LeetCode #1219. Path with Maximum Gold

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本文介绍了一种在网格地图中寻找最大金矿路径的算法。该算法在m*n的网格中，每个单元格代表一定数量的黄金，目标是找到一条路径收集最多的黄金，遵循不重复访问同一单元格且只在非空单元格移动的规则。通过深度优先搜索(DFS)遍历所有可能路径，最终返回最大黄金总量。

题目描述：

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.

class Solution {
public:
    int getMaximumGold(vector<vector<int>>& grid) {
        int m=grid.size(), n=grid[0].size();
        unordered_set<int> visited;
        int result=0;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]>0)
                {
                    visited.insert(i*n+j);
                    DFS(grid,i,j,visited,grid[i][j],result);
                    visited.erase(i*n+j);
                }
            }
        }
        return result;
    }
    
    void DFS(vector<vector<int>>& grid, int i, int j, unordered_set<int>& visited, int cur, int& result)
    {
        int m=grid.size(), n=grid[0].size();
        result=max(cur,result);
        vector<pair<int,int>> dirs={{-1,0},{1,0},{0,-1},{0,1}};
        for(const pair<int,int>& dir:dirs)
        {
            int x=i+dir.first, y=j+dir.second;
            if(x<0||x>=m||y<0||y>=n) continue;
            if(visited.count(x*n+y)||grid[x][y]==0) continue;
            visited.insert(x*n+y);
            DFS(grid,x,y,visited,cur+grid[x][y],result);
            visited.erase(x*n+y);
        }
    }
};