Compass, Map and Spyglass [ ]

最新推荐文章于 2024-06-25 15:52:15 发布

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本文介绍了一种基于坐标差计算迷宫中特定点到目的地最短路径的算法。通过计算各点与目标点坐标的最大差值，实现从海岸线('Y')出发，依次到达城堡('C')、市场('M')和村庄('S')的路径长度计算。

思路: (x, y)坐标之差的最大值为最短路径

def navigation(seaside):
    #replace this for solution
    column_1=len(seaside[0])
    row_1=len(seaside)
    for x in range(row_1):
        for y in range(column_1):
            if 'Y'== seaside[x][y]:yy=[x, y]
            if 'C'== seaside[x][y]:c=[x, y]
            if 'M'== seaside[x][y]:m=[x, y]
            if 'S'== seaside[x][y]:s=[x, y]


    return max(abs(c[0]-yy[0]), abs(c[1]-yy[1]))+max(abs(m[0]-yy[0]), abs(m[1]-yy[1]))+max(abs(s[0]-yy[0]), abs(s[1]-yy[1]))

if __name__ == '__main__':
    print("Example:")
    print(navigation([['Y', 0, 0, 0, 'C'],
                      [ 0,  0, 0, 0,  0],
                      [ 0,  0, 0, 0,  0],
                      ['M', 0, 0, 0, 'S']]))

    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert navigation([['Y', 0, 0, 0, 'C'],
                       [ 0,  0, 0, 0,  0],
                       [ 0,  0, 0, 0,  0],
                       ['M', 0, 0, 0, 'S']]) == 11

    assert navigation([[ 0,  0, 'C'],
                       [ 0, 'S', 0],
                       ['M','Y', 0]]) == 4

    assert navigation([[ 0,  0, 0,  0,  0,  0,  0],
                       [ 0,  0, 0, 'M', 0, 'S', 0],
                       [ 0,  0, 0,  0,  0,  0,  0],
                       [ 0,  0, 0, 'C', 0,  0,  0],
                       [ 0, 'Y',0,  0,  0,  0,  0],
                       [ 0,  0, 0,  0,  0,  0,  0]]) == 9
    print("Coding complete? Click 'Check' to earn cool rewards!")