You are given an m x n integer matrix grid and an array queries of size k.
Find an array answer of size k such that for each integer queries[i] you start in the top left cell of the matrix and repeat the following process:
queries[i] is strictly greater than the value of the current cell that you are in, then you get one point if it is your first time visiting this cell, and you can move to any adjacent cell in all 4 directions: up, down, left, and right.After the process, answer[i] is the maximum number of points you can get. Note that for each query you are allowed to visit the same cell multiple times.
Return the resulting array answer.
\n
Example 1:
\n![\"\"](\"https://assets.leetcode.com/uploads/2022/10/19/yetgriddrawio.png\")
\n\nInput: grid = [[1,2,3],[2,5,7],[3,5,1]], queries = [5,6,2]\nOutput: [5,8,1]\nExplanation: The diagrams above show which cells we visit to get points for each query.\n\n
Example 2:
\n![\"\"](\"https://assets.leetcode.com/uploads/2022/10/20/yetgriddrawio-2.png\")
\n\nInput: grid = [[5,2,1],[1,1,2]], queries = [3]\nOutput: [0]\nExplanation: We can not get any points because the value of the top left cell is already greater than or equal to 3.\n\n\n
\n
Constraints:
\n\nm == grid.lengthn == grid[i].length2 <= m, n <= 10004 <= m * n <= 105k == queries.length1 <= k <= 1041 <= grid[i][j], queries[i] <= 106给你一个大小为 m x n 的整数矩阵 grid 和一个大小为 k 的数组 queries 。
找出一个大小为 k 的数组 answer ，且满足对于每个整数 queries[i] ，你从矩阵 左上角 单元格开始，重复以下过程：
queries[i] 严格 大于你当前所处位置单元格，如果该单元格是第一次访问，则获得 1 分，并且你可以移动到所有 4 个方向（上、下、左、右）上任一 相邻 单元格。在过程结束后，answer[i] 是你可以获得的最大分数。注意，对于每个查询，你可以访问同一个单元格 多次 。
返回结果数组 answer 。
\n\n
示例 1：
\n\n\n输入：grid = [[1,2,3],[2,5,7],[3,5,1]], queries = [5,6,2]\n输出：[5,8,1]\n解释：上图展示了每个查询中访问并获得分数的单元格。\n\n
示例 2：
\n\n\n输入：grid = [[5,2,1],[1,1,2]], queries = [3]\n输出：[0]\n解释：无法获得分数，因为左上角单元格的值大于等于 3 。\n\n\n
\n\n
提示：
\n\nm == grid.lengthn == grid[i].length2 <= m, n <= 10004 <= m * n <= 105k == queries.length1 <= k <= 1041 <= grid[i][j], queries[i] <= 106\\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\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>\"],\"java\":[\"Java\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u88ab\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u5305\\u542b Pair \\u7c7b: https:\\/\\/docs.oracle.com\\/javase\\/8\\/javafx\\/api\\/javafx\\/util\\/Pair.html <\\/p>\"],\"python\":[\"Python\",\" \\u7248\\u672c\\uff1a \\u4e3a\\u4e86\\u65b9\\u4fbf\\u8d77\\u89c1\\uff0c\\u5927\\u90e8\\u5206\\u5e38\\u7528\\u5e93\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8 \\u5bfc\\u5165\\uff0c\\u5982\\uff1aarray<\\/a>, bisect<\\/a>, collections<\\/a>\\u3002\\u5982\\u679c\\u60a8\\u9700\\u8981\\u4f7f\\u7528\\u5176\\u4ed6\\u5e93\\u51fd\\u6570\\uff0c\\u8bf7\\u81ea\\u884c\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n \\u6ce8\\u610f Python 2.7 \\u5c06\\u57282020\\u5e74\\u540e\\u4e0d\\u518d\\u7ef4\\u62a4<\\/a>\\u3002 \\u5982\\u60f3\\u4f7f\\u7528\\u6700\\u65b0\\u7248\\u7684Python\\uff0c\\u8bf7\\u9009\\u62e9Python 3\\u3002<\\/p>\"],\"c\":[\"C\",\" \\u7248\\u672c\\uff1a \\u7f16\\u8bd1\\u65f6\\uff0c\\u5c06\\u4f1a\\u91c7\\u7528 \\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 \\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n 1. 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