leetcode-problemset/leetcode-cn/originData/minimum-cost-of-a-path-with-special-roads.json

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        "question": {
            "questionId": "2686",
            "questionFrontendId": "2662",
            "categoryTitle": "Algorithms",
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            "title": "Minimum Cost of a Path With Special Roads",
            "titleSlug": "minimum-cost-of-a-path-with-special-roads",
            "content": "<p>You are given an array <code>start</code> where <code>start = [startX, startY]</code> represents your initial position <code>(startX, startY)</code> in a 2D space. You are also given the array <code>target</code> where <code>target = [targetX, targetY]</code> represents your target position <code>(targetX, targetY)</code>.</p>\n\n<p>The <strong>cost</strong> of going from a position <code>(x1, y1)</code> to any other position in the space <code>(x2, y2)</code> is <code>|x2 - x1| + |y2 - y1|</code>.</p>\n\n<p>There are also some <strong>special roads</strong>. You are given a 2D array <code>specialRoads</code> where <code>specialRoads[i] = [x1<sub>i</sub>, y1<sub>i</sub>, x2<sub>i</sub>, y2<sub>i</sub>, cost<sub>i</sub>]</code> indicates that the <code>i<sup>th</sup></code> special road goes in <strong>one direction</strong> from <code>(x1<sub>i</sub>, y1<sub>i</sub>)</code> to <code>(x2<sub>i</sub>, y2<sub>i</sub>)</code> with a cost equal to <code>cost<sub>i</sub></code>. You can use each special road any number of times.</p>\n\n<p>Return the <strong>minimum</strong> cost required to go from <code>(startX, startY)</code> to <code>(targetX, targetY)</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\">start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">5</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ol>\n\t<li>(1,1) to (1,2) with a cost of |1 - 1| + |2 - 1| = 1.</li>\n\t<li>(1,2) to (3,3). Use <code><span class=\"example-io\">specialRoads[0]</span></code><span class=\"example-io\"> with</span><span class=\"example-io\"> the cost 2.</span></li>\n\t<li><span class=\"example-io\">(3,3) to (3,4) with a cost of |3 - 3| + |4 - 3| = 1.</span></li>\n\t<li><span class=\"example-io\">(3,4) to (4,5). Use </span><code><span class=\"example-io\">specialRoads[1]</span></code><span class=\"example-io\"> with the cost</span><span class=\"example-io\"> 1.</span></li>\n</ol>\n\n<p><span class=\"example-io\">So the total cost is 1 + 2 + 1 + 1 = 5.</span></p>\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\">start = [3,2], target = [5,7], specialRoads = [[5,7,3,2,1],[3,2,3,4,4],[3,3,5,5,5],[3,4,5,6,6]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">7</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>It is optimal not to use any special edges and go directly from the starting to the ending position with a cost |5 - 3| + |7 - 2| = 7.</p>\n\n<p>Note that the <span class=\"example-io\"><code>specialRoads[0]</code> is directed from (5,7) to (3,2).</span></p>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">start = [1,1], target = [10,4], specialRoads = [[4,2,1,1,3],[1,2,7,4,4],[10,3,6,1,2],[6,1,1,2,3]]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">8</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ol>\n\t<li>(1,1) to (1,2) with a cost of |1 - 1| + |2 - 1| = 1.</li>\n\t<li>(1,2) to (7,4). Use <code><span class=\"example-io\">specialRoads[1]</span></code><span class=\"example-io\"> with the cost</span><span class=\"example-io\"> 4.</span></li>\n\t<li>(7,4) to (10,4) with a cost of |10 - 7| + |4 - 4| = 3.</li>\n</ol>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>start.length == target.length == 2</code></li>\n\t<li><code>1 &lt;= startX &lt;= targetX &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= startY &lt;= targetY &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= specialRoads.length &lt;= 200</code></li>\n\t<li><code>specialRoads[i].length == 5</code></li>\n\t<li><code>startX &lt;= x1<sub>i</sub>, x2<sub>i</sub> &lt;= targetX</code></li>\n\t<li><code>startY &lt;= y1<sub>i</sub>, y2<sub>i</sub> &lt;= targetY</code></li>\n\t<li><code>1 &lt;= cost<sub>i</sub> &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "前往目标的最小代价",
            "translatedContent": "<p>给你一个数组 <code>start</code> ，其中 <code>start = [startX, startY]</code> 表示你的初始位置位于二维空间上的 <code>(startX, startY)</code> 。另给你一个数组 <code>target</code> ，其中 <code>target = [targetX, targetY]</code> 表示你的目标位置 <code>(targetX, targetY)</code> 。</p>\n\n<p>从位置 <code>(x1, y1)</code> 到空间中任一其他位置 <code>(x2, y2)</code> 的 <strong>代价</strong> 是 <code>|x2 - x1| + |y2 - y1|</code> 。</p>\n\n<p>给你一个二维数组 <code>specialRoads</code> ，表示空间中存在的一些 <strong>特殊路径</strong>。其中 <code>specialRoads[i] = [x1<sub>i</sub>, y1<sub>i</sub>, x2<sub>i</sub>, y2<sub>i</sub>, cost<sub>i</sub>]</code> 表示第 <code>i</code> 条特殊路径可以从 <code>(x1<sub>i</sub>, y1<sub>i</sub>)</code> 到 <code>(x2<sub>i</sub>, y2<sub>i</sub>)</code> ，但成本等于 <code>cost<sub>i</sub></code> 。你可以使用每条特殊路径任意次数。</p>\n\n<p>返回从 <code>(startX, startY)</code> 到 <code>(targetX, targetY)</code> 所需的 <strong>最小</strong> 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><strong>输入：</strong><span class=\"example-io\">start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>5</span></p>\n\n<p><b>解释：</b></p>\n\n<ol>\n\t<li>(1,1) 到 (1,2) 花费为 |1 - 1| + |2 - 1| = 1。</li>\n\t<li>(1,2) 到 (3,3)。使用&nbsp;<code><span class=\"example-io\">specialRoads[0]</span></code><span class=\"example-io\">&nbsp;花费为</span><span class=\"example-io\">&nbsp;2。</span></li>\n\t<li><span class=\"example-io\">(3,3) </span>到<span class=\"example-io\"> (3,4) </span>花费为<span class=\"example-io\"> |3 - 3| + |4 - 3| = 1。</span></li>\n\t<li><span class=\"example-io\">(3,4) </span>到<span class=\"example-io\"> (4,5)。</span>使用<span class=\"example-io\"> </span><code><span class=\"example-io\">specialRoads[1]</span></code><span class=\"example-io\"> 花费为</span><span class=\"example-io\"> 1。</span></li>\n</ol>\n\n<p><span class=\"example-io\">所以总花费是 1 + 2 + 1 + 1 = 5。</span></p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">start = [3,2], target = [5,7], specialRoads = [[5,7,3,2,1],[3,2,3,4,4],[3,3,5,5,5],[3,4,5,6,6]]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b></span><span class=\"example-io\">7</span></p>\n\n<p><b>解释：</b></p>\n\n<p>不使用任何特殊路径，直接从开始到结束位置是最优的，花费为&nbsp;|5 - 3| + |7 - 2| = 7。</p>\n\n<p>注意&nbsp;<span class=\"example-io\"><code>specialRoads[0]</code>&nbsp;直接从 (5,7) 到 (3,2)。</span></p>\n</div>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong><span class=\"example-io\">start = [1,1], target = [10,4], specialRoads = [[4,2,1,1,3],[1,2,7,4,4],[10,3,6,1,2],[6,1,1,2,3]]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b></span><span class=\"example-io\">8</span></p>\n\n<p><b>解释：</b></p>\n\n<ol>\n\t<li>(1,1) 到 (1,2) 花费为 |1 - 1| + |2 - 1| = 1。</li>\n\t<li>(1,2) 到 (7,4)。使用&nbsp;<code><span class=\"example-io\">specialRoads[1]</span></code><span class=\"example-io\">&nbsp;花费为</span><span class=\"example-io\">&nbsp;4。</span></li>\n\t<li>(7,4) 到 (10,4) 花费为 |10 - 7| + |4 - 4| = 3。</li>\n</ol>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>start.length == target.length == 2</code></li>\n\t<li><code>1 &lt;= startX &lt;= targetX &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= startY &lt;= targetY &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= specialRoads.length &lt;= 200</code></li>\n\t<li><code>specialRoads[i].length == 5</code></li>\n\t<li><code>startX &lt;= x1<sub>i</sub>, x2<sub>i</sub> &lt;= targetX</code></li>\n\t<li><code>startY &lt;= y1<sub>i</sub>, y2<sub>i</sub> &lt;= targetY</code></li>\n\t<li><code>1 &lt;= cost<sub>i</sub> &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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