leetcode-problemset/leetcode-cn/originData/minimize-the-total-price-of-the-trips.json

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        "question": {
            "questionId": "2739",
            "questionFrontendId": "2646",
            "categoryTitle": "Algorithms",
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            "title": "Minimize the Total Price of the Trips",
            "titleSlug": "minimize-the-total-price-of-the-trips",
            "content": "<p>There exists an undirected and unrooted tree with <code>n</code> nodes indexed from <code>0</code> to <code>n - 1</code>. You are given the integer <code>n</code> and a 2D integer 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.</p>\n\n<p>Each node has an associated price. You are given an integer array <code>price</code>, where <code>price[i]</code> is the price of the <code>i<sup>th</sup></code> node.</p>\n\n<p>The <strong>price sum</strong> of a given path is the sum of the prices of all nodes lying on that path.</p>\n\n<p>Additionally, you are given a 2D integer array <code>trips</code>, where <code>trips[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> indicates that you start the <code>i<sup>th</sup></code> trip from the node <code>start<sub>i</sub></code> and travel to the node <code>end<sub>i</sub></code> by any path you like.</p>\n\n<p>Before performing your first trip, you can choose some <strong>non-adjacent</strong> nodes and halve the prices.</p>\n\n<p>Return <em>the minimum total price sum to perform all the given trips</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/2023/03/16/diagram2.png\" style=\"width: 541px; height: 181px;\" />\n<pre>\n<strong>Input:</strong> n = 4, edges = [[0,1],[1,2],[1,3]], price = [2,2,10,6], trips = [[0,3],[2,1],[2,3]]\n<strong>Output:</strong> 23\n<strong>Explanation:</strong> The diagram above denotes the tree after rooting it at node 2. The first part shows the initial tree and the second part shows the tree after choosing nodes 0, 2, and 3, and making their price half.\nFor the 1<sup>st</sup> trip, we choose path [0,1,3]. The price sum of that path is 1 + 2 + 3 = 6.\nFor the 2<sup>nd</sup> trip, we choose path [2,1]. The price sum of that path is 2 + 5 = 7.\nFor the 3<sup>rd</sup> trip, we choose path [2,1,3]. The price sum of that path is 5 + 2 + 3 = 10.\nThe total price sum of all trips is 6 + 7 + 10 = 23.\nIt can be proven, that 23 is the minimum answer that we can achieve.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/03/16/diagram3.png\" style=\"width: 456px; height: 111px;\" />\n<pre>\n<strong>Input:</strong> n = 2, edges = [[0,1]], price = [2,2], trips = [[0,0]]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> The diagram above denotes the tree after rooting it at node 0. The first part shows the initial tree and the second part shows the tree after choosing node 0, and making its price half.\nFor the 1<sup>st</sup> trip, we choose path [0]. The price sum of that path is 1.\nThe total price sum of all trips is 1. It can be proven, that 1 is the minimum answer that we can achieve.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 50</code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>\n\t<li><code>edges</code> represents a valid tree.</li>\n\t<li><code>price.length == n</code></li>\n\t<li><code>price[i]</code> is an even integer.</li>\n\t<li><code>1 &lt;= price[i] &lt;= 1000</code></li>\n\t<li><code>1 &lt;= trips.length &lt;= 100</code></li>\n\t<li><code>0 &lt;= start<sub>i</sub>, end<sub>i</sub>&nbsp;&lt;= n - 1</code></li>\n</ul>\n",
            "translatedTitle": "最小化旅行的价格总和",
            "translatedContent": "<p>现有一棵无向、无根的树，树中有 <code>n</code> 个节点，按从 <code>0</code> 到 <code>n - 1</code> 编号。给你一个整数 <code>n</code> 和一个长度为 <code>n - 1</code> 的二维整数数组 <code>edges</code> ，其中 <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 表示树中节点 <code>a<sub>i</sub></code> 和 <code>b<sub>i</sub></code> 之间存在一条边。</p>\n\n<p>每个节点都关联一个价格。给你一个整数数组 <code>price</code> ，其中 <code>price[i]</code> 是第 <code>i</code> 个节点的价格。</p>\n\n<p>给定路径的 <strong>价格总和</strong> 是该路径上所有节点的价格之和。</p>\n\n<p>另给你一个二维整数数组 <code>trips</code> ，其中 <code>trips[i] = [start<sub>i</sub>, end<sub>i</sub>]</code> 表示您从节点 <code>start<sub>i</sub></code> 开始第 <code>i</code> 次旅行，并通过任何你喜欢的路径前往节点 <code>end<sub>i</sub></code> 。</p>\n\n<p>在执行第一次旅行之前，你可以选择一些 <strong>非相邻节点</strong> 并将价格减半。</p>\n\n<p>返回执行所有旅行的最小价格总和。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/03/16/diagram2.png\" style=\"width: 541px; height: 181px;\">\n<pre><strong>输入：</strong>n = 4, edges = [[0,1],[1,2],[1,3]], price = [2,2,10,6], trips = [[0,3],[2,1],[2,3]]\n<strong>输出：</strong>23\n<strong>解释：\n</strong>上图表示将节点 2 视为根之后的树结构。第一个图表示初始树，第二个图表示选择节点 0 、2 和 3 并使其价格减半后的树。\n第 1 次旅行，选择路径 [0,1,3] 。路径的价格总和为 1 + 2 + 3 = 6 。\n第 2 次旅行，选择路径 [2,1] 。路径的价格总和为 2 + 5 = 7 。\n第 3 次旅行，选择路径 [2,1,3] 。路径的价格总和为 5 + 2 + 3 = 10 。\n所有旅行的价格总和为 6 + 7 + 10 = 23 。可以证明，23 是可以实现的最小答案。</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2023/03/16/diagram3.png\" style=\"width: 456px; height: 111px;\">\n<pre><strong>输入：</strong>n = 2, edges = [[0,1]], price = [2,2], trips = [[0,0]]\n<strong>输出：</strong>1\n<strong>解释：</strong>\n上图表示将节点 0 视为根之后的树结构。第一个图表示初始树，第二个图表示选择节点 0 并使其价格减半后的树。 \n第 1 次旅行，选择路径 [0] 。路径的价格总和为 1 。 \n所有旅行的价格总和为 1 。可以证明，1 是可以实现的最小答案。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n &lt;= 50</code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt;= n - 1</code></li>\n\t<li><code>edges</code> 表示一棵有效的树</li>\n\t<li><code>price.length == n</code></li>\n\t<li><code>price[i]</code> 是一个偶数</li>\n\t<li><code>1 &lt;= price[i] &lt;= 1000</code></li>\n\t<li><code>1 &lt;= trips.length &lt;= 100</code></li>\n\t<li><code>0 &lt;= start<sub>i</sub>, end<sub>i</sub>&nbsp;&lt;= n - 1</code></li>\n</ul>\n",
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                "The final answer is the price[i] * freq[i], where freq[i] is the number of times node i was visited during the trip, and price[i] is the final price.",
                "To find freq[i] we will use dfs or bfs for each trip and update every node on the path start and end.",
                "Finally, to find the final price[i] we will use dynamic programming on the tree. Let dp(v, 0/1) denote the minimum total price with the node v’s price being halved or not."
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