洛谷 P2495 [SDOI2011]消耗战（虚树，dp）

题面

洛谷

题解

虚树+dp

关于虚树

了解一下

• 具体实现

inline void insert(int x) {
	if (top == 1) {s[++top] = x; return ;}
	int lca = query(x, s[top]);
	while (top > 1 && dfn[s[top-1]] >= dfn[lca]) t[s[top-1]].push_back(s[top]), top--;
	if (lca != s[top]) t[lca].push_back(s[top]), s[top] = lca;
	s[++top] = x;
	return ;
}

bool cmp(int a, int b) {
	return dfn[a] < dfn[b];
}//dfn为在dfs序上的位置

main() {//o为输入的关键点
	sort(o+1, o+1+k, cmp);
	s[top = 1] = 1;
	for (int i = 1; i <= k; i++) insert(o[i]);
    for (int i = 1; i < top; i++) t[s[i]].push_back(s[i+1]);
}

然后对于这个虚树\(dp\)

\(a\)数组表示当前点到根这条链路径最小值

Code

#include<bits/stdc++.h>

#define LL long long
#define RG register

using namespace std;

inline int gi() {
	RG int x = 0; RG char c = getchar(); bool f = 0;
	while (c != '-' && (c < '0' || c > '9')) c = getchar();
	if (c == '-') c = getchar(), f = 1;
	while (c >= '0' && c <= '9') x = x*10+c-'0', c = getchar();
	return f ? -x : x;
}
const int N = 250010, INF = 2147483647;
struct node {
	int to, next, w;
}g[N<<1];
int last[N], gl;
inline void add(int x, int y, int z) {
	g[++gl] = (node) {y, last[x], z};
	last[x] = gl;
	return ;
}
int anc[N][21], dfn[N], cnt, dep[N], a[N];
void init(int u, int fa) {
	anc[u][0] = fa;
	dfn[u] = ++cnt;
	for (int i = 1; i <= 20; i++)
		anc[u][i] = anc[anc[u][i-1]][i-1];
	for (int i = last[u]; i; i = g[i].next) {
		int v = g[i].to;
		if (v == fa) continue;
		dep[v] = dep[u]+1;
		a[v] = min(g[i].w, a[u]);
		init(v, u);
	}
	return ;
}
inline int query(int x, int y) {
	if (dep[x] < dep[y]) swap(x, y);
	for (int i = 20; i >= 0; i--)
		if (dep[anc[x][i]] >= dep[y])
			x = anc[x][i];
	if (x == y) return x;
	for (int i = 20; i >= 0; i--)
		if (anc[x][i] != anc[y][i])
			x = anc[x][i], y = anc[y][i];
	return anc[x][0];
}

vector<int> t[N];
int o[N], s[N], top;

bool cmp(int a, int b) {
	return dfn[a] < dfn[b];
}
inline void insert(int x) {
	if (top == 1) {s[++top] = x; return ;}
	int lca = query(x, s[top]);
	if (lca == s[top]) return ;//祖先都走不到,它肯定也走不到哈
	while (top > 1 && dfn[s[top-1]] >= dfn[lca]) t[s[top-1]].push_back(s[top]), top--;
	if (lca != s[top]) t[lca].push_back(s[top]), s[top] = lca;
	s[++top] = x;
	return ;
}

LL dp(int u) {
	LL S = 0;
	if (!t[u].size()) return a[u];
	for (int i = 0; i < (int)t[u].size(); i++) {
		int v = t[u][i];
		S += dp(v);
	}
	t[u].clear();
	if (u==1) return S;
	return min(S, 1ll*a[u]);
}

int main() {
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	int n = gi();
	for (int i = 1; i < n; i++) {
		int x = gi(), y = gi(), z = gi();
		add(x, y, z); add(y, x, z);
	}
	a[1] = INF;
	init(1, 0);
	int m = gi();
	while (m--) {
		int k = gi();
		for (int i = 1; i <= k; i++)
			o[i] = gi();
		sort(o+1, o+1+k, cmp);
		s[top = 1] = 1;
		for (int i = 1; i <= k; i++) insert(o[i]);
		for (int i = 1; i < top; i++)
			t[s[i]].push_back(s[i+1]);
		printf("%lld\n", dp(1));
	}
	return 0;
}