leetcode-problemset/leetcode-cn/originData/reachable-nodes-in-subdivided-graph.json

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        "question": {
            "questionId": "918",
            "questionFrontendId": "882",
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            "title": "Reachable Nodes In Subdivided Graph",
            "titleSlug": "reachable-nodes-in-subdivided-graph",
            "content": "<p>You are given an undirected graph (the <strong>&quot;original graph&quot;</strong>) with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You decide to <strong>subdivide</strong> each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge.</p>\n\n<p>The graph is given as a 2D array of <code>edges</code> where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, cnt<sub>i</sub>]</code> indicates that there is an edge between nodes <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> in the original graph, and <code>cnt<sub>i</sub></code> is the total number of new nodes that you will <strong>subdivide</strong> the edge into. Note that <code>cnt<sub>i</sub> == 0</code> means you will not subdivide the edge.</p>\n\n<p>To <strong>subdivide</strong> the edge <code>[u<sub>i</sub>, v<sub>i</sub>]</code>, replace it with <code>(cnt<sub>i</sub> + 1)</code> new edges and <code>cnt<sub>i</sub></code> new nodes. The new nodes are <code>x<sub>1</sub></code>, <code>x<sub>2</sub></code>, ..., <code>x<sub>cnt<sub>i</sub></sub></code>, and the new edges are <code>[u<sub>i</sub>, x<sub>1</sub>]</code>, <code>[x<sub>1</sub>, x<sub>2</sub>]</code>, <code>[x<sub>2</sub>, x<sub>3</sub>]</code>, ..., <code>[x<sub>cnt<sub>i</sub>-1</sub>, x<sub>cnt<sub>i</sub></sub>]</code>, <code>[x<sub>cnt<sub>i</sub></sub>, v<sub>i</sub>]</code>.</p>\n\n<p>In this <strong>new graph</strong>, you want to know how many nodes are <strong>reachable</strong> from the node <code>0</code>, where a node is <strong>reachable</strong> if the distance is <code>maxMoves</code> or less.</p>\n\n<p>Given the original graph and <code>maxMoves</code>, return <em>the number of nodes that are <strong>reachable</strong> from node </em><code>0</code><em> in the new graph</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://s3-lc-upload.s3.amazonaws.com/uploads/2018/08/01/origfinal.png\" style=\"width: 600px; height: 247px;\" />\n<pre>\n<strong>Input:</strong> edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3\n<strong>Output:</strong> 13\n<strong>Explanation:</strong> The edge subdivisions are shown in the image above.\nThe nodes that are reachable are highlighted in yellow.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], maxMoves = 10, n = 4\n<strong>Output:</strong> 23\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> edges = [[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]], maxMoves = 17, n = 5\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> Node 0 is disconnected from the rest of the graph, so only node 0 is reachable.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= edges.length &lt;= min(n * (n - 1) / 2, 10<sup>4</sup>)</code></li>\n\t<li><code>edges[i].length == 3</code></li>\n\t<li><code>0 &lt;= u<sub>i</sub> &lt; v<sub>i</sub> &lt; n</code></li>\n\t<li>There are <strong>no multiple edges</strong> in the graph.</li>\n\t<li><code>0 &lt;= cnt<sub>i</sub> &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= maxMoves &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= n &lt;= 3000</code></li>\n</ul>\n",
            "translatedTitle": "细分图中的可到达结点",
            "translatedContent": "<p>给你一个无向图（<strong>原始图</strong>），图中有 <code>n</code> 个节点，编号从 <code>0</code> 到 <code>n - 1</code> 。你决定将图中的每条边 <strong>细分</strong> 为一条节点链，每条边之间的新节点数各不相同。</p>\n\n<p>图用由边组成的二维数组 <code>edges</code> 表示，其中&nbsp;<code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, cnt<sub>i</sub>]</code> 表示原始图中节点&nbsp;<code>u<sub>i</sub></code> 和&nbsp;<code>v<sub>i</sub></code> 之间存在一条边，<code>cnt<sub>i</sub></code> 是将边 <strong>细分</strong> 后的新节点总数。注意，<code>cnt<sub>i</sub> == 0</code> 表示边不可细分。</p>\n\n<p>要 <strong>细分</strong> 边 <code>[ui, vi]</code> ，需要将其替换为 <code>(cnt<sub>i</sub> + 1)</code> 条新边，和&nbsp;<code>cnt<sub>i</sub></code> 个新节点。新节点为 <code>x<sub>1</sub></code>, <code>x<sub>2</sub></code>, ..., <code>x<sub>cnt<sub>i</sub></sub></code> ，新边为 <code>[u<sub>i</sub>, x<sub>1</sub>]</code>, <code>[x<sub>1</sub>, x<sub>2</sub>]</code>, <code>[x<sub>2</sub>, x<sub>3</sub>]</code>, ..., <code>[x<sub>cnt<sub>i</sub>+1</sub>, x<sub>cnt<sub>i</sub></sub>]</code>, <code>[x<sub>cnt<sub>i</sub></sub>, v<sub>i</sub>]</code> 。</p>\n\n<p>现在得到一个&nbsp;<strong>新的细分图</strong> ，请你计算从节点 <code>0</code> 出发，可以到达多少个节点？如果节点间距离是 <code>maxMoves</code> 或更少，则视为 <strong>可以到达</strong> 。</p>\n\n<p>给你原始图和 <code>maxMoves</code> ，返回 <em>新的细分图中从节点 <code>0</code> 出发</em><strong><em> 可到达的节点数</em></strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://s3-lc-upload.s3.amazonaws.com/uploads/2018/08/01/origfinal.png\" style=\"height: 247px; width: 600px;\" />\n<pre>\n<strong>输入：</strong>edges = [[0,1,10],[0,2,1],[1,2,2]], maxMoves = 6, n = 3\n<strong>输出：</strong>13\n<strong>解释：</strong>边的细分情况如上图所示。\n可以到达的节点已经用黄色标注出来。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>edges = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], maxMoves = 10, n = 4\n<strong>输出：</strong>23\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>edges = [[1,2,4],[1,4,5],[1,3,1],[2,3,4],[3,4,5]], maxMoves = 17, n = 5\n<strong>输出：</strong>1\n<strong>解释：</strong>节点 0 与图的其余部分没有连通，所以只有节点 0 可以到达。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= edges.length &lt;= min(n * (n - 1) / 2, 10<sup>4</sup>)</code></li>\n\t<li><code>edges[i].length == 3</code></li>\n\t<li><code>0 &lt;= u<sub>i</sub> &lt; v<sub>i</sub> &lt; n</code></li>\n\t<li>图中 <strong>不存在平行边</strong></li>\n\t<li><code>0 &lt;= cnt<sub>i</sub> &lt;= 10<sup>4</sup></code></li>\n\t<li><code>0 &lt;= maxMoves &lt;= 10<sup>9</sup></code></li>\n\t<li><code>1 &lt;= n &lt;= 3000</code></li>\n</ul>\n",
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href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\\" target=\\\"_blank\\\"> datastructures-js\\/queue<\\/a>\\u3002<\\/p>\"],\"ruby\":[\"Ruby\",\"<p>\\u4f7f\\u7528<code>Ruby 3.1<\\/code>\\u6267\\u884c<\\/p>\\r\\n\\r\\n<p>\\u4e00\\u4e9b\\u5e38\\u7528\\u7684\\u6570\\u636e\\u7ed3\\u6784\\u5df2\\u5728 Algorithms \\u6a21\\u5757\\u4e2d\\u63d0\\u4f9b\\uff1ahttps:\\/\\/www.rubydoc.info\\/github\\/kanwei\\/algorithms\\/Algorithms<\\/p>\"],\"swift\":[\"Swift\",\"<p>\\u7248\\u672c\\uff1a<code>Swift 5.5.2<\\/code><\\/p>\\r\\n\\r\\n<p>\\u6211\\u4eec\\u901a\\u5e38\\u4fdd\\u8bc1\\u66f4\\u65b0\\u5230 <a href=\\\"https:\\/\\/swift.org\\/download\\/\\\" target=\\\"_blank\\\">Apple\\u653e\\u51fa\\u7684\\u6700\\u65b0\\u7248Swift<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u53d1\\u73b0Swift\\u4e0d\\u662f\\u6700\\u65b0\\u7248\\u7684\\uff0c\\u8bf7\\u8054\\u7cfb\\u6211\\u4eec\\uff01\\u6211\\u4eec\\u5c06\\u5c3d\\u5feb\\u66f4\\u65b0\\u3002<\\/p>\"],\"golang\":[\"Go\",\"<p>\\u7248\\u672c\\uff1a<code>Go 1.17<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 <a href=\\\"https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods\\\" target=\\\"_blank\\\">https:\\/\\/godoc.org\\/github.com\\/emirpasic\\/gods<\\/a> \\u7b2c\\u4e09\\u65b9\\u5e93\\u3002<\\/p>\"],\"python3\":[\"Python3\",\"<p>\\u7248\\u672c\\uff1a<code>Python 3.10<\\/code><\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982<a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/array.html\\\" target=\\\"_blank\\\">array<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/bisect.html\\\" target=\\\"_blank\\\">bisect<\\/a>, <a href=\\\"https:\\/\\/docs.python.org\\/3\\/library\\/collections.html\\\" target=\\\"_blank\\\">collections<\\/a>\\u3002 \\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u9700\\u4f7f\\u7528 Map\\/TreeMap \\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"http:\\/\\/www.grantjenks.com\\/docs\\/sortedcontainers\\/\\\" target=\\\"_blank\\\">sortedcontainers<\\/a> \\u5e93\\u3002<\\/p>\"],\"scala\":[\"Scala\",\"<p>\\u7248\\u672c\\uff1a<code>Scala 2.13<\\/code><\\/p>\"],\"kotlin\":[\"Kotlin\",\"<p>\\u7248\\u672c\\uff1a<code>Kotlin 1.3.10<\\/code><\\/p>\"],\"rust\":[\"Rust\",\"<p>\\u7248\\u672c\\uff1a<code>rust 1.58.1<\\/code><\\/p>\\r\\n\\r\\n<p>\\u652f\\u6301 crates.io \\u7684 <a href=\\\"https:\\/\\/crates.io\\/crates\\/rand\\\" target=\\\"_blank\\\">rand<\\/a><\\/p>\"],\"php\":[\"PHP\",\"<p><code>PHP 8.1<\\/code>.<\\/p>\\r\\n\\r\\n<p>With bcmath module.<\\/p>\"],\"typescript\":[\"TypeScript\",\"<p>TypeScript 4.5.4<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2020<\\/p>\"],\"racket\":[\"Racket\",\"<p><a href=\\\"https:\\/\\/docs.racket-lang.org\\/guide\\/performance.html#%28tech._c%29\\\" target=\\\"_blank\\\">Racket CS<\\/a> v8.3<\\/p>\\r\\n\\r\\n<p>\\u4f7f\\u7528 #lang racket<\\/p>\\r\\n\\r\\n<p>\\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\"]}",
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