【BZOJ1758】【WC2010】重建计划

【题目链接】

• 点击打开链接
  

【思路要点】

• 首先，题目要求求解平均值的最大值，考虑分数规划。二分答案\(Mid\)，将树上每条边权减去\(Mid\)，问题变为树上是否存在边数在\(L\)到\(R\)之间的权值之和非负的路径。考虑点分治。
• 记录\(Dist\)数组，\(Dist_{i}\)表示边数为\(i\)的从分治根节点出发的路径的最大权值。
• 对于每一个分治子树进行BFS，记当前BFS深度为\(i\)，在\(Dist\)数组上使用单点队列维护区间\([max(0,L-i),min(R-i,Depth_{root})]\)最大值。
• 若如此做，每个分治重心处处理的时间复杂度为\(O(\sum_{i=1}^{子树数量}(Size_{i}+max_{j=1}^{i-1}\{Depth_{j}\}))\)，该复杂度可能退化为平方级别，导致超时。
• 考虑对每个子树根据其深度排序（基数排序），这样每个分治重心处处理的时间复杂度变为\(O(\sum_{i=1}^{子树数量}(Size_{i}+Depth_{i-1}))=O(Size_{分治重心})\)。即可通过。
• 另外，由于时限较紧，需要先将点分树构建出来才能够通过。

【代码】

#include<bits/stdc++.h>
using namespace std; 
#define MAXN	100005
#define INF	1e9
#define EPS	1e-5
struct edge {int dest; double len; }; 
vector <edge> a[MAXN], b[MAXN]; 
vector <int> c[MAXN]; 
int n, L, R, root, Root, ArrSize; 
double delta, value[MAXN]; 
bool found, visited[MAXN]; 
int size[MAXN], weight[MAXN], depth[MAXN], Depth[MAXN]; 
void getroot(int pos, int father, int total) {
	size[pos] = 1; 
	depth[pos] = 0; 
	weight[pos] = 0; 
	for (unsigned i = 0; i<a[pos].size(); i++)
		if (!visited[a[pos][i].dest] && a[pos][i].dest != father) {
			getroot(a[pos][i].dest, pos, total); 
			size[pos] += size[a[pos][i].dest]; 
			weight[pos] = max(weight[pos], size[a[pos][i].dest]); 
			depth[pos] = max(depth[pos], depth[a[pos][i].dest]); 
		}
	depth[pos]++; 
	weight[pos] = max(weight[pos], total - size[pos]); 
	if (weight[pos] < weight[root]) root = pos; 
}
void prework(int pos, int total) {
	visited[pos] = true; 
	getroot(pos, 0, total); 
	static int cnt[MAXN], rank[MAXN]; 
	int maxnum = 0; 
	for (unsigned i = 0; i<a[pos].size(); i++)
		if (!visited[a[pos][i].dest]) {
			maxnum = max(maxnum, depth[a[pos][i].dest]); 
			cnt[depth[a[pos][i].dest]]++; 
		}
	Depth[pos] = maxnum; 
	for (int i = 1; i <= maxnum; i++) cnt[i] += cnt[i - 1]; 
	int curr = 0; 
	for (unsigned i = 0; i<a[pos].size(); i++)
		if (!visited[a[pos][i].dest]) {
			rank[cnt[depth[a[pos][i].dest]]--] = i; 
			curr++; 
		}
	for (int i = 1; i <= curr; i++) b[pos].push_back(a[pos][rank[i]]); 
	for (int i = 1; i <= maxnum; i++) cnt[i] = 0; 
	for (unsigned i = 0; i<a[pos].size(); i++)
		if (!visited[a[pos][i].dest]) {
			root = 0; 
			getroot(a[pos][i].dest, 0, size[a[pos][i].dest]); 
			c[pos].push_back(root); 
			prework(root, size[a[pos][i].dest]); 
		}
}
void getans(int pos, double len) {
	static int q[MAXN], f[MAXN], Q[MAXN], dist[MAXN]; 
	static double Len[MAXN]; 
	int l = 0, r = 0, QL = 0, QR =  - 1, point = min(R, ArrSize); 
	while (point >= L) {
		while (QR >= QL && value[point]>value[Q[QR]]) QR--; 
		Q[++QR] = point; 
		point--; 
	}
	if (QL <= QR && value[Q[QL]] >= 0) found = true; 
	q[0] = pos; f[0] = 0; dist[0] = 1; Len[0] = len; 
	while (l <= r) {
		while (QL <= QR && dist[l] + Q[QL]>R) QL++;  
		while (point >= 0 && point + dist[l] >= L) {
			while (QR >= QL && value[point]>value[Q[QR]]) QR--; 
			Q[++QR] = point; 
			point--; 
		}
		if (QL <= QR && Len[l] + value[Q[QL]] >= 0) found = true; 
		int tmp = q[l]; 
		for (unsigned i = 0; i<a[tmp].size(); i++)
			if (!visited[a[tmp][i].dest] && a[tmp][i].dest != f[l]) {
				int next = a[tmp][i].dest; 
				r++; 
				q[r] = next; 
				f[r] = tmp; 
				dist[r] = dist[l] + 1; 
				Len[r] = Len[l] + a[tmp][i].len - delta; 
			}
		l++; 
	}
	ArrSize = dist[r]; 
	for (int i = 0; i <= r; i++)
		value[dist[i]] = max(value[dist[i]], Len[i]); 
}
void work(int pos) {
	visited[pos] = true; 
	ArrSize = 0; 
	for (unsigned i = 0; i<b[pos].size(); i++)
		getans(b[pos][i].dest, b[pos][i].len - delta); 
	for (int i = 1; i <= ArrSize; i++)
		value[i] = -INF; 
	for (unsigned i = 0; i<c[pos].size(); i++)
		work(c[pos][i]); 
}
void init() {
	memset(visited, false, sizeof(visited)); 
	size[0] = weight[0] = n; 
	root = 0; getroot(1, 0, n); 
	Root = root; 
	prework(root, n); 
	value[0] = 0; 
	for (int i = 1; i <= n; i++)
		value[i] = -INF; 
}
bool check(double x) {
	delta = x; 
	memset(visited, false, sizeof(visited)); 
	found = false; 
	work(Root); 
	return found; 
}
int main() {
	scanf("%d%d%d", &n, &L, &R); 
	double l = INF, r = 0; 
	for (int i = 1; i<n; i++) {
		int x, y; 
		double z; 
		scanf("%d%d%lf", &x, &y, &z); 
		a[x].push_back((edge){y, z}); 
		a[y].push_back((edge){x, z}); 
		r = max(r, z); l = min(l, z); 
	}
	init(); 
	while (l + EPS < r) {
		double mid = (l + r) / 2; 
		if (check(mid)) l = mid; 
		else r = mid; 
	}
	printf("%.3lf\n", (l + r) / 2 + EPS); 
	return 0; 
}