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.
Constraints:
m == grid.length
n == grid[i].length
2 <= m, n <= 50
grid
consists of distinct integers from0
tom * n - 1
.moveCost.length == m * n
moveCost[i].length == n
1 <= moveCost[i][j] <= 100
Solution: DP
Let dp[i][j] := min cost to reach grid[i][j] from the first row.
dp[i][j] = min{grid[i][j] + dp[i – 1][k] + moveCost[grid[i – 1][k]][j]} 0 <= k < n
For each node, try all possible nodes from the previous row.
Time complexity: O(m*n2)
Space complexity: O(m*n) -> O(n)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
// Author: Huahua class Solution { public: int minPathCost(vector<vector<int>>& grid, vector<vector<int>>& moveCost) { const int m = grid.size(); const int n = grid[0].size(); vector<vector<int>> dp(m, vector<int>(n, INT_MAX)); for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) for (int k = 0; k < n; ++k) dp[i][j] = min(dp[i][j], grid[i][j] + (i ? dp[i - 1][k] + moveCost[grid[i - 1][k]][j] : 0)); return *min_element(begin(dp[m - 1]), end(dp[m - 1])); } }; |
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