mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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183 lines
25 KiB
JSON
183 lines
25 KiB
JSON
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"title": "Minimum Path Cost in a Grid",
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"content": "<p>You are given a <strong>0-indexed</strong> <code>m x n</code> integer matrix <code>grid</code> consisting of <strong>distinct</strong> integers from <code>0</code> to <code>m * n - 1</code>. You can move in this matrix from a cell to any other cell in the <strong>next</strong> row. That is, if you are in cell <code>(x, y)</code> such that <code>x < m - 1</code>, you can move to any of the cells <code>(x + 1, 0)</code>, <code>(x + 1, 1)</code>, ..., <code>(x + 1, n - 1)</code>. <strong>Note</strong> that it is not possible to move from cells in the last row.</p>\n\n<p>Each possible move has a cost given by a <strong>0-indexed</strong> 2D array <code>moveCost</code> of size <code>(m * n) x n</code>, where <code>moveCost[i][j]</code> is the cost of moving from a cell with value <code>i</code> to a cell in column <code>j</code> of the next row. The cost of moving from cells in the last row of <code>grid</code> can be ignored.</p>\n\n<p>The cost of a path in <code>grid</code> is the <strong>sum</strong> of all values of cells visited plus the <strong>sum</strong> of costs of all the moves made. Return <em>the <strong>minimum</strong> cost of a path that starts from any cell in the <strong>first</strong> row and ends at any cell in the <strong>last</strong> row.</em></p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/04/28/griddrawio-2.png\" style=\"width: 301px; height: 281px;\" />\n<pre>\n<strong>Input:</strong> grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]]\n<strong>Output:</strong> 17\n<strong>Explanation: </strong>The path with the minimum possible cost is the path 5 -> 0 -> 1.\n- The sum of the values of cells visited is 5 + 0 + 1 = 6.\n- The cost of moving from 5 to 0 is 3.\n- The cost of moving from 0 to 1 is 8.\nSo the total cost of the path is 6 + 3 + 8 = 17.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> grid = [[5,1,2],[4,0,3]], moveCost = [[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> The path with the minimum possible cost is the path 2 -> 3.\n- The sum of the values of cells visited is 2 + 3 = 5.\n- The cost of moving from 2 to 3 is 1.\nSo the total cost of this path is 5 + 1 = 6.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == grid.length</code></li>\n\t<li><code>n == grid[i].length</code></li>\n\t<li><code>2 <= m, n <= 50</code></li>\n\t<li><code>grid</code> consists of distinct integers from <code>0</code> to <code>m * n - 1</code>.</li>\n\t<li><code>moveCost.length == m * n</code></li>\n\t<li><code>moveCost[i].length == n</code></li>\n\t<li><code>1 <= moveCost[i][j] <= 100</code></li>\n</ul>\n",
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"translatedTitle": "网格中的最小路径代价",
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"translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始的整数矩阵 <code>grid</code> ,矩阵大小为 <code>m x n</code> ,由从 <code>0</code> 到 <code>m * n - 1</code> 的不同整数组成。你可以在此矩阵中,从一个单元格移动到 <strong>下一行</strong> 的任何其他单元格。如果你位于单元格 <code>(x, y)</code> ,且满足 <code>x < m - 1</code> ,你可以移动到 <code>(x + 1, 0)</code>, <code>(x + 1, 1)</code>, ..., <code>(x + 1, n - 1)</code><strong> </strong>中的任何一个单元格。<strong>注意:</strong> 在最后一行中的单元格不能触发移动。</p>\n\n<p>每次可能的移动都需要付出对应的代价,代价用一个下标从 <strong>0</strong> 开始的二维数组 <code>moveCost</code> 表示,该数组大小为 <code>(m * n) x n</code> ,其中 <code>moveCost[i][j]</code> 是从值为 <code>i</code> 的单元格移动到下一行第 <code>j</code> 列单元格的代价。从 <code>grid</code> 最后一行的单元格移动的代价可以忽略。</p>\n\n<p><code>grid</code> 一条路径的代价是:所有路径经过的单元格的 <strong>值之和</strong> 加上 所有移动的 <strong>代价之和 </strong>。从 <strong>第一行</strong> 任意单元格出发,返回到达 <strong>最后一行</strong> 任意单元格的最小路径代价<em>。</em></p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/04/28/griddrawio-2.png\" style=\"width: 301px; height: 281px;\" /></p>\n\n<pre>\n<strong>输入:</strong>grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]]\n<strong>输出:</strong>17\n<strong>解释:</strong>最小代价的路径是 5 -> 0 -> 1 。\n- 路径途经单元格值之和 5 + 0 + 1 = 6 。\n- 从 5 移动到 0 的代价为 3 。\n- 从 0 移动到 1 的代价为 8 。\n路径总代价为 6 + 3 + 8 = 17 。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>grid = [[5,1,2],[4,0,3]], moveCost = [[12,10,15],[20,23,8],[21,7,1],[8,1,13],[9,10,25],[5,3,2]]\n<strong>输出:</strong>6\n<strong>解释:</strong>\n最小代价的路径是 2 -> 3 。 \n- 路径途经单元格值之和 2 + 3 = 5 。 \n- 从 2 移动到 3 的代价为 1 。 \n路径总代价为 5 + 1 = 6 。</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>m == grid.length</code></li>\n\t<li><code>n == grid[i].length</code></li>\n\t<li><code>2 <= m, n <= 50</code></li>\n\t<li><code>grid</code> 由从 <code>0</code> 到 <code>m * n - 1</code> 的不同整数组成</li>\n\t<li><code>moveCost.length == m * n</code></li>\n\t<li><code>moveCost[i].length == n</code></li>\n\t<li><code>1 <= moveCost[i][j] <= 100</code></li>\n</ul>\n",
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"What is the optimal cost to get to each of the cells in the second row? What about the third row?",
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