mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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178 lines
25 KiB
JSON
178 lines
25 KiB
JSON
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"title": "Graph Connectivity With Threshold",
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"content": "<p>We have <code>n</code> cities labeled from <code>1</code> to <code>n</code>. Two different cities with labels <code>x</code> and <code>y</code> are directly connected by a bidirectional road if and only if <code>x</code> and <code>y</code> share a common divisor <strong>strictly greater</strong> than some <code>threshold</code>. More formally, cities with labels <code>x</code> and <code>y</code> have a road between them if there exists an integer <code>z</code> such that all of the following are true:</p>\n\n<ul>\n\t<li><code>x % z == 0</code>,</li>\n\t<li><code>y % z == 0</code>, and</li>\n\t<li><code>z > threshold</code>.</li>\n</ul>\n\n<p>Given the two integers, <code>n</code> and <code>threshold</code>, and an array of <code>queries</code>, you must determine for each <code>queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> if cities <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> are connected directly or indirectly. (i.e. there is some path between them).</p>\n\n<p>Return <em>an array </em><code>answer</code><em>, where </em><code>answer.length == queries.length</code><em> and </em><code>answer[i]</code><em> is </em><code>true</code><em> if for the </em><code>i<sup>th</sup></code><em> query, there is a path between </em><code>a<sub>i</sub></code><em> and </em><code>b<sub>i</sub></code><em>, or </em><code>answer[i]</code><em> is </em><code>false</code><em> if there is no path.</em></p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/09/ex1.jpg\" style=\"width: 382px; height: 181px;\" />\n<pre>\n<strong>Input:</strong> n = 6, threshold = 2, queries = [[1,4],[2,5],[3,6]]\n<strong>Output:</strong> [false,false,true]\n<strong>Explanation:</strong> The divisors for each number:\n1: 1\n2: 1, 2\n3: 1, <u>3</u>\n4: 1, 2, <u>4</u>\n5: 1, <u>5</u>\n6: 1, 2, <u>3</u>, <u>6</u>\nUsing the underlined divisors above the threshold, only cities 3 and 6 share a common divisor, so they are the\nonly ones directly connected. The result of each query:\n[1,4] 1 is not connected to 4\n[2,5] 2 is not connected to 5\n[3,6] 3 is connected to 6 through path 3--6\n</pre>\n\n<p><strong>Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/10/tmp.jpg\" style=\"width: 532px; height: 302px;\" />\n<pre>\n<strong>Input:</strong> n = 6, threshold = 0, queries = [[4,5],[3,4],[3,2],[2,6],[1,3]]\n<strong>Output:</strong> [true,true,true,true,true]\n<strong>Explanation:</strong> The divisors for each number are the same as the previous example. However, since the threshold is 0,\nall divisors can be used. Since all numbers share 1 as a divisor, all cities are connected.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/10/17/ex3.jpg\" style=\"width: 282px; height: 282px;\" />\n<pre>\n<strong>Input:</strong> n = 5, threshold = 1, queries = [[4,5],[4,5],[3,2],[2,3],[3,4]]\n<strong>Output:</strong> [false,false,false,false,false]\n<strong>Explanation:</strong> Only cities 2 and 4 share a common divisor 2 which is strictly greater than the threshold 1, so they are the only ones directly connected.\nPlease notice that there can be multiple queries for the same pair of nodes [x, y], and that the query [x, y] is equivalent to the query [y, x].\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= n <= 10<sup>4</sup></code></li>\n\t<li><code>0 <= threshold <= n</code></li>\n\t<li><code>1 <= queries.length <= 10<sup>5</sup></code></li>\n\t<li><code>queries[i].length == 2</code></li>\n\t<li><code>1 <= a<sub>i</sub>, b<sub>i</sub> <= cities</code></li>\n\t<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>\n</ul>\n",
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"translatedTitle": "带阈值的图连通性",
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"translatedContent": "<p>有 <code>n</code> 座城市，编号从 <code>1</code> 到 <code>n</code> 。编号为 <code>x</code> 和 <code>y</code> 的两座城市直接连通的前提是： <code>x</code> 和 <code>y</code> 的公因数中，至少有一个 <strong>严格大于</strong> 某个阈值 <code>threshold</code> 。更正式地说，如果存在整数 <code>z</code> ，且满足以下所有条件，则编号 <code>x</code> 和 <code>y</code> 的城市之间有一条道路：</p>\n\n<ul>\n\t<li><code>x % z == 0</code></li>\n\t<li><code>y % z == 0</code></li>\n\t<li><code>z > threshold</code></li>\n</ul>\n\n<p>给你两个整数 <code>n</code> 和 <code>threshold</code> ，以及一个待查询数组，请你判断每个查询<code> queries[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> 指向的城市 <code>a<sub>i</sub></code> 和 <code>b<sub>i</sub></code> 是否连通（即，它们之间是否存在一条路径）。</p>\n\n<p>返回数组 <code>answer</code> ，其中<code>answer.length == queries.length</code> 。如果第 <code>i</code> 个查询中指向的城市 <code>a<sub>i</sub></code> 和 <code>b<sub>i</sub></code> 连通，则 <code>answer[i]</code> 为 <code>true</code> ；如果不连通，则 <code>answer[i]</code> 为 <code>false</code> 。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/10/18/ex1.jpg\" style=\"width: 382px; height: 181px;\" /></p>\n\n<p> </p>\n\n<pre>\n<strong>输入：</strong>n = 6, threshold = 2, queries = [[1,4],[2,5],[3,6]]\n<strong>输出：</strong>[false,false,true]\n<strong>解释：</strong>每个数的因数如下：\n1: 1\n2: 1, 2\n3: 1, <strong>3</strong>\n4: 1, 2, <strong>4</strong>\n5: 1, <strong>5</strong>\n6: 1, 2, <strong>3</strong>, <strong>6</strong>\n所有大于阈值的的因数已经加粗标识，只有城市 3 和 6 共享公约数 3 ，因此结果是： \n[1,4] 1 与 4 不连通\n[2,5] 2 与 5 不连通\n[3,6] 3 与 6 连通，存在路径 3--6\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/10/18/tmp.jpg\" style=\"width: 532px; height: 302px;\" /></p>\n\n<p> </p>\n\n<pre>\n<strong>输入：</strong>n = 6, threshold = 0, queries = [[4,5],[3,4],[3,2],[2,6],[1,3]]\n<strong>输出：</strong>[true,true,true,true,true]\n<strong>解释：</strong>每个数的因数与上一个例子相同。但是，由于阈值为 0 ，所有的因数都大于阈值。因为所有的数字共享公因数 1 ，所以所有的城市都互相连通。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/10/16/ex3.jpg\" style=\"width: 282px; height: 282px;\" /></p>\n\n<p> </p>\n\n<pre>\n<strong>输入：</strong>n = 5, threshold = 1, queries = [[4,5],[4,5],[3,2],[2,3],[3,4]]\n<strong>输出：</strong>[false,false,false,false,false]\n<strong>解释：</strong>只有城市 2 和 4 共享的公约数 2 严格大于阈值 1 ，所以只有这两座城市是连通的。\n注意，同一对节点 [x, y] 可以有多个查询，并且查询 [x，y] 等同于查询 [y，x] 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 <= n <= 10<sup>4</sup></code></li>\n\t<li><code>0 <= threshold <= n</code></li>\n\t<li><code>1 <= queries.length <= 10<sup>5</sup></code></li>\n\t<li><code>queries[i].length == 2</code></li>\n\t<li><code>1 <= a<sub>i</sub>, b<sub>i</sub> <= cities</code></li>\n\t<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>\n</ul>\n",
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