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）

 1 class Solution {
 2 private:
 3     //int dx[4] = {-1, 0, 1, 0};
 4     int d[5] = {0, 1, 0, -1, 0};
 5 public:
 6     int getMaximumGold(vector<vector<int>>& grid) {
 7         int m = grid.size();
 8         if (m == 0)
 9             return 0;
10         int n = grid[0].size();
11         int maxGold = 0;
12         for (int i = 0; i < m; ++i) {
13             for (int j = 0; j < n; ++j) {
14                 if (grid[i][j] > 0) {
15                     maxGold = max(maxGold, dfs(grid, i, j));
16                 }
17             }
18         }
19         return maxGold;
20     }
21     int dfs(vector<vector<int>> &grid, int i, int j) {
22         int result = 0;
23         int temp = grid[i][j];
24         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
25         for (int k = 0; k < 4; ++k) { //遍历四个方向
26             int x = i + d[k], y = j + d[k + 1];
27             if (x >= 0 && x < grid.size() && y >= 0 && y <  grid[0].size() && grid[x][y] != 0) {
28                 result = max(result, dfs(grid, x, y));
29             }
30         }
31         grid[i][j] = temp; //将grid[i][j]恢复，使得不同路径还能访问这个点
32         return result + grid[i][j];
33     }
34 };