leetcode-problemset/leetcode-cn/originData/maximum-points-after-collecting-coins-from-all-nodes.json

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        "question": {
            "questionId": "3179",
            "questionFrontendId": "2920",
            "categoryTitle": "Algorithms",
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            "title": "Maximum Points After Collecting Coins From All Nodes",
            "titleSlug": "maximum-points-after-collecting-coins-from-all-nodes",
            "content": "<p>There exists an undirected tree rooted at node <code>0</code> with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You are given a 2D <strong>integer</strong> array <code>edges</code> of length <code>n - 1</code>, where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree. You are also given a <strong>0-indexed</strong> array <code>coins</code> of size <code>n</code> where <code>coins[i]</code> indicates the number of coins in the vertex <code>i</code>, and an integer <code>k</code>.</p>\n\n<p>Starting from the root, you have to collect all the coins such that the coins at a node can only be collected if the coins of its ancestors have been already collected.</p>\n\n<p>Coins at <code>node<sub>i</sub></code> can be collected in one of the following ways:</p>\n\n<ul>\n\t<li>Collect all the coins, but you will get <code>coins[i] - k</code> points. If <code>coins[i] - k</code> is negative then you will lose <code>abs(coins[i] - k)</code> points.</li>\n\t<li>Collect all the coins, but you will get <code>floor(coins[i] / 2)</code> points. If this way is used, then for all the <code>node<sub>j</sub></code> present in the subtree of <code>node<sub>i</sub></code>, <code>coins[j]</code> will get reduced to <code>floor(coins[j] / 2)</code>.</li>\n</ul>\n\n<p>Return <em>the <strong>maximum points</strong> you can get after collecting the coins from <strong>all</strong> the tree nodes.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/09/18/ex1-copy.png\" style=\"width: 60px; height: 316px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;\" />\n<pre>\n<strong>Input:</strong> edges = [[0,1],[1,2],[2,3]], coins = [10,10,3,3], k = 5\n<strong>Output:</strong> 11                        \n<strong>Explanation:</strong> \nCollect all the coins from node 0 using the first way. Total points = 10 - 5 = 5.\nCollect all the coins from node 1 using the first way. Total points = 5 + (10 - 5) = 10.\nCollect all the coins from node 2 using the second way so coins left at node 3 will be floor(3 / 2) = 1. Total points = 10 + floor(3 / 2) = 11.\nCollect all the coins from node 3 using the second way. Total points = 11 + floor(1 / 2) = 11.\nIt can be shown that the maximum points we can get after collecting coins from all the nodes is 11. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<strong class=\"example\"> <img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/09/18/ex2.png\" style=\"width: 140px; height: 147px; padding: 10px; background: #fff; border-radius: .5rem;\" /></strong>\n\n<pre>\n<strong>Input:</strong> edges = [[0,1],[0,2]], coins = [8,4,4], k = 0\n<strong>Output:</strong> 16\n<strong>Explanation:</strong> \nCoins will be collected from all the nodes using the first way. Therefore, total points = (8 - 0) + (4 - 0) + (4 - 0) = 16.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == coins.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= coins[i] &lt;= 10<sup>4</sup></font></code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= edges[i][0], edges[i][1] &lt; n</font></code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= k &lt;= 10<sup>4</sup></font></code></li>\n</ul>\n",
            "translatedTitle": "收集所有金币可获得的最大积分",
            "translatedContent": "<p>有一棵由 <code>n</code> 个节点组成的无向树，以&nbsp;<code>0</code>&nbsp; 为根节点，节点编号从 <code>0</code> 到 <code>n - 1</code> 。给你一个长度为 <code>n - 1</code> 的二维 <strong>整数</strong> 数组 <code>edges</code> ，其中 <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示在树上的节点 <code>a<sub>i</sub></code> 和 <code>b<sub>i</sub></code> 之间存在一条边。另给你一个下标从 <strong>0</strong> 开始、长度为 <code>n</code> 的数组 <code>coins</code> 和一个整数 <code>k</code> ，其中 <code>coins[i]</code> 表示节点 <code>i</code> 处的金币数量。</p>\n\n<p>从根节点开始，你必须收集所有金币。要想收集节点上的金币，必须先收集该节点的祖先节点上的金币。</p>\n\n<p>节点 <code>i</code> 上的金币可以用下述方法之一进行收集：</p>\n\n<ul>\n\t<li>收集所有金币，得到共计 <code>coins[i] - k</code> 点积分。如果 <code>coins[i] - k</code> 是负数，你将会失去 <code>abs(coins[i] - k)</code> 点积分。</li>\n\t<li>收集所有金币，得到共计 <code>floor(coins[i] / 2)</code> 点积分。如果采用这种方法，节点 <code>i</code> 子树中所有节点 <code>j</code> 的金币数 <code>coins[j]</code> 将会减少至 <code>floor(coins[j] / 2)</code> 。</li>\n</ul>\n\n<p>返回收集 <strong>所有</strong> 树节点的金币之后可以获得的最大积分。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/09/18/ex1-copy.png\" style=\"width: 60px; height: 316px; padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem;\" />\n<pre>\n<strong>输入：</strong>edges = [[0,1],[1,2],[2,3]], coins = [10,10,3,3], k = 5\n<strong>输出：</strong>11                        \n<strong>解释：</strong>\n使用第一种方法收集节点 0 上的所有金币。总积分 = 10 - 5 = 5 。\n使用第一种方法收集节点 1 上的所有金币。总积分 = 5 + (10 - 5) = 10 。\n使用第二种方法收集节点 2 上的所有金币。所以节点 3 上的金币将会变为 floor(3 / 2) = 1 ，总积分 = 10 + floor(3 / 2) = 11 。\n使用第二种方法收集节点 3 上的所有金币。总积分 =  11 + floor(1 / 2) = 11.\n可以证明收集所有节点上的金币能获得的最大积分是 11 。 \n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n<strong class=\"example\"> <img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/09/18/ex2.png\" style=\"width: 140px; height: 147px; padding: 10px; background: #fff; border-radius: .5rem;\" /></strong>\n\n<pre>\n<strong>输入：</strong>edges = [[0,1],[0,2]], coins = [8,4,4], k = 0\n<strong>输出：</strong>16\n<strong>解释：</strong>\n使用第一种方法收集所有节点上的金币，因此，总积分 = (8 - 0) + (4 - 0) + (4 - 0) = 16 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == coins.length</code></li>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= coins[i] &lt;= 10<sup>4</sup></font></code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= edges[i][0], edges[i][1] &lt; n</font></code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= k &lt;= 10<sup>4</sup></font></code></li>\n</ul>\n",
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                "Let <code>dp[x][t]</code> be the maximum points we can get from the subtree rooted at node <code>x</code> and the second operation has been used <code>t</code> times in its ancestors.",
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