【LG4103】[HEOI2014]大工程

题面

洛谷

题解

先建虚树，下面所有讨论均是在虚树上的。

对于第一问：直接统计所有树边对答案的贡献即可。

对于第\(2,3\)问：记\(f[x]\)表示在\(x\)的子树内离\(x\)距离最远的关键点的距离，\(g[x]\)表示在\(x\)的子树内离\(x\)距离最近的关键点的距离。

具体更新以\(f[x]\)为例：

访问到\(v\in son_x\)，

如果以前访问过的点中有关键点，则有\(f[x]=max(f[x],f[v]+dis(u,v)+f[x])\)，

每次还要向上传递，即\(f[x]=max(f[x],f[v]+dis(u,v))\)。

代码

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <algorithm>

using namespace std;

inline int gi() {

    register int data = 0, w = 1;

    register char ch = 0;

    while (!isdigit(ch) && ch != '-') ch = getchar();

    if (ch == '-') w = -1, ch = getchar();

    while (isdigit(ch)) data = 10 * data + ch - '0', ch = getchar();

    return w * data;

}

const int MAX_N = 1e6 + 5;

struct Graph { int to, next; } e[MAX_N << 2];

int fir1[MAX_N], fir2[MAX_N], e_cnt;

void clearGraph() {

	memset(fir1, -1, sizeof(fir1));

	memset(fir2, -1, sizeof(fir2));

}

void Add_Edge(int *fir, int u, int v) {

	e[e_cnt] = (Graph){v, fir[u]};

	fir[u] = e_cnt++;

}

namespace Tree {

	int fa[MAX_N], dep[MAX_N], size[MAX_N], top[MAX_N], son[MAX_N], dfn[MAX_N], tim;

	void dfs1(int x) {

		dfn[x] = ++tim;

	    size[x] = 1, dep[x] = dep[fa[x]] + 1;

		for (int i = fir1[x]; ~i; i = e[i].next) {

			int v = e[i].to; if (v == fa[x]) continue;

			fa[v] = x; dfs1(v); size[x] += size[v];

			if (size[v] > size[son[x]]) son[x] = v;

		}

	}

	void dfs2(int x, int tp) {

		top[x] = tp;

	    if (son[x]) dfs2(son[x], tp);

		for (int i = fir1[x]; ~i; i = e[i].next) {

			int v = e[i].to; if (v == fa[x] || v == son[x]) continue;

			dfs2(v, v);

		}

	}

	int LCA(int x, int y) {

		while (top[x] != top[y]) {

			if (dep[top[x]] < dep[top[y]]) swap(x, y);

			x = fa[top[x]];

		}

		return dep[x] < dep[y] ? x : y;

	}

}

using Tree::LCA; using Tree::dfn; using Tree::dep;

int N, M, K, a[MAX_N];

bool key[MAX_N];

int f[MAX_N], g[MAX_N], s[MAX_N];

bool cmp(int i, int j) { return dfn[i] < dfn[j]; }

void build() {

	static int stk[MAX_N], top;

	sort(&a[1], &a[K + 1], cmp);

	stk[top = 1] = 1; fir2[1] = -1;

	e_cnt = 0;

	for (int i = 1; i <= K; i++) {

		key[a[i]] = 1;

		if (a[i] == 1) continue;

		int lca = LCA(stk[top], a[i]);

		if (lca != stk[top]) {

			while (dfn[lca] < dfn[stk[top - 1]]) {

				int u = stk[top], v = stk[top - 1];

				Add_Edge(fir2, u, v), Add_Edge(fir2, v, u);

				--top;

			}

			if (dfn[lca] > dfn[stk[top - 1]]) {

				fir2[lca] = -1; int u = stk[top], v = lca;

				Add_Edge(fir2, u, v), Add_Edge(fir2, v, u);

				stk[top] = lca;

			}

			else {

				int u = lca, v = stk[top--];

				Add_Edge(fir2, u, v), Add_Edge(fir2, v, u);

			}

		}

		fir2[a[i]] = -1, stk[++top] = a[i];

	}

	for (int i = 1; i < top; i++) {

		int u = stk[i], v = stk[i + 1];

		Add_Edge(fir2, u, v), Add_Edge(fir2, v, u);

	}

}

long long ans1;

int ans2, ans3;

void Dp(int x, int fa) {

	s[x] = key[x], f[x] = 0, g[x] = (key[x] ? 0 : 1e9);

	for (int i = fir2[x]; ~i; i = e[i].next) {

		int v = e[i].to; if (v == fa) continue;

		Dp(v, x);

	}

	for (int i = fir2[x]; ~i; i = e[i].next) {

		int v = e[i].to, w = dep[v] - dep[x];

		if (v == fa) continue;

		ans1 += 1ll * (K - s[v]) * s[v] * w;

		if (s[x] > 0) {

			ans2 = min(ans2, g[x] + w + g[v]);

			ans3 = max(ans3, f[x] + w + f[v]);

		}

		g[x] = min(g[x], g[v] + w);

	    f[x] = max(f[x], f[v] + w);

		s[x] += s[v];

	}

	key[x] = 0;

}

int main () {

#ifndef ONLINE_JUDGE

    freopen("cpp.in", "r", stdin);

#endif

	clearGraph();

	N = gi();

	for (int i = 1; i < N; i++) {

		int u = gi(), v = gi();

		Add_Edge(fir1, u, v), Add_Edge(fir1, v, u);

	}

	Tree::dfs1(1), Tree::dfs2(1, 1);

	M = gi();

	while (M--) {

		ans1 = 0, ans2 = 1e9, ans3 = 0;

		K = gi(); for (int i = 1; i <= K; i++) a[i] = gi();

		build();

		Dp(1, 0);

		printf("%lld %d %d\n", ans1, ans2, ans3);

	}

	return 0;

}