Luckily, there are a number of options that are available that are worth

本文介绍了多种飓风防护措施,包括便宜且易于安装的胶合板护窗、符合建筑规范的风暴面板、无需额外护窗的风暴玻璃、多功能巴哈马护窗、美观的殖民地式护窗、可折叠的百叶窗式护窗以及昂贵但高效的滚动式护窗。

Protection of your home or business from the potential damage of a hurricane or other tropical storm is extremely important and you should consider the various protections that are available to you. Luckily, there are a number of options that are available that are worth consideration and each vary in the level of cost and quality of protection. In addition, sum of these systems provide additional benefits to your home such as additional security and even decorative qualities.

Plywood

Even if this technique is not going to satisfy building codes, for home owners who have no additional defense device, plywood hurricane shutters are among the cheapest type of protection against storm damage. In order to work, plywood window shutters needs to be secured properly by utilizing a slab of wood at least 5/8 inches thick that overlaps your windows by no less than 4 inches on all sides. Wooden shutters also need to be mounted securely by utilizing barrel mounting bolts and tapping anchoring screws to help keep the blowing wind from sneaking under the wooden shutters.

Storm Panels

One of the most affordable, code-legal storm damage safeguard is available in the kind of aluminum, metal or the more pricey polycarbonate panels. Hurricane storm panels are usually fixed onto pre-installed mounting bolts or tracks well before a storm and then stacked away when they are not necessary. Mainly because hurricane panels are relatively cheap and work effectively, they really are a favorite of quite a few home owners. The installation of these panels are not always extraordinarily difficult but they will usually take more than one person to finish.

Storm Glass

Hurricane glass, which protects just like the plastic-layered automobile car window glass, is growing in reputation for the home market since no window shutters are required. However, installing hurricane glass is not easy and, in most cases, must be installed by a contractor. Hurricane glass supplies full time hurricane defense but the price can be huge for those who own older homes since they will often require complete window replacements.

Bahama Hurricane Window Shutters

Using Bahama window shutter systems is as simple as lowering and locking them in place. This all year round instant drop-down defense serves a dual function as hurricane damage protection and then a sunshade, working into a house's overall look instead of taking away from it.

Colonial Hurricane Shades

Just like the Bahama shades, the permanently installed colonial hurricane shutters are usually designed to improve the outward appearance of the home. In contrast to the Bahama window shutters, colonial shutters fold in from the ends of a window, where they are snapped, closed, and locked during stormy weather.

Accordion Hurricane Window Shutters

The accordion style hurricane shutters are able to hidden when not needed. They are folded alongside the windows on which they are installed. Depending on the size of the windows on which they are installed, the accordion shutters can be made up of one or more pieces. The shutters are usually manufactured from durable aluminum which can either be secured from the inside or outside. Accordion shutters operate on a wheel-gliding device which is often vulnerable to breakage. Being a constant fixture, accordion shutters do alter the outdoor style of the property.

Roll Down Hurricane Shutters

Roll down hurricane shutters are the most expensive type of hurricane shutter system. When installed, they completely cover the window that is being protected but are nearly invisible when not in use. When there is an impending storm, the shutters can be instantly closed. The more expensive models are lowered via an electric motor controller by a switch or button. Other style roll down shutters are closed or lowered with a manual hand crank. Roll down hurricane shutters are very effective in securing your home or business from vandalism or theft as well as being extremely effective against storm damage.

这是一个经典的**树上的最小覆盖路径问题**: 给定一棵 $ N $ 个节点的无根树(边权为1),对于每个查询 $ K $,要求选出 $ K $ 个点,使得 Claire 从某个入口出发,走完这 $ K $ 个点的**最短路径长度最小**。 她可以选择任意一个点作为起点,并且只需要访问这 $ K $ 个点(不要求返回)。 --- ## ✅ 问题转化 我们要找的是:在树中选择 $ K $ 个点,使得访问它们的最短路径长度最小。 等价于:**找出一个包含 $ K $ 个节点的连通子树,其“遍历距离”最小**。 而遍历一棵树的最小距离 = $ 2 \times (\text{边数}) - (\text{最长链长度}) $ 但注意:我们不需要回到起点 ⇒ 最优路径是沿着子树的一条路径走,不回头。 实际上,**访问一个点集的最短路径 = 该点集诱导子树的边数 × 2 - 直径** 更简单的方法是: > 在树中,访问 $ K $ 个点的最短路径长度 = $ 2 \times (K - 1) - \max_{\text{diameter of K-node subtree}} (\text{diam} - 1) $? 不直观。 --- ## ✅ 正确思路:贪心构造最长可能路径 关键观察: - Claire 可以从任意点开始 - 她要访问 $ K $ 个点,最优方式是选择一个**直径尽可能长的连通子图**,然后从一端走到另一端 - 实际上,最优策略是:**选一条路径(链)作为主干,然后尽量少走回头路** 但更好的方法来自经典结论: > 访问 $ K $ 个点的最短路径长度 = $ 2 \times (K - 1) - L + 1 $,其中 $ L $ 是所选 $ K $ 个点构成的子树中最长路径(即直径) 但这复杂。 --- ## ✅ 经典结论(重要!) > **最小行走距离 = $ 2 \times (K - 1) - \max_{T' \subseteq T, |T'|=K} \text{diameter}(T') $ 的最大值?不对** 反例:如果选的是一条链,则距离 = $ K - 1 $ 如果选的是星型,则距离 = $ 2(K-1) $(必须来回) 所以我们应该**最大化所选 $ K $ 个点之间的连通性和路径长度** ### 🌟 正确结论: > 要使行走距离最短,应选择一个**包含最多“连续路径”的子树**,即:**优先选择一条长链,再添加靠近它的点** 但实际上,最优解是: > **最小路径长度 = $ 2 \times (K - 1) - \max_{P} (|P| - 1) $**,其中 $ P $ 是所选子树中的最长路径? No. --- ## ✅ 真正正确的思路:**删边法 + 树的直径** 来自 ACM/ICPC 经典题: > 在一棵树中,访问 $ K $ 个点的最短路径 = $ 2 \times (K - 1) - \max_{S \subseteq V, |S|=K} \text{diameter}(S) + 1 $? 还是否 等等! ### ✅ 正确模型: 想象你构建了一个包含 $ K $ 个点的子树 $ T_K $。 遍历这个子树所需的最短路径是:**从某点出发,走完所有边至少一次?不,可以路过而不回溯** 但 Claire 不需要走所有边,只需访问这些点。所以最优路径是:**在这棵子树中找一条最长路径(直径),然后从一端走到另一端,途中会经过所有点吗?不一定** 但她可以走一条路径,覆盖尽可能多的点。 然而,为了最小化行走距离,我们应该让这 $ K $ 个点**尽可能集中在一个链上**。 --- ## ✅ 最佳策略:**答案 = $ K - \max_{\text{path in tree}} \min(K, \text{number of nodes on path}) $ ? No** --- ## ✅ 正解来源:CodeForces / ICPC 类似题 —— “Tree and $ K $ vertices” ### 🔥 关键洞察: > **最小行走距离 = $ 2K - 2 - \max_{u,v} \min(K, \text{dist}(u,v)) $**? 不对 --- ## ✅ 查阅类似题目后得出:这是「选择 $ K $ 个点,使其诱导子树的边权和最小」的问题 但我们是要最小化**行走路径**,不是子树大小。 ### 🚀 正确模型: - 一旦选定 $ K $ 个点,最优行走路径是从某个点出发,DFS 遍历整个连通部分,但可以最后停在叶子。 - 这样的路径长度 = $ 2 \times (内部边数) - (最长路径)$ - 更准确地说:**任何树的遍历,最少行走距离 = $ 2 \times (边数) - \text{diameter}$** 因为你可以从直径一端走到另一端,中间分支都走两次(去回),唯独直径只走一次。 设选中的 $ K $ 个点形成一棵子树 $ T_K $,有 $ E = K - 1 $ 条边。 则遍历它所需距离 = $ 2(K - 1) - D $,其中 $ D $ 是该子树的直径。 要最小化这个值 ⇒ 就要**最大化 $ D $**,即让这 $ K $ 个点尽可能分布在一条长链上。 所以: > 最小距离 = $ 2(K - 1) - \max D $ 但 $ D \leq \min(K, \text{tree diameter}) $ 而且我们能构造出 $ D = \min(K, \text{longest path in tree}) $ 但 not always. --- ## ✅ 正确做法(Accepted 思路): > **答案 = $ 2K - 2 - \ell $**,其中 $ \ell $ 是能在树中找到的、长度为 $ K $ 的连通子图中最长的路径长度(即最大可能直径) 但我们无法枚举所有子图。 --- ## ✅ 实际正确解法(基于“重心扩展”或“直径中心”) ### 🌟 经典结论(来自 CF 和 ICPC): > 在树中,访问 $ K $ 个点的最短路径长度 = $ 2K - 2 - \max_{v} \min(K, \text{longest path through } v) $ ? No --- ## ✅ 换个角度:**答案 = $ K - \max_{\text{chain}} \min(K, \text{length of chain}) $**? No --- ## ✅ 观察样例 ``` 输入: 1 4 2 3 2 1 2 4 2 2 4 ``` 树结构: ``` 3 | 2 / \ 1 4 ``` 形状是:2 是中心,连接 1,3,4 - 查询 K=2:选哪两个点? - 选相邻两点,如 2 和 1 ⇒ 路径长 = 1 - 输出 1 ✅ - 查询 K=4:必须 visit all - 例如:1→2→3→4,距离 = 3?但输出是 4 ❌ Wait! Output is: ``` 1 4 ``` So for K=4, answer is 4? How? Let’s simulate: To visit all 4 nodes starting from any point. Best route: start at 1 → 2 → 3 → 2 → 4, distance = 4 Yes! Because you have to go back to 2 after visiting 3, then go to 4. You cannot avoid backtracking. Number of edges traversed = 4 In a tree with K=4 nodes, if the structure forces backtracking, the minimum walk = 2*(K - 1) - (diameter - 1) Here: - K = 4 - Edges = 3 - But you must traverse some edge twice Specifically: any traversal that visits all nodes in a tree must use each edge at least once; if it's not on the main path, it may be used twice. The minimum walking distance to visit K nodes = $ 2 \times (K - 1) - \text{length of the longest simple path among the selected K nodes} $ Because: - You can save one traversal for each edge on the final path (you don't backtrack) - All other branches are entered and left ⇒ cost 2 per edge - The unique path from start to end is only walked once So total = $ 2(E) - (L - 1) $ where $ E = K - 1 $, $ L = \text{number of nodes on the path} $, so $ L - 1 = \text{length in edges} $ Let $ D $ = number of edges in the longest path within the selected set ⇒ saves $ D $ steps Then total distance = $ 2(K-1) - D $ We want to minimize this ⇒ maximize $ D $ So: > Answer = $ 2(K - 1) - \max D $ And $ \max D $ over all K-node subtrees = the maximum possible diameter of any connected K-node subtree But how to compute it fast? --- ## ✅ Insight: the optimal set is always a union of a long path plus nearby nodes But even better: **the maximum possible diameter among any K-node connected subtree is min(K-1, global_diameter)** No. For example, if K=2, max diameter = 1 If K=3, we can get diameter 2 if there is a path of length 2 So yes, max possible D = min(K-1, global_tree_diameter) Wait: in our sample: - Global tree: 1-2-3 or 1-2-4 or 3-2-4, diameter = 2 (e.g., 3-2-4 has length 2) - For K=4: max possible diameter = 2 (since no three-edge path) - So answer = 2*(4-1) - 2 = 6 - 2 = 4 ✅ - For K=2: max possible diameter = 1 ⇒ answer = 2*1 - 1 = 1 ✅ Perfect! So algorithm: 1. Find the **diameter of the tree** (in terms of number of edges) - Use two BFS/DFS: pick any node, find farthest, then from that find farthest again ⇒ distance is diameter (edge count) 2. Precompute for each K from 1 to N: - `ans[K] = 2*(K - 1) - min(K - 1, diam)` - because max achievable diameter in K-node subtree is `min(K-1, diam)` - when K=1, ans=0 But wait: can we always achieve diameter = min(K-1, diam)? Yes! Because we can take a segment of the diameter path of length `min(K, diam+1)` nodes, which gives diameter `min(K-1, diam)` And adding extra nodes won’t increase the diameter beyond `diam`, but we can always select a path of length `min(K-1, diam)` So yes. Therefore: > **Answer for query K = 2*(K - 1) - min(K - 1, diameter)** = $$ \begin{cases} 2K - 2 - (K - 1) = K - 1 & \text{if } K - 1 \leq \text{diam} \\ 2K - 2 - \text{diam} & \text{otherwise} \end{cases} $$ But since `min(K-1, diam)` is just `K-1` if `K-1 <= diam`, else `diam` So: ```cpp ans = 2*(K-1) - min(K-1, diam); ``` Which simplifies to: - If K-1 <= diam: ans = 2*(K-1) - (K-1) = K-1 - Else: ans = 2*(K-1) - diam But in our case: - diam = 2 (edges), e.g., 3-2-4 - K=2: 2-1=1 <= 2 ⇒ ans = 1 ✅ - K=4: 4-1=3 > 2 ⇒ ans = 2*3 - 2 = 6 - 2 = 4 ✅ Perfect. --- ## ✅ C++ 实现代码 ```cpp #include <iostream> #include <vector> #include <queue> #include <algorithm> #include <cstring> using namespace std; typedef pair<int, int> pii; // BFS 返回 {farthest_node, distance} pii bfs(const vector<vector<int>>& graph, int start) { int n = graph.size(); vector<int> dist(n, -1); queue<int> q; q.push(start); dist[start] = 0; int u = start; while (!q.empty()) { u = q.front(); q.pop(); for (int v : graph[v]) { if (dist[v] == -1) { dist[v] = dist[u] + 1; q.push(v); } } } int maxDist = 0; int farthest = start; for (int i = 1; i < n; ++i) { if (dist[i] > maxDist) { maxDist = dist[i]; farthest = i; } } return {farthest, maxDist}; } int main() { ios::sync_with_stdio(false); cin.tie(0); int T; cin >> T; while (T--) { int N, M; cin >> N >> M; vector<vector<int>> graph(N + 1); for (int i = 0; i < N - 1; ++i) { int u, v; cin >> u >> v; graph[u].push_back(v); graph[v].push_back(u); } // Step 1: find diameter using two BFS int start = 1; pii p1 = bfs(graph, start); pii p2 = bfs(graph, p1.first); int diameter = p2.second; // in terms of number of edges // Process queries while (M--) { int K; cin >> K; if (K == 1) { cout << 0 << '\n'; } else { int max_achievable_diameter = min(K - 1, diameter); int ans = 2 * (K - 1) - max_achievable_diameter; cout << ans << '\n'; } } } return 0; } ``` ⚠️ Fix bug in BFS loop: ```cpp for (int v : graph[u]) { // was 'graph[v]' ``` --- ## ✅ Fixed BFS function ```cpp pii bfs(const vector<vector<int>>& graph, int start) { int n = graph.size(); vector<int> dist(n, -1); queue<int> q; q.push(start); dist[start] = 0; int u = start; while (!q.empty()) { u = q.front(); q.pop(); for (int neighbor : graph[u]) { // fixed: graph[u], not graph[v] if (dist[neighbor] == -1) { dist[neighbor] = dist[u] + 1; q.push(neighbor); } } } int maxDist = 0; int farthest = start; for (int i = 1; i < n; ++i) { if (dist[i] > maxDist) { maxDist = dist[i]; farthest = i; } } return {farthest, maxDist}; } ``` --- ## ✅ Summary - Diameter of tree found by two BFS - For each query K: - If K == 1: 0 - Else: `ans = 2*(K-1) - min(K-1, diameter)` - This works because the best way to minimize walking is to choose a set of K nodes with the largest possible internal diameter, saving backtracking on that path. ---
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