leetcode-problemset/leetcode-cn/originData/find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance.json

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            "questionId": "1456",
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            "categoryTitle": "Algorithms",
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            "title": "Find the City With the Smallest Number of Neighbors at a Threshold Distance",
            "titleSlug": "find-the-city-with-the-smallest-number-of-neighbors-at-a-threshold-distance",
            "content": "<p>There are <code>n</code> cities numbered from <code>0</code> to <code>n-1</code>. Given the array <code>edges</code> where <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>, weight<sub>i</sub>]</code> represents a bidirectional and weighted edge between cities <code>from<sub>i</sub></code> and <code>to<sub>i</sub></code>, and given the integer <code>distanceThreshold</code>.</p>\n\n<p>Return the city with the smallest number of cities that are reachable through some path and whose distance is <strong>at most</strong> <code>distanceThreshold</code>, If there are multiple such cities, return the city with the greatest number.</p>\n\n<p>Notice that the distance of a path connecting cities <em><strong>i</strong></em> and <em><strong>j</strong></em> is equal to the sum of the edges&#39; weights along that path.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/16/find_the_city_01.png\" style=\"width: 300px; height: 225px;\" />\n<pre>\n<strong>Input:</strong> n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4\n<strong>Output:</strong> 3\n<strong>Explanation: </strong>The figure above describes the graph.&nbsp;\nThe neighboring cities at a distanceThreshold = 4 for each city are:\nCity 0 -&gt; [City 1, City 2]&nbsp;\nCity 1 -&gt; [City 0, City 2, City 3]&nbsp;\nCity 2 -&gt; [City 0, City 1, City 3]&nbsp;\nCity 3 -&gt; [City 1, City 2]&nbsp;\nCities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2020/01/16/find_the_city_02.png\" style=\"width: 300px; height: 225px;\" />\n<pre>\n<strong>Input:</strong> n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2\n<strong>Output:</strong> 0\n<strong>Explanation: </strong>The figure above describes the graph.&nbsp;\nThe neighboring cities at a distanceThreshold = 2 for each city are:\nCity 0 -&gt; [City 1]&nbsp;\nCity 1 -&gt; [City 0, City 4]&nbsp;\nCity 2 -&gt; [City 3, City 4]&nbsp;\nCity 3 -&gt; [City 2, City 4]\nCity 4 -&gt; [City 1, City 2, City 3]&nbsp;\nThe city 0 has 1 neighboring city at a distanceThreshold = 2.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 100</code></li>\n\t<li><code>1 &lt;= edges.length &lt;= n * (n - 1) / 2</code></li>\n\t<li><code>edges[i].length == 3</code></li>\n\t<li><code>0 &lt;= from<sub>i</sub> &lt; to<sub>i</sub> &lt; n</code></li>\n\t<li><code>1 &lt;= weight<sub>i</sub>,&nbsp;distanceThreshold &lt;= 10^4</code></li>\n\t<li>All pairs <code>(from<sub>i</sub>, to<sub>i</sub>)</code> are distinct.</li>\n</ul>\n",
            "translatedTitle": "阈值距离内邻居最少的城市",
            "translatedContent": "<p>有 <code>n</code> 个城市，按从 <code>0</code> 到 <code>n-1</code> 编号。给你一个边数组 <code>edges</code>，其中 <code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>, weight<sub>i</sub>]</code> 代表 <code>from<sub>i</sub></code> 和 <code>to<sub>i</sub></code><sub> </sub>两个城市之间的双向加权边，距离阈值是一个整数 <code>distanceThreshold</code>。</p>\n\n<p>返回能通过某些路径到达其他城市数目最少、且路径距离 <strong>最大</strong> 为 <code>distanceThreshold</code> 的城市。如果有多个这样的城市，则返回编号最大的城市。</p>\n\n<p>注意，连接城市 <em><strong>i</strong></em> 和 <em><strong>j</strong></em> 的路径的距离等于沿该路径的所有边的权重之和。</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/2020/01/26/find_the_city_01.png\" style=\"height: 225px; width: 300px;\" /></p>\n\n<pre>\n<strong>输入：</strong>n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4\n<strong>输出：</strong>3\n<strong>解释：</strong>城市分布图如上。\n每个城市阈值距离 distanceThreshold = 4 内的邻居城市分别是：\n城市 0 -> [城市 1, 城市 2] \n城市 1 -> [城市 0, 城市 2, 城市 3] \n城市 2 -> [城市 0, 城市 1, 城市 3] \n城市 3 -> [城市 1, 城市 2] \n城市 0 和 3 在阈值距离 4 以内都有 2 个邻居城市，但是我们必须返回城市 3，因为它的编号最大。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/01/26/find_the_city_02.png\" style=\"height: 225px; width: 300px;\" /></strong></p>\n\n<pre>\n<strong>输入：</strong>n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2\n<strong>输出：</strong>0\n<strong>解释：</strong>城市分布图如上。 \n每个城市阈值距离 distanceThreshold = 2 内的邻居城市分别是：\n城市 0 -> [城市 1] \n城市 1 -> [城市 0, 城市 4] \n城市 2 -> [城市 3, 城市 4] \n城市 3 -> [城市 2, 城市 4]\n城市 4 -> [城市 1, 城市 2, 城市 3] \n城市 0 在阈值距离 2 以内只有 1 个邻居城市。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 <= n <= 100</code></li>\n\t<li><code>1 <= edges.length <= n * (n - 1) / 2</code></li>\n\t<li><code>edges[i].length == 3</code></li>\n\t<li><code>0 <= from<sub>i</sub> < to<sub>i</sub> < n</code></li>\n\t<li><code>1 <= weight<sub>i</sub>, distanceThreshold <= 10^4</code></li>\n\t<li>所有 <code>(from<sub>i</sub>, to<sub>i</sub>)</code> 都是不同的。</li>\n</ul>\n",
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                "Use Floyd-Warshall's algorithm to compute any-point to any-point distances. (Or can also do Dijkstra from every node due to the weights are non-negative).",
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