Question

我一直致力于骑士巡回演出模拟，而且我对如何将Arnd Roth改变应用于我的python程序感到难过。这里的程序片段是大部分计算，并通过电路板，前往最不可能移动的位置。我想要改变的是，如果有多个位置具有最少的移动次数（即它们都有2次移动），我想通过测量它们的位置和中心之间的距离并选择一个来打破它们之间的联系。离它最远的。我该怎么做呢？

    dx = [-2, -1, 1, 2, -2, -1, 1, 2]
dy = [1, 2, 2, 1, -1, -2, -2, -1]
# start the Knight from a random position
while tourTotal < tourMax:
    chessBoardX = chessX; chessBoardY = chessY # width and height of the chessboard
    chessBoard = [[0 for x in range(chessBoardX)] for y in range(chessBoardY)] # chessboard
# directions the Knight can move on the chessboard
    currentX = random.randint(0, chessBoardX - 1)
    currentY = random.randint(0, chessBoardY - 1)
    currentFailures = 0

    for k in range(chessBoardX * chessBoardY):
        chessBoard[currentY][currentX] = k + 1
        priorityQueue = [] # priority queue of available neighbors
        for i in range(8):
            newX = currentX + dx[i]; newY = currentY + dy[i]
            if newX >= 0 and newX < chessBoardX and newY >= 0 and newY < chessBoardY:
                if chessBoard[newY][newX] == 0:#if not visited
                    # count the available neighbors of the neighbor
                    counter = 0#counter is 0
                    for j in range(8):#max 8 moves
                        eX = newX + dx[j]; eY = newY + dy[j] #shows 1 move
                        if eX >= 0 and eX < chessBoardX and eY >= 0 and eY < chessBoardY:#if move is in parameters
                            if chessBoard[eY][eX] == 0: counter += 1 #if move is not visited, count is added
                    heappush(priorityQueue, (counter, i))#the amount of moves is pushed, along with the corresponding direction value
        # move to the neighbor that has min number of available neighbors
        if len(priorityQueue) > 0:
            (p, m) = heappop(priorityQueue)
            currentX += dx[m]; currentY += dy[m]
        else: break

Answer 1

位置（x，y）与nx*ny矩形板中心（基于0的索引）的距离{Manhattan}只是{{ 1}}。

如果你想要Euclidean距离，它将非常相似。

寻找离棋盘中心最远的移动 - 蟒蛇2

1 个答案: