mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
202 lines
24 KiB
JSON
202 lines
24 KiB
JSON
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"categoryTitle": "Algorithms",
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"title": "Count Ways to Build Rooms in an Ant Colony",
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"titleSlug": "count-ways-to-build-rooms-in-an-ant-colony",
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"content": "<p>You are an ant tasked with adding <code>n</code> new rooms numbered <code>0</code> to <code>n-1</code> to your colony. You are given the expansion plan as a <strong>0-indexed</strong> integer array of length <code>n</code>, <code>prevRoom</code>, where <code>prevRoom[i]</code> indicates that you must build room <code>prevRoom[i]</code> before building room <code>i</code>, and these two rooms must be connected <strong>directly</strong>. Room <code>0</code> is already built, so <code>prevRoom[0] = -1</code>. The expansion plan is given such that once all the rooms are built, every room will be reachable from room <code>0</code>.</p>\r\n\r\n<p>You can only build <strong>one room</strong> at a time, and you can travel freely between rooms you have <strong>already built</strong> only if they are <strong>connected</strong>. You can choose to build <strong>any room</strong> as long as its <strong>previous room</strong> is already built.</p>\r\n\r\n<p>Return <em>the <strong>number of different orders</strong> you can build all the rooms in</em>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\r\n\r\n<p> </p>\r\n<p><strong class=\"example\">Example 1:</strong></p>\r\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/19/d1.JPG\" style=\"width: 200px; height: 212px;\" />\r\n<pre>\r\n<strong>Input:</strong> prevRoom = [-1,0,1]\r\n<strong>Output:</strong> 1\r\n<strong>Explanation:</strong> There is only one way to build the additional rooms: 0 → 1 → 2\r\n</pre>\r\n\r\n<p><strong class=\"example\">Example 2:</strong></p>\r\n<strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/19/d2.JPG\" style=\"width: 200px; height: 239px;\" /></strong>\r\n\r\n<pre>\r\n<strong>Input:</strong> prevRoom = [-1,0,0,1,2]\r\n<strong>Output:</strong> 6\r\n<strong>Explanation:\r\n</strong>The 6 ways are:\r\n0 → 1 → 3 → 2 → 4\r\n0 → 2 → 4 → 1 → 3\r\n0 → 1 → 2 → 3 → 4\r\n0 → 1 → 2 → 4 → 3\r\n0 → 2 → 1 → 3 → 4\r\n0 → 2 → 1 → 4 → 3\r\n</pre>\r\n\r\n<p> </p>\r\n<p><strong>Constraints:</strong></p>\r\n\r\n<ul>\r\n\t<li><code>n == prevRoom.length</code></li>\r\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\r\n\t<li><code>prevRoom[0] == -1</code></li>\r\n\t<li><code>0 <= prevRoom[i] < n</code> for all <code>1 <= i < n</code></li>\r\n\t<li>Every room is reachable from room <code>0</code> once all the rooms are built.</li>\r\n</ul>",
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"translatedTitle": "统计为蚁群构筑房间的不同顺序",
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"translatedContent": "<p>你是一只蚂蚁,负责为蚁群构筑 <code>n</code> 间编号从 <code>0</code> 到 <code>n-1</code> 的新房间。给你一个 <strong>下标从 0 开始</strong> 且长度为 <code>n</code> 的整数数组 <code>prevRoom</code> 作为扩建计划。其中,<code>prevRoom[i]</code> 表示在构筑房间 <code>i</code> 之前,你必须先构筑房间 <code>prevRoom[i]</code> ,并且这两个房间必须 <strong>直接</strong> 相连。房间 <code>0</code> 已经构筑完成,所以 <code>prevRoom[0] = -1</code> 。扩建计划中还有一条硬性要求,在完成所有房间的构筑之后,从房间 <code>0</code> 可以访问到每个房间。</p>\n\n<p>你一次只能构筑 <strong>一个</strong> 房间。你可以在 <strong>已经构筑好的</strong> 房间之间自由穿行,只要这些房间是 <strong>相连的</strong> 。如果房间 <code>prevRoom[i]</code> 已经构筑完成,那么你就可以构筑房间 <code>i</code>。</p>\n\n<p>返回你构筑所有房间的 <strong>不同顺序的数目</strong> 。由于答案可能很大,请返回对 <code>10<sup>9</sup> + 7</code> <strong>取余</strong> 的结果。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/19/d1.JPG\" style=\"width: 200px; height: 212px;\" />\n<pre>\n<strong>输入:</strong><code>prevRoom</code> = [-1,0,1]\n<strong>输出:</strong>1\n<strong>解释:</strong>仅有一种方案可以完成所有房间的构筑:0 → 1 → 2\n</pre>\n\n<p><strong>示例 2:</strong></p>\n<strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/19/d2.JPG\" style=\"width: 200px; height: 239px;\" /></strong>\n\n<pre>\n<strong>输入:</strong><code>prevRoom</code> = [-1,0,0,1,2]\n<strong>输出:</strong>6\n<strong>解释:\n</strong>有 6 种不同顺序:\n0 → 1 → 3 → 2 → 4\n0 → 2 → 4 → 1 → 3\n0 → 1 → 2 → 3 → 4\n0 → 1 → 2 → 4 → 3\n0 → 2 → 1 → 3 → 4\n0 → 2 → 1 → 4 → 3\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>n == prevRoom.length</code></li>\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>prevRoom[0] == -1</code></li>\n\t<li>对于所有的 <code>1 <= i < n</code> ,都有 <code>0 <= prevRoom[i] < n</code></li>\n\t<li>题目保证所有房间都构筑完成后,从房间 <code>0</code> 可以访问到每个房间</li>\n</ul>\n",
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"Use dynamic programming.",
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"Let dp[i] be the number of ways to solve the problem for the subtree of node i.",
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"Imagine you are trying to fill an array with the order of traversal, dp[i] equals the multiplications of the number of ways to distribute the subtrees of the children of i on the array using combinatorics, multiplied bu their dp values."
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