update

leetcode-cn/originData/network-delay-time.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>有 <code>n</code> 个网络节点，标记为&nbsp;<code>1</code>&nbsp;到 <code>n</code>。</p>\n\n<p>给你一个列表&nbsp;<code>times</code>，表示信号经过 <strong>有向</strong> 边的传递时间。&nbsp;<code>times[i] = (u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>)</code>，其中&nbsp;<code>u<sub>i</sub></code>&nbsp;是源节点，<code>v<sub>i</sub></code>&nbsp;是目标节点， <code>w<sub>i</sub></code>&nbsp;是一个信号从源节点传递到目标节点的时间。</p>\n\n<p>现在，从某个节点&nbsp;<code>K</code>&nbsp;发出一个信号。需要多久才能使所有节点都收到信号？如果不能使所有节点收到信号，返回&nbsp;<code>-1</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2019/05/23/931_example_1.png\" style=\"height: 220px; width: 200px;\" /></p>\n\n<pre>\n<strong>输入：</strong>times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2\n<strong>输出：</strong>2\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>times = [[1,2,1]], n = 2, k = 1\n<strong>输出：</strong>1\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>times = [[1,2,1]], n = 2, k = 2\n<strong>输出：</strong>-1\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= k &lt;= n &lt;= 100</code></li>\n\t<li><code>1 &lt;= times.length &lt;= 6000</code></li>\n\t<li><code>times[i].length == 3</code></li>\n\t<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>\n\t<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>\n\t<li><code>0 &lt;= w<sub>i</sub> &lt;= 100</code></li>\n\t<li>所有 <code>(u<sub>i</sub>, v<sub>i</sub>)</code> 对都 <strong>互不相同</strong>（即，不含重复边）</li>\n</ul>\n",
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             "difficulty": "Medium",
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             "hints": [
                 "We visit each node at some time, and if that time is better than the fastest time we've reached this node, we travel along outgoing edges in sorted order.  Alternatively, we could use Dijkstra's algorithm."
             ],