leetcode-problemset/leetcode-cn/originData/time-to-cross-a-bridge.json

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            "questionId": "2642",
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            "categoryTitle": "Algorithms",
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            "title": "Time to Cross a Bridge",
            "titleSlug": "time-to-cross-a-bridge",
            "content": "<p>There are <code>k</code> workers who want to move <code>n</code> boxes from the right (old) warehouse to the left (new) warehouse. You are given the two integers <code>n</code> and <code>k</code>, and a 2D integer array <code>time</code> of size <code>k x 4</code> where <code>time[i] = [right<sub>i</sub>, pick<sub>i</sub>, left<sub>i</sub>, put<sub>i</sub>]</code>.</p>\n\n<p>The warehouses are separated by a river and connected by a bridge. Initially, all <code>k</code> workers are waiting on the left side of the bridge. To move the boxes, the <code>i<sup>th</sup></code> worker can do the following:</p>\n\n<ul>\n\t<li>Cross the bridge to the right side in <code>right<sub>i</sub></code> minutes.</li>\n\t<li>Pick a box from the right warehouse in <code>pick<sub>i</sub></code> minutes.</li>\n\t<li>Cross the bridge to the left side in <code>left<sub>i</sub></code> minutes.</li>\n\t<li>Put the box into the left warehouse in <code>put<sub>i</sub></code> minutes.</li>\n</ul>\n\n<p>The <code>i<sup>th</sup></code> worker is <strong>less efficient</strong> than the j<code><sup>th</sup></code> worker if either condition is met:</p>\n\n<ul>\n\t<li><code>left<sub>i</sub> + right<sub>i</sub> &gt; left<sub>j</sub> + right<sub>j</sub></code></li>\n\t<li><code>left<sub>i</sub> + right<sub>i</sub> == left<sub>j</sub> + right<sub>j</sub></code> and <code>i &gt; j</code></li>\n</ul>\n\n<p>The following rules regulate the movement of the workers through the bridge:</p>\n\n<ul>\n\t<li>Only one worker can use the bridge at a time.</li>\n\t<li>When the bridge is unused prioritize the <strong>least efficient</strong> worker (who have picked up the box) on the right side to cross. If not,&nbsp;prioritize the <strong>least efficient</strong> worker on the left side to cross.</li>\n\t<li>If enough workers have already been dispatched from the left side to pick up all the remaining boxes, <strong>no more</strong> workers will be sent from the left side.</li>\n</ul>\n\n<p>Return the <strong>elapsed minutes</strong> at which the last box reaches the <strong>left side of the bridge</strong>.</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 = 1, k = 3, time = [[1,1,2,1],[1,1,3,1],[1,1,4,1]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">6</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<pre>\nFrom 0 to 1 minutes: worker 2 crosses the bridge to the right.\nFrom 1 to 2 minutes: worker 2 picks up a box from the right warehouse.\nFrom 2 to 6 minutes: worker 2 crosses the bridge to the left.\nFrom 6 to 7 minutes: worker 2 puts a box at the left warehouse.\nThe whole process ends after 7 minutes. We return 6 because the problem asks for the instance of time at which the last worker reaches the left side of the bridge.\n</pre>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">n = 3, k = 2, time =</span> [[1,5,1,8],[10,10,10,10]]</p>\n\n<p><strong>Output:</strong> 37</p>\n\n<p><strong>Explanation:</strong></p>\n\n<pre>\n<img src=\"https://assets.leetcode.com/uploads/2024/11/21/378539249-c6ce3c73-40e7-4670-a8b5-7ddb9abede11.png\" style=\"width: 450px; height: 176px;\" />\n</pre>\n\n<p>The last box reaches the left side at 37 seconds. Notice, how we <strong>do not</strong> put the last boxes down, as that would take more time, and they are already on the left with the workers.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n, k &lt;= 10<sup>4</sup></code></li>\n\t<li><code>time.length == k</code></li>\n\t<li><code>time[i].length == 4</code></li>\n\t<li><code>1 &lt;= left<sub>i</sub>, pick<sub>i</sub>, right<sub>i</sub>, put<sub>i</sub> &lt;= 1000</code></li>\n</ul>\n",
            "translatedTitle": "过桥的时间",
            "translatedContent": "<p>共有 <code>k</code> 位工人计划将 <code>n</code> 个箱子从右侧的（旧）仓库移动到左侧的（新）仓库。给你两个整数 <code>n</code> 和 <code>k</code>，以及一个二维整数数组 <code>time</code> ，数组的大小为 <code>k x 4</code> ，其中 <code>time[i] = [right<sub>i</sub>, pick<sub>i</sub>, left<sub>i</sub>, put<sub>i</sub>]</code> 。</p>\n\n<p>一条河将两座仓库分隔，只能通过一座桥通行。旧仓库位于河的右岸，新仓库在河的左岸。开始时，所有 <code>k</code> 位工人都在桥的左侧等待。为了移动这些箱子，第 <code>i</code> 位工人（下标从 <strong>0</strong> 开始）可以：</p>\n\n<ul>\n\t<li>从左岸（新仓库）跨过桥到右岸（旧仓库），用时 <code>right<sub>i</sub></code> 分钟。</li>\n\t<li>从旧仓库选择一个箱子，并返回到桥边，用时 <code>pick<sub>i</sub></code> 分钟。不同工人可以同时搬起所选的箱子。</li>\n\t<li>从右岸（旧仓库）跨过桥到左岸（新仓库），用时 <code>left<sub>i</sub></code> 分钟。</li>\n\t<li>将箱子放入新仓库，并返回到桥边，用时 <code>put<sub>i</sub></code> 分钟。不同工人可以同时放下所选的箱子。</li>\n</ul>\n\n<p>如果满足下面任一条件，则认为工人 <code>i</code> 的 <strong>效率低于</strong> 工人 <code>j</code> ：</p>\n\n<ul>\n\t<li><code>left<sub>i</sub> + right<sub>i</sub> &gt; left<sub>j</sub> + right<sub>j</sub></code></li>\n\t<li><code>left<sub>i</sub> + right<sub>i</sub> == left<sub>j</sub> + right<sub>j</sub></code> 且 <code>i &gt; j</code></li>\n</ul>\n\n<p>工人通过桥时需要遵循以下规则：</p>\n\n<ul>\n\t<li>同时只能有一名工人过桥。</li>\n\t<li>当桥梁未被使用时，优先让右侧 <strong>效率最低</strong> 的工人（已经拿起盒子的工人）过桥。如果不是，优先让左侧 <strong>效率最低</strong> 的工人通过。</li>\n\t<li>如果左侧已经派出足够的工人来拾取所有剩余的箱子，则 <strong>不会</strong> 再从左侧派出工人。</li>\n</ul>\n\n<p><span class=\"text-only\" data-eleid=\"8\" style=\"white-space: pre;\">请你返回最后一个箱子 </span><strong><span class=\"text-only\" data-eleid=\"9\" style=\"white-space: pre;\">到达桥左侧</span></strong><span class=\"text-only\" data-eleid=\"10\" style=\"white-space: pre;\"> 的时间。</span></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 = 1, k = 3, time = [[1,1,2,1],[1,1,3,1],[1,1,4,1]]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>6</span></p>\n\n<p><b>解释：</b></p>\n\n<pre>\n从 0 到 1 分钟：工人 2 通过桥到达右侧。\n从 1 到 2 分钟：工人 2 从右侧仓库拿起箱子。\n从 2 到 6 分钟：工人 2 通过桥到达左侧。\n从 6 到 7 分钟：工人 2 向左侧仓库放下箱子。\n整个过程在 7 分钟后结束。我们返回 6 因为该问题要求的是最后一名工人到达桥梁左侧的时间。\n</pre>\n</div>\n\n<p><strong class=\"example\">示例&nbsp;2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">n = 3, k = 2, time =</span> [[1,5,1,8],[10,10,10,10]]</p>\n\n<p><b>输出：</b>37</p>\n\n<p><strong>解释：</strong></p>\n\n<pre>\n<img src=\"https://assets.leetcode.com/uploads/2024/11/21/378539249-c6ce3c73-40e7-4670-a8b5-7ddb9abede11.png\" style=\"width: 450px; height: 176px;\" />\n</pre>\n\n<p>最后一个盒子在37秒时到达左侧。请注意，我们并 <strong>没有</strong> 放下最后一个箱子，因为那样会花费更多时间，而且它们已经和工人们一起在左边。</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, k &lt;= 10<sup>4</sup></code></li>\n\t<li><code>time.length == k</code></li>\n\t<li><code>time[i].length == 4</code></li>\n\t<li><code>1 &lt;= left<sub>i</sub>, pick<sub>i</sub>, right<sub>i</sub>, put<sub>i</sub> &lt;= 1000</code></li>\n</ul>\n",
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<code>use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/troydhanson.github.io\\/uthash\\/\\\" target=\\\"_blank\\\">uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n<p><b>1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n    int id;            \\/* we'll use this field as the key *\\/\\r\\n    char name[10];\\r\\n    UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry 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5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/blob\\/v5\\/README.md\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.4.0<\\/a>\\uff0c<a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/v4.2.3\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.3<\\/a> \\u4ee5\\u53ca <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/deque\\/tree\\/v1.0.4\\\" target=\\\"_blank\\\"> datastructures-js\\/deque@1.0.4<\\/a>\\u3002<\\/p>\"],\"racket\":[\"Racket\",\"<p><a 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