leetcode 2304. Minimum Path Cost in a Grid（python）

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時間 2022-06-20 11:15:09

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描述

You are given a 0-indexed m x n integer matrix grid consisting of distinct integers from 0 to m * n - 1. You can move in this matrix from a cell to any other cell in the next row. That is, if you are in cell (x, y) such that x < m - 1, you can move to any of the cells (x + 1, 0), (x + 1, 1), ..., (x + 1, n - 1). Note that it is not possible to move from cells in the last row.

Each possible move has a cost given by a 0-indexed 2D array moveCost of size (m * n) x n, where moveCost[i][j] is the cost of moving from a cell with value i to a cell in column j of the next row. The cost of moving from cells in the last row of grid can be ignored.

The cost of a path in grid is the sum of all values of cells visited plus the sum of costs of all the moves made. Return the minimum cost of a path that starts from any cell in the first row and ends at any cell in the last row.

Example 1:

Input: grid = [[5,3],[4,0],[2,1]], moveCost = [[9,8],[1,5],[10,12],[18,6],[2,4],[14,3]]
Output: 17
Explanation: The path with the minimum possible cost is the path 5 -> 0 -> 1.
- The sum of the values of cells visited is 5 + 0 + 1 = 6.
- The cost of moving from 5 to 0 is 3.
- The cost of moving from 0 to 1 is 8.
So the total cost of the path is 6 + 3 + 8 = 17.

Example 2:

Input: 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]]
Output: 6
Explanation: The path with the minimum possible cost is the path 2 -> 3.
- The sum of the values of cells visited is 2 + 3 = 5.
- The cost of moving from 2 to 3 is 1.
So the total cost of this path is 5 + 1 = 6.

Note:

m == grid.length
n == grid[i].length
2 <= m, n <= 50
grid consists of distinct integers from 0 to m * n - 1.
moveCost.length == m * n
moveCost[i].length == n
1 <= moveCost[i][j] <= 100

解析

根據題意，給定一個 0 索引的 m x n 整數矩陣 grid ，由 0 到 m * n - 1 的不同整陣列成。您可以在此矩陣中從一個單元格移動到下一行的任何其他單元格。 需要注意的是，到最後一行的單元格停止移動。 每個可能的移動都有一個由大小為 (m * n) x n 的 0 索引 2D 陣列 moveCost 給出的代價，其中 moveCost[i][j] 是從值為 i 的單元格移動到第 j 列中的單元格的代價。網格中路徑的成本是訪問的所有單元格值的總和加上所有移動的成本總和。 返回從第一行的任何單元格開始到最後一行的任何單元格結束的路徑的最小代價。

解答

class Solution(object):
    def minPathCost(self, grid, moveCost):
        """
        :type grid: List[List[int]]
        :type moveCost: List[List[int]]
        :rtype: int
        """
        M = len(grid)
        N = len(grid[0])
        dp = [[float('inf')] * N for _ in range(M)]
        dp[0] = grid[0]
        for i in range(1, M):
            for j in range(N):
                for k in range(N):
                    dp[i][j] = min(dp[i][j], grid[i][j] + dp[i-1][k] + moveCost[grid[i-1][k]][j])
        return min(dp[-1])

執行結果

34 / 34 test cases passed.
Status: Accepted
Runtime: 2733 ms
Memory Usage: 18.2 MB

原題連結

http://leetcode.com/contest/weekly-contest-297/problems/minimum-path-cost-in-a-grid/

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