【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