update

leetcode-cn/originData/minimum-weighted-subgraph-with-the-required-paths.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个整数&nbsp;<code>n</code>&nbsp;，它表示一个 <strong>带权有向</strong> 图的节点数，节点编号为&nbsp;<code>0</code> 到&nbsp;<code>n - 1</code>&nbsp;。</p>\n\n<p>同时给你一个二维整数数组&nbsp;<code>edges</code>&nbsp;，其中&nbsp;<code>edges[i] = [from<sub>i</sub>, to<sub>i</sub>, weight<sub>i</sub>]</code>&nbsp;，表示从&nbsp;<code>from<sub>i</sub></code>&nbsp;到&nbsp;<code>to<sub>i</sub></code>&nbsp;有一条边权为&nbsp;<code>weight<sub>i</sub></code>&nbsp;的 <strong>有向</strong> 边。</p>\n\n<p>最后，给你三个 <strong>互不相同</strong>&nbsp;的整数&nbsp;<code>src1</code>&nbsp;，<code>src2</code>&nbsp;和&nbsp;<code>dest</code>&nbsp;，表示图中三个不同的点。</p>\n\n<p>请你从图中选出一个 <b>边权和最小</b>&nbsp;的子图，使得从 <code>src1</code>&nbsp;和 <code>src2</code>&nbsp;出发，在这个子图中，都 <strong>可以</strong>&nbsp;到达 <code>dest</code>&nbsp;。如果这样的子图不存在，请返回 <code>-1</code>&nbsp;。</p>\n\n<p><strong>子图</strong>&nbsp;中的点和边都应该属于原图的一部分。子图的边权和定义为它所包含的所有边的权值之和。</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/2022/02/17/example1drawio.png\" style=\"width: 263px; height: 250px;\" /></p>\n\n<pre>\n<b>输入：</b>n = 6, edges = [[0,2,2],[0,5,6],[1,0,3],[1,4,5],[2,1,1],[2,3,3],[2,3,4],[3,4,2],[4,5,1]], src1 = 0, src2 = 1, dest = 5\n<b>输出：</b>9\n<strong>解释：</strong>\n上图为输入的图。\n蓝色边为最优子图之一。\n注意，子图 [[1,0,3],[0,5,6]] 也能得到最优解，但无法在满足所有限制的前提下，得到更优解。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/02/17/example2-1drawio.png\" style=\"width: 350px; height: 51px;\" /></p>\n\n<pre>\n<b>输入：</b>n = 3, edges = [[0,1,1],[2,1,1]], src1 = 0, src2 = 1, dest = 2\n<b>输出：</b>-1\n<strong>解释：</strong>\n上图为输入的图。\n可以看到，不存在从节点 1 到节点 2 的路径，所以不存在任何子图满足所有限制。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>3 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= edges.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>edges[i].length == 3</code></li>\n\t<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub>, src1, src2, dest &lt;= n - 1</code></li>\n\t<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>\n\t<li><code>src1</code>&nbsp;，<code>src2</code>&nbsp;和&nbsp;<code>dest</code>&nbsp;两两不同。</li>\n\t<li><code>1 &lt;= weight[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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             "difficulty": "Hard",
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@@ -143,7 +143,7 @@
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             "hints": [
                 "Consider what the paths from src1 to dest and src2 to dest would look like in the optimal solution.",
                 "It can be shown that in an optimal solution, the two paths from src1 and src2 will coincide at one node, and the remaining part to dest will be the same for both paths. Now consider how to find the node where the paths will coincide.",