leetcode-problemset/leetcode-cn/originData/minimum-edge-reversals-so-every-node-is-reachable.json

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        "question": {
            "questionId": "3105",
            "questionFrontendId": "2858",
            "categoryTitle": "Algorithms",
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            "title": "Minimum Edge Reversals So Every Node Is Reachable",
            "titleSlug": "minimum-edge-reversals-so-every-node-is-reachable",
            "content": "<p>There is a <strong>simple directed graph</strong> with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. The graph would form a <strong>tree</strong> if its edges were bi-directional.</p>\n\n<p>You are given an integer <code>n</code> and a <strong>2D</strong> integer array <code>edges</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> represents a <strong>directed edge</strong> going from node <code>u<sub>i</sub></code> to node <code>v<sub>i</sub></code>.</p>\n\n<p>An <strong>edge reversal</strong> changes the direction of an edge, i.e., a directed edge going from node <code>u<sub>i</sub></code> to node <code>v<sub>i</sub></code> becomes a directed edge going from node <code>v<sub>i</sub></code> to node <code>u<sub>i</sub></code>.</p>\n\n<p>For every node <code>i</code> in the range <code>[0, n - 1]</code>, your task is to <strong>independently</strong> calculate the <strong>minimum</strong> number of <strong>edge reversals</strong> required so it is possible to reach any other node starting from node <code>i</code> through a <strong>sequence</strong> of <strong>directed edges</strong>.</p>\n\n<p>Return <em>an integer array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> is the</em><em> </em> <em><strong>minimum</strong> number of <strong>edge reversals</strong> required so it is possible to reach any other node starting from node </em><code>i</code><em> through a <strong>sequence</strong> of <strong>directed edges</strong>.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<p><img height=\"246\" src=\"https://assets.leetcode.com/uploads/2023/08/26/image-20230826221104-3.png\" width=\"312\" /></p>\n\n<pre>\n<strong>Input:</strong> n = 4, edges = [[2,0],[2,1],[1,3]]\n<strong>Output:</strong> [1,1,0,2]\n<strong>Explanation:</strong> The image above shows the graph formed by the edges.\nFor node 0: after reversing the edge [2,0], it is possible to reach any other node starting from node 0.\nSo, answer[0] = 1.\nFor node 1: after reversing the edge [2,1], it is possible to reach any other node starting from node 1.\nSo, answer[1] = 1.\nFor node 2: it is already possible to reach any other node starting from node 2.\nSo, answer[2] = 0.\nFor node 3: after reversing the edges [1,3] and [2,1], it is possible to reach any other node starting from node 3.\nSo, answer[3] = 2.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<p><img height=\"217\" src=\"https://assets.leetcode.com/uploads/2023/08/26/image-20230826225541-2.png\" width=\"322\" /></p>\n\n<pre>\n<strong>Input:</strong> n = 3, edges = [[1,2],[2,0]]\n<strong>Output:</strong> [2,0,1]\n<strong>Explanation:</strong> The image above shows the graph formed by the edges.\nFor node 0: after reversing the edges [2,0] and [1,2], it is possible to reach any other node starting from node 0.\nSo, answer[0] = 2.\nFor node 1: it is already possible to reach any other node starting from node 1.\nSo, answer[1] = 0.\nFor node 2: after reversing the edge [1, 2], it is possible to reach any other node starting from node 2.\nSo, answer[2] = 1.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code>edges[i].length == 2</code></li>\n\t<li><code>0 &lt;= u<sub>i</sub> == edges[i][0] &lt; n</code></li>\n\t<li><code>0 &lt;= v<sub>i</sub> == edges[i][1] &lt; n</code></li>\n\t<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>\n\t<li>The input is generated such&nbsp;that if the edges were bi-directional, the graph would be a tree.</li>\n</ul>\n",
            "translatedTitle": "可以到达每一个节点的最少边反转次数",
            "translatedContent": "<p>给你一个 <code>n</code>&nbsp;个点的 <strong>简单有向图</strong>&nbsp;（没有重复边的有向图），节点编号为 <code>0</code>&nbsp;到 <code>n - 1</code>&nbsp;。如果这些边是双向边，那么这个图形成一棵&nbsp;<strong>树</strong>&nbsp;。</p>\n\n<p>给你一个整数&nbsp;<code>n</code>&nbsp;和一个 <strong>二维</strong>&nbsp;整数数组&nbsp;<code>edges</code>&nbsp;，其中&nbsp;<code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code>&nbsp;表示从节点&nbsp;<code>u<sub>i</sub></code>&nbsp;到节点&nbsp;<code>v<sub>i</sub></code>&nbsp;有一条&nbsp;<strong>有向边</strong>&nbsp;。</p>\n\n<p><strong>边反转</strong>&nbsp;指的是将一条边的方向反转，也就是说一条从节点&nbsp;<code>u<sub>i</sub></code>&nbsp;到节点&nbsp;<code>v<sub>i</sub></code>&nbsp;的边会变为一条从节点&nbsp;<code>v<sub>i</sub></code>&nbsp;到节点&nbsp;<code>u<sub>i</sub></code>&nbsp;的边。</p>\n\n<p>对于范围&nbsp;<code>[0, n - 1]</code>&nbsp;中的每一个节点 <code>i</code>&nbsp;，你的任务是分别 <strong>独立</strong> 计算 <strong>最少</strong>&nbsp;需要多少次 <strong>边反转</strong>&nbsp;，从节点 <code>i</code>&nbsp;出发经过 <strong>一系列有向边</strong>&nbsp;，可以到达所有的节点。</p>\n\n<p>请你返回一个长度为 <code>n</code>&nbsp;的整数数组<em>&nbsp;</em><code>answer</code><em>&nbsp;</em>，其中<em>&nbsp;</em><code>answer[i]</code>表示从节点&nbsp;<code>i</code>&nbsp;出发，可以到达所有节点的&nbsp;<strong>最少边反转</strong>&nbsp;次数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<p><img height=\"246\" src=\"https://assets.leetcode.com/uploads/2023/08/26/image-20230826221104-3.png\" width=\"312\" /></p>\n\n<pre>\n<b>输入：</b>n = 4, edges = [[2,0],[2,1],[1,3]]\n<b>输出：</b>[1,1,0,2]\n<b>解释：</b>上图表示了与输入对应的简单有向图。\n对于节点 0 ：反转 [2,0] ，从节点 0 出发可以到达所有节点。\n所以 answer[0] = 1 。\n对于节点 1 ：反转 [2,1] ，从节点 1 出发可以到达所有节点。\n所以 answer[1] = 1 。\n对于节点 2 ：不需要反转就可以从节点 2 出发到达所有节点。\n所以 answer[2] = 0 。\n对于节点 3 ：反转 [1,3] 和 [2,1] ，从节点 3 出发可以到达所有节点。\n所以 answer[3] = 2 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<p><img height=\"217\" src=\"https://assets.leetcode.com/uploads/2023/08/26/image-20230826225541-2.png\" width=\"322\" /></p>\n\n<pre>\n<b>输入：</b>n = 3, edges = [[1,2],[2,0]]\n<b>输出：</b>[2,0,1]\n<b>解释：</b>上图表示了与输入对应的简单有向图。\n对于节点 0 ：反转 [2,0] 和 [1,2] ，从节点 0 出发可以到达所有节点。\n所以 answer[0] = 2 。\n对于节点 1 ：不需要反转就可以从节点 2 出发到达所有节点。\n所以 answer[1] = 0 。\n对于节点 2 ：反转 [1,2] ，从节点 2 出发可以到达所有节点。\n所以 answer[2] = 1 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code>edges[i].length == 2</code></li>\n\t<li><code>0 &lt;= u<sub>i</sub> == edges[i][0] &lt; n</code></li>\n\t<li><code>0 &lt;= v<sub>i</sub> == edges[i][1] &lt; n</code></li>\n\t<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>\n\t<li>输入保证如果边是双向边，可以得到一棵树。</li>\n</ul>\n",
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                "The problem can be solved using tree DP.",
                "Using node <code>0</code> as the root, let <code>dp[x]</code> be the minimum number of edge reversals so node <code>x</code> can reach every node in its subtree.",
                "Using a DFS traversing the edges bidirectionally, we can compute <code>dp</code>.<br />\r\n<code>dp[x] = dp[y] +</code> (<code>1</code> if the edge between <code>x</code> and <code>y</code> is going from <code>y</code> to <code>x</code>; <code>0</code> otherwise), where <code>x</code> is the parent of <code>y</code>.",
                "Let <code>answer[x]</code> be the minimum number of edge reversals so it is possible to reach any other node starting from node <code>x</code>.",
                "Using another DFS starting from node <code>0</code> and traversing the edges bidirectionally, we can compute <code>answer</code>.<br />\r\n<code>answer[0] = dp[0]</code><br />\r\n<code>answer[y] = answer[x] +</code> (<code>1</code> if the edge between <code>x</code> and <code>y</code> is going from <code>x</code> to <code>y</code>; <code>-1</code> otherwise), where <code>x</code> is the parent of <code>y</code>."
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