Luogu5024 [NOIP2018]保卫王国

题目蓝链

Description

给定一棵树，树上每一个点都有一个点权。你要在这棵树上选择一个点集，需要满足树上任意一条边所连的两个端点中至少有一个端点被选择，现在你需要最小化这个点集的点权和

有多次询问，每次询问钦定两个点分别选或不选，整棵树的代价是多少

Solution

我们需要维护以下三个数组

\(f[i][0/1]\)表示以\(i\)为根的子树中所有节点，\(i\)号点不选\(/\)选，所花费的最小代价是多少

\(F[i][j][0/1][0/1]\)表示\(i\)号点向上跳\(2^j\)的父亲\(f\)的子树(去掉以\(i\)为根的子树)，\(i\)号点不选\(/\)选，\(f\)号点不选\(/\)选，所花费的最小代价

\(rf[i][0/1]\)表示以\(i\)号点不选\(/\)选，整棵树的最小代价

然后询问的时候分两种情况讨论

一种情况就是一个点为另一个点的祖先，就把较深的节点\(a\)倍增到另一个点\(b\)的儿子处，然后讨论一下\(b\)选\(/\)不选

否则，就先把两个点都倍增到它们\(lca\)的儿子节点处，然后讨论\(lca\)选\(/\)不选就可以了

Code

#include <bits/stdc++.h>

using namespace std;

#define fst first
#define snd second
#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long LL;
typedef pair<int, int> pii;

inline int read() {
	int sum = 0, fg = 1; char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;
	for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);
	return fg * sum;
}

const int maxn = 1e5 + 10;
const LL inf = 0x3f3f3f3f3f3f3f3f;

vector<int> g[maxn];

struct node {
	LL v[2][2];
	node() { memset(v, 0x3f, sizeof v); }
}F[maxn][17];

int n, m, d[maxn], fa[maxn][17], w[maxn];
LL f[maxn][2], rf[maxn][2];

void dfs(int now, int _f) {
	d[now] = d[_f] + 1, fa[now][0] = _f;
	for (int i = 1; i <= 16; i++) fa[now][i] = fa[fa[now][i - 1]][i - 1];
	f[now][0] = 0, f[now][1] = w[now];
	for (int i = 0; i < g[now].size(); i++) {
		int son = g[now][i];
		if (son == _f) continue;
		dfs(son, now);
		f[now][0] += f[son][1];
		f[now][1] += min(f[son][0], f[son][1]);
	}
}

node merge(const node &a, const node &b) {
	node res;
	for (int i = 0; i < 2; i++)
		for (int j = 0; j < 2; j++)
			for (int k = 0; k < 2; k++)
				res.v[i][j] = min(res.v[i][j], a.v[i][k] + b.v[k][j]);
	return res;
}

void Dfs(int now, int f0, int f1) {
	rf[now][0] = f[now][0] + f1;
	rf[now][1] = f[now][1] + min(f0, f1);
	for (int i = 1; i <= 16; i++) F[now][i] = merge(F[now][i - 1], F[fa[now][i - 1]][i - 1]);
	for (int i = 0; i < g[now].size(); i++) {
		int son = g[now][i];
		if (son == fa[now][0]) continue;
		F[son][0].v[0][0] = inf;
		F[son][0].v[1][0] = f[now][0] - f[son][1];
		F[son][0].v[0][1] = F[son][0].v[1][1] = f[now][1] - min(f[son][0], f[son][1]);
		Dfs(son, rf[now][0] - f[son][1], rf[now][1] - min(f[son][0], f[son][1]));
	}
}

int main() {
#ifdef xunzhen
	freopen("defense.in", "r", stdin);
	freopen("defense.out", "w", stdout);
#endif

	static char Tmp[10];
	n = read(), m = read(), scanf("%s", Tmp);
	for (int i = 1; i <= n; i++) w[i] = read();
	for (int i = 1; i < n; i++) {
		int x = read(), y = read();
		g[x].push_back(y);
		g[y].push_back(x);
	}

	dfs(1, 0);
	Dfs(1, 0, 0);

	while (m--) {
		int a = read(), x = read(), b = read(), y = read();
		if (!(x | y) && (fa[a][0] == b || fa[b][0] == a)) {
			printf("-1\n"); continue;
		}
		if (d[a] < d[b]) swap(a, b), swap(x, y);
		node A, B;
		A.v[x][x] = f[a][x];
		for (int i = 16; ~i; i--)
			if (d[fa[a][i]] > d[b])
				A = merge(A, F[a][i]), a = fa[a][i];
		if (fa[a][0] == b) {
			LL ans0 = rf[b][0] - f[a][1], ans1 = rf[b][1] - min(f[a][0], f[a][1]);
			ans0 += A.v[x][1], ans1 += min(A.v[x][0], A.v[x][1]);
			printf("%lld\n", y ? ans1 : ans0);
		} else {
			if (d[a] > d[b]) A = merge(A, F[a][0]), a = fa[a][0];
			B.v[y][y] = f[b][y];
			for (int i = 16; ~i; i--)
				if (fa[a][i] != fa[b][i]) {
					A = merge(A, F[a][i]), B = merge(B, F[b][i]);
					a = fa[a][i], b = fa[b][i];
				}
			int lca = fa[a][0];
			LL ans0 = rf[lca][0] - f[a][1] - f[b][1], ans1 = rf[lca][1] - min(f[a][0], f[a][1]) - min(f[b][0], f[b][1]);
			ans0 += A.v[x][1] + B.v[y][1], ans1 += min(A.v[x][0], A.v[x][1]) + min(B.v[y][0], B.v[y][1]);
			printf("%lld\n", min(ans0, ans1));
		}
	}

	return 0;
}