1219. 黄金矿工

class Solution:
    def getMaximumGold(self, grid: List[List[int]]) -> int:
        def dfs(i, j):
            if not (0 <= i < m and 0 <= j < n and grid[i][j]):
                return 0
            t = grid[i][j]
            grid[i][j] = 0
            ans = t + max(
                dfs(i + a, j + b) for a, b in [[0, 1], [0, -1], [-1, 0], [1, 0]]
            )
            grid[i][j] = t
            return ans

        m, n = len(grid), len(grid[0])
        return max(dfs(i, j) for i in range(m) for j in range(n))

class Solution {
    private int[][] grid;
    private int m;
    private int n;

    public int getMaximumGold(int[][] grid) {
        m = grid.length;
        n = grid[0].length;
        this.grid = grid;
        int ans = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                ans = Math.max(ans, dfs(i, j));
            }
        }
        return ans;
    }

    private int dfs(int i, int j) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
            return 0;
        }
        int t = grid[i][j];
        grid[i][j] = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        int ans = 0;
        for (int k = 0; k < 4; ++k) {
            ans = Math.max(ans, t + dfs(i + dirs[k], j + dirs[k + 1]));
        }
        grid[i][j] = t;
        return ans;
    }
}

class Solution {
public:
    vector<int> dirs = {-1, 0, 1, 0, -1};

    int getMaximumGold(vector<vector<int>>& grid) {
        int ans = 0;
        for (int i = 0; i < grid.size(); ++i)
            for (int j = 0; j < grid[0].size(); ++j)
                ans = max(ans, dfs(i, j, grid));
        return ans;
    }

    int dfs(int i, int j, vector<vector<int>>& grid) {
        if (i < 0 || i >= grid.size() || j < 0 || j >= grid[0].size() || grid[i][j] == 0) return 0;
        int t = grid[i][j];
        grid[i][j] = 0;
        int ans = 0;
        for (int k = 0; k < 4; ++k) ans = max(ans, t + dfs(i + dirs[k], j + dirs[k + 1], grid));
        grid[i][j] = t;
        return ans;
    }
};

func getMaximumGold(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	var dfs func(i, j int) int
	dfs = func(i, j int) int {
		if i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0 {
			return 0
		}
		t := grid[i][j]
		grid[i][j] = 0
		ans := 0
		dirs := []int{-1, 0, 1, 0, -1}
		for k := 0; k < 4; k++ {
			ans = max(ans, t+dfs(i+dirs[k], j+dirs[k+1]))
		}
		grid[i][j] = t
		return ans
	}
	ans := 0
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			ans = max(ans, dfs(i, j))
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

/**
 * @param {number[][]} grid
 * @return {number}
 */
var getMaximumGold = function (grid) {
    const m = grid.length;
    const n = grid[0].length;
    function dfs(i, j) {
        if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) {
            return 0;
        }
        const t = grid[i][j];
        grid[i][j] = 0;
        let ans = 0;
        const dirs = [-1, 0, 1, 0, -1];
        for (let k = 0; k < 4; ++k) {
            ans = Math.max(ans, t + dfs(i + dirs[k], j + dirs[k + 1]));
        }
        grid[i][j] = t;
        return ans;
    }
    let ans = 0;
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            ans = Math.max(ans, dfs(i, j));
        }
    }
    return ans;
};