Question

我正在尝试查找从0、0到N，N矩阵总和最小的路径。我从技术上解决了它，但这并不能扩展。它可以在较小的N x N矩阵上正常工作，但是它会在我正在为其构建的80x80矩阵上放下床，因为它具有（2 * 80）！ /（80！* 80！）可能的路径。正在考虑添加另一个内核，但不确定如何在所有递归中使用（我几乎不了解如何添加处理器，所以我不想搞砸）。规则是您只能向下移动或移至右侧列。我将进行一些小调整，例如跟踪最小路径，因此我不会遍历所有路径以求和，然后找到最小路径。但是我怀疑这会对运行时产生影响。有谁知道找到此最小路径的更有效方法？

有关该问题的更多详细信息，请访问https://projecteuler.net/problem=81。

def matrix_path_traverser():
    file = 'p081_matrix.txt'

    with open(file, 'r') as f:
        matrix = []

        for line in f:
            line = line.rstrip('\n')
            line = line.split(',')
            numerized_line = [int(x) for x in line]
            matrix.append(numerized_line)

    class Paths:
        paths = []
        sums = []

    def paths(matrix, row, column, n, path_object, prior_path):
        # Append the new matrix value to the path of previous values
        # Prior_path is a copy from the previous iteration. This is useful for bifurcating
        prior_path.append(matrix[row][column])

        # If the traverser hits the far right column, append all values directly below 
        if column == n - 1:
            for y in range(row + 1, n):
                prior_path.append(matrix[y][column])
            # Append the completed path to the list of possible paths
            path_object.paths.append(prior_path)
            return
        # If the traverser hits the bottom row, append all values directly to the right
        elif row == n - 1:
            for x in range(column + 1, n):
                prior_path.append(matrix[row][x])
            path_object.paths.append(prior_path)
            return
        else:
            # If the traverser isn't at the far right column or bottom row, make two moves. One to the right, one down
            paths(matrix, row, column + 1, n, path_object, prior_path.copy())
            paths(matrix, row + 1, column, n, path_object, prior_path.copy())

    # initialize Paths object to collect all paths that will be generated
    p = Paths()
    # Begin recursion
    paths(matrix, 0, 0, len(matrix), p, [])
    # Sum all paths
    p.sums = [sum(path) for path in p.paths]
    # Extract minimum
    return min(p.sums)

大规模遍历矩阵路径

0 个答案: