leetcode 1368. Minimum Cost to Make at Least One Valid Path in a Grid

Given a m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be:

• 1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1])
• 2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1])
• 3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j])
• 4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j])

Notice that there could be some invalid signs on the cells of the grid which points outside the grid.

You will initially start at the upper left cell (0,0). A valid path in the grid is a path which starts from the upper left cell (0,0) and ends at the bottom-right cell (m - 1, n - 1) following the signs on the grid. The valid path doesn't have to be the shortest.

You can modify the sign on a cell with cost = 1. You can modify the sign on a cell one time only.

Return the minimum cost to make the grid have at least one valid path.

 

Example 1:

 

Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
Output: 3
Explanation: You will start at point (0, 0).
The path to (3, 3) is as follows. (0, 0) --> (0, 1) --> (0, 2) --> (0, 3) change the arrow to down with cost = 1 --> (1, 3) --> (1, 2) --> (1, 1) --> (1, 0) change the arrow to down with cost = 1 --> (2, 0) --> (2, 1) --> (2, 2) --> (2, 3) change the arrow to down with cost = 1 --> (3, 3)
The total cost = 3.

Example 2:

Input: grid = [[1,1,3],[3,2,2],[1,1,4]]
Output: 0
Explanation: You can follow the path from (0, 0) to (2, 2).

Example 3:

 

Input: grid = [[1,2],[4,3]]
Output: 1

Example 4:

Input: grid = [[2,2,2],[2,2,2]]
Output: 3

Example 5:

Input: grid = [[4]]
Output: 0

 

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 100

题目根据单元格内的指示牌行走。求最少更改几次指示牌可以从grid[0][0]走到grid[-1][-1](右下角)

简单题。DP，一轮一轮的计算更改step次能走到的所有点。代码如下：

class Solution(object):
    def minCost(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        row = len(grid)
        col = len(grid[0])
        status = [[False for _ in range(col)] for _ in range(row)]
        l = [[0,0]]
        step = 0
        while True:
            t = []
            for x,y in l: # 遍历所有可以走到的位置
                while not status[x][y]:
                    status[x][y] = True
                    d = grid[x][y]
                    t.append([x,y])
                    if d == 1 and y < col-1:
                        y += 1
                    elif d == 2 and y > 0:
                        y -= 1
                    elif d == 3 and x < row-1:
                        x += 1
                    elif d == 4 and x > 0:
                        x -= 1
            if status[-1][-1]:#没有走到
                return step
            step += 1 #增加一次更改的机会，将这一轮新增的点上下左右的点加入到下一轮起点列表中
            l = []
            for x,y in t:
                if x > 0:
                    l.append([x-1,y])
                if x < row-1:
                    l.append([x+1,y])
                if y > 0:
                    l.append([x,y-1])
                if y < col-1:
                    l.append([x,y+1])