Question

我有一个矩阵：

GroupBy

1-障碍物

0-常规单元格

我想实现一种算法，用于找到两个选定像元之间的最短路径（如果不能对角线的话）。我尝试了A *算法，但未给出正确的结果：

from cmath import rect, phase
from math import radians, degrees

def meanAngle(deg):
    complexDegree = sum(rect(1, radians(d)) for d in deg) / len(deg)
    argument = phase(complexDegree)
    meanAngle = degrees(argument)
    return meanAngle

def meanTime(times):
    t = (time.split(':') for time in times)
    seconds = ((float(s) + int(m) * 60 + int(h) * 3600) 
               for h, m, s in t)
    day = 24 * 60 * 60
    toAngles = [s * 360. / day for s in seconds]
    meanAsAngle = meanAngle(toAngles)
    meanSeconds = meanAsAngle * day / 360.
    if meanSeconds < 0:
        meanSeconds += day
    h, m = divmod(meanSeconds, 3600)
    m, s = divmod(m, 60)
    return('%02i:%02i:%02i' % (h, m, s))

print(meanTime(["15:00:00", "21:00:00"]))
# 18:00:00
print(meanTime(["23:00:00", "01:00:00"]))
# 00:00:00

请告诉我如何用Python语言实现它，以便其正常工作。

Answer 1

以下是您的问题的BFS实现：https://ideone.com/tuBu3G 我们从感兴趣的起点开始排队，并在到达终点后停止。在迭代的每个步骤中，我们的目标都是探索新的未探索状态，并将该新点的距离设置为 1 +该点到其探索位置的距离。

from collections import deque


graph = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
         [0, 0, 0, 0, 1, 0, 0, 0, 0, 0]]

# To move left, right, up and down
delta_x = [-1, 1, 0, 0]
delta_y = [0, 0, 1, -1]

def valid(x, y):
    if x < 0 or x >= len(graph) or y < 0 or y >= len(graph[x]):
        return False
    return (graph[x][y] != 1)

def solve(start, end):
    Q = deque([start])
    dist = {start: 0}
    while len(Q):
        curPoint = Q.popleft()
        curDist = dist[curPoint]
        if curPoint == end:
            return curDist
        for dx, dy in zip(delta_x, delta_y):
            nextPoint = (curPoint[0] + dx, curPoint[1] + dy)
            if not valid(nextPoint[0], nextPoint[1]) or nextPoint in dist.keys():
                continue
            dist[nextPoint] = curDist + 1
            Q.append(nextPoint)

print(solve((0,0), (6,7)))

打印：＃13

Answer 2

这里是一个如何在python 3中实现BFS的示例：

import collections


    def breadth_first_search(graph, root): 
        visited, queue = set(), collections.deque([root])
        while queue: 
        vertex = queue.popleft()
        for neighbour in graph[vertex]: 
            if neighbour not in visited: 
                visited.add(neighbour) 
                queue.append(neighbour) 


if __name__ == '__main__':
    graph = [[0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
            [0, 0, 0, 0, 1, 0, 0, 0, 0, 0]]
    breadth_first_search(graph, 0)

我希望这能对您有所帮助，请让我知道它的进展！

查找两个选定单元格之间的最短路径（如果不能对角线移动）

2 个答案: