leetcode-problemset/leetcode-cn/originData/maximum-spending-after-buying-items.json

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            "questionId": "3107",
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            "categoryTitle": "Algorithms",
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            "title": "Maximum Spending After Buying Items",
            "titleSlug": "maximum-spending-after-buying-items",
            "content": "<p>You are given a <strong>0-indexed</strong> <code>m * n</code> integer matrix <code>values</code>, representing the values of <code>m * n</code> different items in <code>m</code> different shops. Each shop has <code>n</code> items where the <code>j<sup>th</sup></code> item in the <code>i<sup>th</sup></code> shop has a value of <code>values[i][j]</code>. Additionally, the items in the <code>i<sup>th</sup></code> shop are sorted in non-increasing order of value. That is, <code>values[i][j] &gt;= values[i][j + 1]</code> for all <code>0 &lt;= j &lt; n - 1</code>.</p>\n\n<p>On each day, you would like to buy a single item from one of the shops. Specifically, On the <code>d<sup>th</sup></code> day you can:</p>\n\n<ul>\n\t<li>Pick any shop <code>i</code>.</li>\n\t<li>Buy the rightmost available item <code>j</code> for the price of <code>values[i][j] * d</code>. That is, find the greatest index <code>j</code> such that item <code>j</code> was never bought before, and buy it for the price of <code>values[i][j] * d</code>.</li>\n</ul>\n\n<p><strong>Note</strong> that all items are pairwise different. For example, if you have bought item <code>0</code> from shop <code>1</code>, you can still buy item <code>0</code> from any other shop.</p>\n\n<p>Return <em>the <strong>maximum amount of money that can be spent</strong> on buying all </em> <code>m * n</code> <em>products</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> values = [[8,5,2],[6,4,1],[9,7,3]]\n<strong>Output:</strong> 285\n<strong>Explanation:</strong> On the first day, we buy product 2 from shop 1 for a price of values[1][2] * 1 = 1.\nOn the second day, we buy product 2 from shop 0 for a price of values[0][2] * 2 = 4.\nOn the third day, we buy product 2 from shop 2 for a price of values[2][2] * 3 = 9.\nOn the fourth day, we buy product 1 from shop 1 for a price of values[1][1] * 4 = 16.\nOn the fifth day, we buy product 1 from shop 0 for a price of values[0][1] * 5 = 25.\nOn the sixth day, we buy product 0 from shop 1 for a price of values[1][0] * 6 = 36.\nOn the seventh day, we buy product 1 from shop 2 for a price of values[2][1] * 7 = 49.\nOn the eighth day, we buy product 0 from shop 0 for a price of values[0][0] * 8 = 64.\nOn the ninth day, we buy product 0 from shop 2 for a price of values[2][0] * 9 = 81.\nHence, our total spending is equal to 285.\nIt can be shown that 285 is the maximum amount of money that can be spent buying all m * n products. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> values = [[10,8,6,4,2],[9,7,5,3,2]]\n<strong>Output:</strong> 386\n<strong>Explanation:</strong> On the first day, we buy product 4 from shop 0 for a price of values[0][4] * 1 = 2.\nOn the second day, we buy product 4 from shop 1 for a price of values[1][4] * 2 = 4.\nOn the third day, we buy product 3 from shop 1 for a price of values[1][3] * 3 = 9.\nOn the fourth day, we buy product 3 from shop 0 for a price of values[0][3] * 4 = 16.\nOn the fifth day, we buy product 2 from shop 1 for a price of values[1][2] * 5 = 25.\nOn the sixth day, we buy product 2 from shop 0 for a price of values[0][2] * 6 = 36.\nOn the seventh day, we buy product 1 from shop 1 for a price of values[1][1] * 7 = 49.\nOn the eighth day, we buy product 1 from shop 0 for a price of values[0][1] * 8 = 64\nOn the ninth day, we buy product 0 from shop 1 for a price of values[1][0] * 9 = 81.\nOn the tenth day, we buy product 0 from shop 0 for a price of values[0][0] * 10 = 100.\nHence, our total spending is equal to 386.\nIt can be shown that 386 is the maximum amount of money that can be spent buying all m * n products.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= m == values.length &lt;= 10</code></li>\n\t<li><code>1 &lt;= n == values[i].length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= values[i][j] &lt;= 10<sup>6</sup></code></li>\n\t<li><code>values[i]</code> are sorted in non-increasing order.</li>\n</ul>\n",
            "translatedTitle": "购买物品的最大开销",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始大小为&nbsp;<code>m * n</code>&nbsp;的整数矩阵&nbsp;<code>values</code>&nbsp;，表示&nbsp;<code>m</code>&nbsp;个不同商店里&nbsp;<code>m * n</code>&nbsp;件不同的物品。每个商店有 <code>n</code>&nbsp;件物品，第&nbsp;<code>i</code>&nbsp;个商店的第 <code>j</code>&nbsp;件物品的价值为&nbsp;<code>values[i][j]</code>&nbsp;。除此以外，第&nbsp;<code>i</code>&nbsp;个商店的物品已经按照价值非递增排好序了，也就是说对于所有&nbsp;<code>0 &lt;= j &lt; n - 1</code>&nbsp;都有&nbsp;<code>values[i][j] &gt;= values[i][j + 1]</code>&nbsp;。</p>\n\n<p>每一天，你可以在一个商店里购买一件物品。具体来说，在第&nbsp;<code>d</code>&nbsp;天，你可以：</p>\n\n<ul>\n\t<li>选择商店&nbsp;<code>i</code>&nbsp;。</li>\n\t<li>购买数组中最右边的物品&nbsp;<code>j</code>&nbsp;，开销为&nbsp;<code>values[i][j] * d</code>&nbsp;。换句话说，选择该商店中还没购买过的物品中最大的下标&nbsp;<code>j</code>&nbsp;，并且花费&nbsp;<code>values[i][j] * d</code>&nbsp;去购买。</li>\n</ul>\n\n<p><strong>注意</strong>，所有物品都视为不同的物品。比方说如果你已经从商店 <code>1</code>&nbsp;购买了物品&nbsp;<code>0</code>&nbsp;，你还可以在别的商店里购买其他商店的物品&nbsp;<code>0</code>&nbsp;。</p>\n\n<p>请你返回购买所有 <code>m * n</code>&nbsp;件物品需要的 <strong>最大开销</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>values = [[8,5,2],[6,4,1],[9,7,3]]\n<b>输出：</b>285\n<b>解释：</b>第一天，从商店 1 购买物品 2 ，开销为 values[1][2] * 1 = 1 。\n第二天，从商店 0 购买物品 2 ，开销为 values[0][2] * 2 = 4 。\n第三天，从商店 2 购买物品 2 ，开销为 values[2][2] * 3 = 9 。\n第四天，从商店 1 购买物品 1 ，开销为 values[1][1] * 4 = 16 。\n第五天，从商店 0 购买物品 1 ，开销为 values[0][1] * 5 = 25 。\n第六天，从商店 1 购买物品 0 ，开销为 values[1][0] * 6 = 36 。\n第七天，从商店 2 购买物品 1 ，开销为 values[2][1] * 7 = 49 。\n第八天，从商店 0 购买物品 0 ，开销为 values[0][0] * 8 = 64 。\n第九天，从商店 2 购买物品 0 ，开销为 values[2][0] * 9 = 81 。\n所以总开销为 285 。\n285 是购买所有 m * n 件物品的最大总开销。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>values = [[10,8,6,4,2],[9,7,5,3,2]]\n<b>输出：</b>386\n<b>解释：</b>第一天，从商店 0 购买物品 4 ，开销为 values[0][4] * 1 = 2 。\n第二天，从商店 1 购买物品 4 ，开销为 values[1][4] * 2 = 4 。\n第三天，从商店 1 购买物品 3 ，开销为 values[1][3] * 3 = 9 。\n第四天，从商店 0 购买物品 3 ，开销为 values[0][3] * 4 = 16 。\n第五天，从商店 1 购买物品 2 ，开销为 values[1][2] * 5 = 25 。\n第六天，从商店 0 购买物品 2 ，开销为 values[0][2] * 6 = 36 。\n第七天，从商店 1 购买物品 1 ，开销为 values[1][1] * 7 = 49 。\n第八天，从商店 0 购买物品 1 ，开销为 values[0][1] * 8 = 64 。\n第九天，从商店 1 购买物品 0 ，开销为 values[1][0] * 9 = 81 。\n第十天，从商店 0 购买物品 0 ，开销为 values[0][0] * 10 = 100 。\n所以总开销为 386 。\n386 是购买所有 m * n 件物品的最大总开销。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= m == values.length &lt;= 10</code></li>\n\t<li><code>1 &lt;= n == values[i].length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= values[i][j] &lt;= 10<sup>6</sup></code></li>\n\t<li><code>values[i]</code>&nbsp;按照非递增顺序排序。</li>\n</ul>\n",
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