[top], 0), top--;
		}
		s[++top] = q[i];
	}
	while(top > 1) add(s[top - 1], s[top], 0), top--;
	dfs2(s[1]);
	printf("%lld\n", ans[s[1]]);
	for(i = 1; i <= m; i++) flag[q[i]] = 0;
}

int main()
{
	int i, x, y, z;
	n = read();
	memset(head, -1, sizeof(head));
	for(i = 1; i < n; i++)
	{
		x = read();
		y = read();
		z = read();
		add(x, y, z);
		add(y, x, z);
	}
	dp[1] = 1ll * 1e9 * 1e9;
	dfs1(1);
	T = read();
	while(T--) solve();
	return 0;
}