mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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193 lines
20 KiB
JSON
193 lines
20 KiB
JSON
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"title": "Maximum Points After Collecting Coins From All Nodes",
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"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> </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> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == coins.length</code></li>\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\n\t<li><code><font face=\"monospace\">0 <= coins[i] <= 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 <= edges[i][0], edges[i][1] < n</font></code></li>\n\t<li><code><font face=\"monospace\">0 <= k <= 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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"Note that the value of each <code>node <= 10<sup>4</sup></code>, so when <code>t >= 14</code> <code>dp[x][t]</code> is always <code>0</code>.",
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"General equation will be: <code>dp[x][t] = max((coins[x] >> t) - k + sigma(dp[y][t]), (coins[x] >> (t + 1)) + sigma(dp[y][t + 1]))</code> where nodes denoted by <code>y</code> in the sigma, are the direct children of node <code>x</code>."
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