leetcode 1219. Path with Maximum Gold

In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.

Return the maximum amount of gold you can collect under the conditions:

• Every time you are located in a cell you will collect all the gold in that cell.
• From your position you can walk one step to the left, right, up or down.
• You can't visit the same cell more than once.
• Never visit a cell with 0 gold.
• You can start and stop collecting gold from any position in the grid that has some gold.

Example 1:

Input: grid = [[0,6,0],[5,8,7],[0,9,0]]
Output: 24
Explanation:
[[0,6,0],
 [5,8,7],
 [0,9,0]]
Path to get the maximum gold, 9 -> 8 -> 7.

Example 2:

Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]]
Output: 28
Explanation:
[[1,0,7],
 [2,0,6],
 [3,4,5],
 [0,3,0],
 [9,0,20]]
Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.

Constraints:

• 1 <= grid.length, grid[i].length <= 15
• 0 <= grid[i][j] <= 100
• There are at most 25 cells containing gold.

思路：常规题（回溯法）

class Solution {
private:
    //int dx[4] = {-1, 0, 1, 0};
    int d[5] = {0, 1, 0, -1, 0};
public:
    int getMaximumGold(vector<vector<int>>& grid) {
        int m = grid.size();
        if (m == 0)
            return 0;
        int n = grid[0].size();
        int maxGold = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] > 0) {
                    maxGold = max(maxGold, dfs(grid, i, j));
                }
            }
        }
        return maxGold;
    }
    int dfs(vector<vector<int>> &grid, int i, int j) {
        int result = 0;
        int temp = grid[i][j];
        grid[i][j] = 0; //将grid[i][j]设为0，使得深层的dfs（同一条路径）不会再遍历这个点
        for (int k = 0; k < 4; ++k) { //遍历四个方向
            int x = i + d[k], y = j + d[k + 1];
            if (x >= 0 && x < grid.size() && y >= 0 && y <  grid[0].size() && grid[x][y] != 0) {
                result = max(result, dfs(grid, x, y));
            }
        }
        grid[i][j] = temp; //将grid[i][j]恢复，使得不同路径还能访问这个点
        return result + grid[i][j];
    }
};