mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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204 lines
29 KiB
JSON
204 lines
29 KiB
JSON
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"title": "Cheapest Flights Within K Stops",
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"content": "<p>There are <code>n</code> cities connected by some number of flights. You are given an array <code>flights</code> where <code>flights[i] = [from<sub>i</sub>, to<sub>i</sub>, price<sub>i</sub>]</code> indicates that there is a flight from city <code>from<sub>i</sub></code> to city <code>to<sub>i</sub></code> with cost <code>price<sub>i</sub></code>.</p>\n\n<p>You are also given three integers <code>src</code>, <code>dst</code>, and <code>k</code>, return <em><strong>the cheapest price</strong> from </em><code>src</code><em> to </em><code>dst</code><em> with at most </em><code>k</code><em> stops. </em>If there is no such route, return<em> </em><code>-1</code>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-3drawio.png\" style=\"width: 332px; height: 392px;\" />\n<pre>\n<strong>Input:</strong> n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1\n<strong>Output:</strong> 700\n<strong>Explanation:</strong>\nThe graph is shown above.\nThe optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.\nNote that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-1drawio.png\" style=\"width: 332px; height: 242px;\" />\n<pre>\n<strong>Input:</strong> n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1\n<strong>Output:</strong> 200\n<strong>Explanation:</strong>\nThe graph is shown above.\nThe optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-2drawio.png\" style=\"width: 332px; height: 242px;\" />\n<pre>\n<strong>Input:</strong> n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0\n<strong>Output:</strong> 500\n<strong>Explanation:</strong>\nThe graph is shown above.\nThe optimal path with no stops from city 0 to 2 is marked in red and has cost 500.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= n <= 100</code></li>\n\t<li><code>0 <= flights.length <= (n * (n - 1) / 2)</code></li>\n\t<li><code>flights[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>from<sub>i</sub> != to<sub>i</sub></code></li>\n\t<li><code>1 <= price<sub>i</sub> <= 10<sup>4</sup></code></li>\n\t<li>There will not be any multiple flights between two cities.</li>\n\t<li><code>0 <= src, dst, k < n</code></li>\n\t<li><code>src != dst</code></li>\n</ul>\n",
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"translatedTitle": "K 站中转内最便宜的航班",
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"translatedContent": "<p>有 <code>n</code> 个城市通过一些航班连接。给你一个数组 <code>flights</code> ，其中 <code>flights[i] = [from<sub>i</sub>, to<sub>i</sub>, price<sub>i</sub>]</code> ，表示该航班都从城市 <code>from<sub>i</sub></code> 开始，以价格 <code>price<sub>i</sub></code> 抵达 <code>to<sub>i</sub></code>。</p>\n\n<p>现在给定所有的城市和航班，以及出发城市 <code>src</code> 和目的地 <code>dst</code>，你的任务是找到出一条最多经过 <code>k</code> 站中转的路线，使得从 <code>src</code> 到 <code>dst</code> 的 <strong>价格最便宜</strong> ，并返回该价格。 如果不存在这样的路线，则输出 <code>-1</code>。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-3drawio.png\" style=\"width: 332px; height: 392px;\" />\n<pre>\n<strong>输入:</strong> \nn = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1\n<strong>输出:</strong> 700 \n<strong>解释:</strong> 城市航班图如上\n从城市 0 到城市 3 经过最多 1 站的最佳路径用红色标记，费用为 100 + 600 = 700。\n请注意，通过城市 [0, 1, 2, 3] 的路径更便宜，但无效，因为它经过了 2 站。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-1drawio.png\" style=\"width: 332px; height: 242px;\" />\n<pre>\n<strong>输入:</strong> \nn = 3, edges = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1\n<strong>输出:</strong> 200\n<strong>解释:</strong> \n城市航班图如上\n从城市 0 到城市 2 经过最多 1 站的最佳路径标记为红色，费用为 100 + 100 = 200。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-2drawio.png\" style=\"width: 332px; height: 242px;\" />\n<pre>\n<b>输入：</b>n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0\n<b>输出：</b>500\n<strong>解释：</strong>\n城市航班图如上\n从城市 0 到城市 2 不经过站点的最佳路径标记为红色，费用为 500。\n</pre>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= n <= 100</code></li>\n\t<li><code>0 <= flights.length <= (n * (n - 1) / 2)</code></li>\n\t<li><code>flights[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>from<sub>i</sub> != to<sub>i</sub></code></li>\n\t<li><code>1 <= price<sub>i</sub> <= 10<sup>4</sup></code></li>\n\t<li>航班没有重复，且不存在自环</li>\n\t<li><code>0 <= src, dst, k < n</code></li>\n\t<li><code>src != dst</code></li>\n</ul>\n",
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