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
0gold. - 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 <= 150 <= 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 };