{ "data": { "question": { "questionId": "3105", "questionFrontendId": "2858", "categoryTitle": "Algorithms", "boundTopicId": 2442880, "title": "Minimum Edge Reversals So Every Node Is Reachable", "titleSlug": "minimum-edge-reversals-so-every-node-is-reachable", "content": "

There is a simple directed graph with n nodes labeled from 0 to n - 1. The graph would form a tree if its edges were bi-directional.

\n\n

You are given an integer n and a 2D integer array edges, where edges[i] = [ui, vi] represents a directed edge going from node ui to node vi.

\n\n

An edge reversal changes the direction of an edge, i.e., a directed edge going from node ui to node vi becomes a directed edge going from node vi to node ui.

\n\n

For every node i in the range [0, n - 1], your task is to independently calculate the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges.

\n\n

Return an integer array answer, where answer[i] is the minimum number of edge reversals required so it is possible to reach any other node starting from node i through a sequence of directed edges.

\n\n

 

\n

Example 1:

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\n\n
\nInput: n = 4, edges = [[2,0],[2,1],[1,3]]\nOutput: [1,1,0,2]\nExplanation: 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
\n\n

Example 2:

\n\n

\n\n
\nInput: n = 3, edges = [[1,2],[2,0]]\nOutput: [2,0,1]\nExplanation: 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
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "可以到达每一个节点的最少边反转次数", "translatedContent": "

给你一个 n 个点的 简单有向图 （没有重复边的有向图），节点编号为 0 到 n - 1 。如果这些边是双向边，那么这个图形成一棵 树 。

\n\n

给你一个整数 n 和一个 二维 整数数组 edges ，其中 edges[i] = [ui, vi] 表示从节点 ui 到节点 vi 有一条 有向边 。

\n\n

边反转 指的是将一条边的方向反转，也就是说一条从节点 ui 到节点 vi 的边会变为一条从节点 vi 到节点 ui 的边。

\n\n

对于范围 [0, n - 1] 中的每一个节点 i ，你的任务是分别 独立 计算 最少 需要多少次 边反转 ，从节点 i 出发经过 一系列有向边 ，可以到达所有的节点。

\n\n

请你返回一个长度为 n 的整数数组 answer ，其中 answer[i]表示从节点 i 出发，可以到达所有节点的 最少边反转 次数。

\n\n

 

\n\n

示例 1：

\n\n

\n\n
\n输入：n = 4, edges = [[2,0],[2,1],[1,3]]\n输出：[1,1,0,2]\n解释：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
\n\n

示例 2：

\n\n

\n\n
\n输入：n = 3, edges = [[1,2],[2,0]]\n输出：[2,0,1]\n解释：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
\n\n

 

\n\n

提示：

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(listof (listof exact-integer?)) (listof exact-integer?))\n\n )", "__typename": "CodeSnippetNode" }, { "lang": "Erlang", "langSlug": "erlang", "code": "-spec min_edge_reversals(N :: integer(), Edges :: [[integer()]]) -> [integer()].\nmin_edge_reversals(N, Edges) ->\n .", "__typename": "CodeSnippetNode" }, { "lang": "Elixir", "langSlug": "elixir", "code": "defmodule Solution do\n @spec min_edge_reversals(n :: integer, edges :: [[integer]]) :: [integer]\n def min_edge_reversals(n, edges) do\n\n end\nend", "__typename": "CodeSnippetNode" } ], "stats": "{\"totalAccepted\": \"1.3K\", \"totalSubmission\": \"1.9K\", \"totalAcceptedRaw\": 1349, \"totalSubmissionRaw\": 1904, \"acRate\": \"70.9%\"}", "hints": [ "The problem can be solved using tree DP.", "Using node 0 as the root, let dp[x] be the minimum number of edge reversals so node x can reach every node in its subtree.", "Using a DFS traversing the edges bidirectionally, we can compute dp.
\r\ndp[x] = dp[y] + (1 if the edge between x and y is going from y to x; 0 otherwise), where x is the parent of y.", "Let answer[x] be the minimum number of edge reversals so it is possible to reach any other node starting from node x.", "Using another DFS starting from node 0 and traversing the edges bidirectionally, we can compute answer.
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\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 24.2\"],\"elixir\":[\"Elixir\",\"Elixir 1.13.0 with Erlang\\/OTP 24.2\"],\"dart\":[\"Dart\",\"
Dart 2.17.3<\\/p>\\r\\n\\r\\n
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