mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
162 lines
20 KiB
JSON
162 lines
20 KiB
JSON
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"questionId": "1000434",
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"categoryTitle": "Algorithms",
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"title": "烹饪料理",
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"translatedContent": "欢迎各位勇者来到力扣城，城内设有烹饪锅供勇者制作料理，为自己恢复状态。\n\n勇者背包内共有编号为 `0 ~ 4` 的五种食材，其中 `materials[j]` 表示第 `j` 种食材的数量。通过这些食材可以制作若干料理，`cookbooks[i][j]` 表示制作第 `i` 种料理需要第 `j` 种食材的数量，而 `attribute[i] = [x,y]` 表示第 `i` 道料理的美味度 `x` 和饱腹感 `y`。\n\n在饱腹感不小于 `limit` 的情况下，请返回勇者可获得的最大美味度。如果无法满足饱腹感要求，则返回 `-1`。\n\n**注意：**\n- 每种料理只能制作一次。\n\n\n**示例 1：**\n>输入：`materials = [3,2,4,1,2]`\n>`cookbooks = [[1,1,0,1,2],[2,1,4,0,0],[3,2,4,1,0]]`\n>`attribute = [[3,2],[2,4],[7,6]]`\n>`limit = 5`\n>\n>输出：`7`\n>\n>解释：\n>食材数量可以满足以下两种方案：\n>方案一：制作料理 0 和料理 1，可获得饱腹感 2+4、美味度 3+2\n>方案二：仅制作料理 2， 可饱腹感为 6、美味度为 7\n>因此在满足饱腹感的要求下，可获得最高美味度 7\n\n**示例 2：**\n>输入：`materials = [10,10,10,10,10]`\n>`cookbooks = [[1,1,1,1,1],[3,3,3,3,3],[10,10,10,10,10]]`\n>`attribute = [[5,5],[6,6],[10,10]]`\n>`limit = 1`\n>\n>输出：`11`\n>\n>解释：通过制作料理 0 和 1，可满足饱腹感，并获得最高美味度 11\n\n**提示：**\n+ `materials.length == 5`\n+ `1 <= cookbooks.length == attribute.length <= 8`\n+ `cookbooks[i].length == 5`\n+ `attribute[i].length == 2`\n+ `0 <= materials[i], cookbooks[i][j], attribute[i][j] <= 20`\n+ `1 <= limit <= 100`\n",
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