如何跟踪矩阵中的路径？答案

【问题标题】：How do I trace a path in matrix?如何跟踪矩阵中的路径？
【发布时间】：2021-04-19 22:59:18
【问题描述】：

我有如下矩阵：

[[0,0,0,0,0,0,0,0,0,0,0,0,0,0],
 [0,0,0,0,0,0,1,0,0,1,1,1,1,0],
 [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
 [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
 [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
 [0,0,0,0,0,0,1,0,0,1,1,1,1,0],
 [0,0,0,0,0,0,1,0,0,1,0,0,0,0],
 [0,0,0,0,0,0,1,0,0,1,0,0,0,0],
 [0,0,0,0,0,0,1,1,1,1,1,1,1,0],
 [0,0,0,0,0,0,0,0,0,0,0,0,0,0]]

如何在矩阵中找到跟踪 1 的路径？正如您所看到的，矩阵的 1 代表类似于单词“U2”的东西，我想返回一条类似于它应该遵循的像素的路径来绘制它。如果它已经向前移动了，它可以向后移动一条路径，但它不应该跳一个 0，它应该总是在 1 上移动。

【问题讨论】：

在理解您的需求时遇到了一些麻烦。 1s不代表这样的路径吗？如果要绘制它，可以将背景转换为黑色，并将 1 转换为 RGB 中的白色。
您想要做的与做图形seed or flood fill 非常相似，因此您可以调整算法以产生您想要的结果。
您的问题是如果有超过 1 个路径可用，如何选择要走的路径。这使得它依赖于一些寻路方法。您可以将所有 0 设置为墙壁，然后使用 Bellman Ford method 进行扩散寻路。

标签： python arrays algorithm matrix path-finding

【解决方案1】：

如果您只想要一个包含数字 1 位置的列表列表，您可以迭代，您可以执行以下操作：

>>>block = [[0,0,0,0,0,0,0,0,0,0,0,0,0,0],
...         [0,0,0,0,0,0,1,0,0,1,1,1,1,0],
...         [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
...         [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
...         [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
...         [0,0,0,0,0,0,1,0,0,1,1,1,1,0],
...         [0,0,0,0,0,0,1,0,0,1,0,0,0,0],
...         [0,0,0,0,0,0,1,0,0,1,0,0,0,0],
...         [0,0,0,0,0,0,1,1,1,1,1,1,1,0],
...         [0,0,0,0,0,0,0,0,0,0,0,0,0,0]]

>>> answer = []
>>> for i in block:
...    answer.append([j for j, e in enumerate(i) if e == 1])

>>>answer
[[], [6, 9, 10, 11, 12], [6, 9, 12], [6, 9, 12], [6, 9, 12], [6, 9, 10, 11, 12], [6, 9], [6, 9], [6, 7, 8, 9, 10, 11, 12], []]

【讨论】：

【解决方案2】：

a = [[0,0,0,0,0,0,0,0,0,0,0,0,0,0],
     [0,0,0,0,0,0,1,0,0,1,1,1,1,0],
     [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
     [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
     [0,0,0,0,0,0,1,0,0,1,0,0,1,0],
     [0,0,0,0,0,0,1,0,0,1,1,1,1,0],
     [0,0,0,0,0,0,1,0,0,1,0,0,0,0],
     [0,0,0,0,0,0,1,0,0,1,0,0,0,0],
     [0,0,0,0,0,0,1,1,1,1,1,1,1,0],
     [0,0,0,0,0,0,0,0,0,0,0,0,0,0]]


class U2:

    def __init__(self, matrix):
        self.matrix = matrix
        self.solved = []

    def find_path(self):
        for line in self.matrix:
            comprehension = ["___" if item==1 else item=="x" for item in line]             
            self.solved.append(comprehension)
        return self.solved



a = U2(a)
solved = a.find_path()
for i in solved:
    print(i)

这样的？这是输出：

[False, False, False, False, False, False, False, False, False, False, False, False, False, False]
[False, False, False, False, False, False, '___', False, False, '___', '___', '___', '___', False]
[False, False, False, False, False, False, '___', False, False, '___', False, False, '___', False]
[False, False, False, False, False, False, '___', False, False, '___', False, False, '___', False]
[False, False, False, False, False, False, '___', False, False, '___', False, False, '___', False]
[False, False, False, False, False, False, '___', False, False, '___', '___', '___', '___', False]
[False, False, False, False, False, False, '___', False, False, '___', False, False, False, False]
[False, False, False, False, False, False, '___', False, False, '___', False, False, False, False]
[False, False, False, False, False, False, '___', '___', '___', '___', '___', '___', '___', False]
[False, False, False, False, False, False, False, False, False, False, False, False, False, False]

【讨论】：