leetcode-problemset/leetcode-cn/originData/minimum-cost-walk-in-weighted-graph.json

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        "question": {
            "questionId": "3348",
            "questionFrontendId": "100244",
            "categoryTitle": "Algorithms",
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            "title": "Minimum Cost Walk in Weighted Graph",
            "titleSlug": "minimum-cost-walk-in-weighted-graph",
            "content": "<p>There is an undirected weighted graph with <code>n</code> vertices labeled from <code>0</code> to <code>n - 1</code>.</p>\n\n<p>You are given the integer <code>n</code> and an array <code>edges</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between vertices <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code> with a weight of <code>w<sub>i</sub></code>.</p>\n\n<p>A walk on a graph is a sequence of vertices and edges. The walk starts and ends with a vertex, and each edge connects the vertex that comes before it and the vertex that comes after it. It&#39;s important to note that a walk may visit the same edge or vertex more than once.</p>\n\n<p>The <strong>cost</strong> of a walk starting at node <code>u</code> and ending at node <code>v</code> is defined as the bitwise <code>AND</code> of the weights of the edges traversed during the walk. In other words, if the sequence of edge weights encountered during the walk is <code>w<sub>0</sub>, w<sub>1</sub>, w<sub>2</sub>, ..., w<sub>k</sub></code>, then the cost is calculated as <code>w<sub>0</sub> &amp; w<sub>1</sub> &amp; w<sub>2</sub> &amp; ... &amp; w<sub>k</sub></code>, where <code>&amp;</code> denotes the bitwise <code>AND</code> operator.</p>\n\n<p>You are also given a 2D array <code>query</code>, where <code>query[i] = [s<sub>i</sub>, t<sub>i</sub>]</code>. For each query, you need to find the minimum cost of the walk starting at vertex <code>s<sub>i</sub></code> and ending at vertex <code>t<sub>i</sub></code>. If there exists no such walk, the answer is <code>-1</code>.</p>\n\n<p>Return <em>the array </em><code>answer</code><em>, where </em><code>answer[i]</code><em> denotes the <strong>minimum</strong> cost of a walk for query </em><code>i</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[1,-1]</span></p>\n\n<p><strong>Explanation:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/01/31/q4_example1-1.png\" style=\"padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 351px; height: 141px;\" />\n<p>To achieve the cost of 1 in the first query, we need to move on the following edges: <code>0-&gt;1</code> (weight 7), <code>1-&gt;2</code> (weight 1), <code>2-&gt;1</code> (weight 1), <code>1-&gt;3</code> (weight 7).</p>\n\n<p>In the second query, there is no walk between nodes 3 and 4, so the answer is -1.</p>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n</div>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[0]</span></p>\n\n<p><strong>Explanation:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/01/31/q4_example2e.png\" style=\"padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 211px; height: 181px;\" />\n<p>To achieve the cost of 0 in the first query, we need to move on the following edges: <code>1-&gt;2</code> (weight 1), <code>2-&gt;1</code> (weight 6), <code>1-&gt;2</code> (weight 1).</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &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;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</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;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= query.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>query[i].length == 2</code></li>\n\t<li><code>0 &lt;= s<sub>i</sub>, t<sub>i</sub> &lt;= n - 1</code></li>\n</ul>\n",
            "translatedTitle": "带权图里旅途的最小代价",
            "translatedContent": "<p>给你一个 <code>n</code>&nbsp;个节点的带权无向图，节点编号为 <code>0</code>&nbsp;到 <code>n - 1</code>&nbsp;。</p>\n\n<p>给你一个整数 <code>n</code>&nbsp;和一个数组&nbsp;<code>edges</code>&nbsp;，其中&nbsp;<code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>]</code>&nbsp;表示节点&nbsp;<code>u<sub>i</sub></code> 和&nbsp;<code>v<sub>i</sub></code>&nbsp;之间有一条权值为&nbsp;<code>w<sub>i</sub></code>&nbsp;的无向边。</p>\n\n<p>在图中，一趟旅途包含一系列节点和边。旅途开始和结束点都是图中的节点，且图中存在连接旅途中相邻节点的边。注意，一趟旅途可能访问同一条边或者同一个节点多次。</p>\n\n<p>如果旅途开始于节点 <code>u</code>&nbsp;，结束于节点 <code>v</code>&nbsp;，我们定义这一趟旅途的 <strong>代价</strong>&nbsp;是经过的边权按位与 <code>AND</code>&nbsp;的结果。换句话说，如果经过的边对应的边权为&nbsp;<code>w<sub>0</sub>, w<sub>1</sub>, w<sub>2</sub>, ..., w<sub>k</sub></code>&nbsp;，那么代价为<code>w<sub>0</sub> &amp; w<sub>1</sub> &amp; w<sub>2</sub> &amp; ... &amp; w<sub>k</sub></code>&nbsp;，其中&nbsp;<code>&amp;</code>&nbsp;表示按位与&nbsp;<code>AND</code>&nbsp;操作。</p>\n\n<p>给你一个二维数组&nbsp;<code>query</code>&nbsp;，其中&nbsp;<code>query[i] = [s<sub>i</sub>, t<sub>i</sub>]</code>&nbsp;。对于每一个查询，你需要找出从节点开始&nbsp;<code>s<sub>i</sub></code>&nbsp;，在节点&nbsp;<code>t<sub>i</sub></code>&nbsp;处结束的旅途的最小代价。如果不存在这样的旅途，答案为&nbsp;<code>-1</code>&nbsp;。</p>\n\n<p>返回数组<em>&nbsp;</em><code>answer</code>&nbsp;，其中<em>&nbsp;</em><code>answer[i]</code><em>&nbsp;</em>表示对于查询 <code>i</code>&nbsp;的&nbsp;<strong>最小</strong>&nbsp;旅途代价。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>n = 5, edges = [[0,1,7],[1,3,7],[1,2,1]], query = [[0,3],[3,4]]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>[1,-1]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/01/31/q4_example1-1.png\" style=\"padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 351px; height: 141px;\" /></p>\n\n<p>第一个查询想要得到代价为 1 的旅途，我们依次访问：<code>0-&gt;1</code>（边权为 7 ）<code>1-&gt;2</code>&nbsp;（边权为 1 ）<code>2-&gt;1</code>（边权为 1 ）<code>1-&gt;3</code>&nbsp;（边权为 7 ）。</p>\n\n<p>第二个查询中，无法从节点 3 到节点 4 ，所以答案为 -1 。</p>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n</div>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>n = 3, edges = [[0,2,7],[0,1,15],[1,2,6],[1,2,1]], query = [[1,2]]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>[0]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2024/01/31/q4_example2e.png\" style=\"padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 211px; height: 181px;\" /></p>\n\n<p>第一个查询想要得到代价为 0 的旅途，我们依次访问：<code>1-&gt;2</code>（边权为 1 ），<code>2-&gt;1</code>（边权 为 6 ），<code>1-&gt;2</code>（边权为 1 ）。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &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;= u<sub>i</sub>, v<sub>i</sub> &lt;= n - 1</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;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= query.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>query[i].length == 2</code></li>\n\t<li><code>0 &lt;= s<sub>i</sub>, t<sub>i</sub> &lt;= n - 1</code></li>\n</ul>\n",
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