Fetch the repository succeeded.
This action will force synchronization from 小墨/力扣题库(完整版), which will overwrite any changes that you have made since you forked the repository, and can not be recovered!!!
Synchronous operation will process in the background and will refresh the page when finishing processing. Please be patient.
<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>
<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>
<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' weights along that path.</p>
<p> </p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/16/find_the_city_01.png" style="width: 300px; height: 225px;" />
<pre>
<strong>Input:</strong> n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
<strong>Output:</strong> 3
<strong>Explanation: </strong>The figure above describes the graph.
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2]
City 1 -> [City 0, City 2, City 3]
City 2 -> [City 0, City 1, City 3]
City 3 -> [City 1, City 2]
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.
</pre>
<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/01/16/find_the_city_02.png" style="width: 300px; height: 225px;" />
<pre>
<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
<strong>Output:</strong> 0
<strong>Explanation: </strong>The figure above describes the graph.
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1]
City 1 -> [City 0, City 4]
City 2 -> [City 3, City 4]
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3]
The city 0 has 1 neighboring city at a distanceThreshold = 2.
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>2 <= n <= 100</code></li>
<li><code>1 <= edges.length <= n * (n - 1) / 2</code></li>
<li><code>edges[i].length == 3</code></li>
<li><code>0 <= from<sub>i</sub> < to<sub>i</sub> < n</code></li>
<li><code>1 <= weight<sub>i</sub>, distanceThreshold <= 10^4</code></li>
<li>All pairs <code>(from<sub>i</sub>, to<sub>i</sub>)</code> are distinct.</li>
</ul>
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。