【leetcode】1219. Path with Maximum Gold

题目如下：

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.

解题思路：DFS或者BFS都可以。本题主要是需要记录遍历过的节点，防止重复遍历陷入死循环。我的记录方法是利用整数的位操作，给grid中每个节点都分配一个序号，按从左往右从上往下的顺序，(0,0)是2^0，(0,1)是2^1次方，依次类推。

代码如下：

class Solution(object):
    def getMaximumGold(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        def getNumber(x,y):
            v = x*len(grid[0]) + y
            return 2**v
        res = 0
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if grid[i][j] == 0:continue
                count = grid[i][j]
                flag = 0
                queue = [(i,j,count,flag | getNumber(i,j))]
                direction = [(0,1),(0,-1),(1,0),(-1,0)]
                while len(queue) > 0:
                    x,y,count,flag = queue.pop(0)
                    res = max(res,count)
                    for (x1,y1) in direction:
                        if x1 + x >= 0 and x1+x < len(grid) and y+y1 >=0 and y+y1 < len(grid[0]) and grid[x+x1][y+y1] != 0 \
                            and flag & getNumber(x1+x,y1+y) == 0:
                            new_count = count + grid[x1+x][y1+y]
                            queue.append((x+x1,y+y1,new_count,flag | getNumber(x1+x,y1+y)))
        return res