布线问题分支限界法python

根据提供的引用内容，这是一个布线问题的分支限界法的Python实现。布线问题是指在电路板上布置电线，使得电线的长度最短，且电线之间不会相互干扰。分支限界法是一种搜索算法，用于在所有可能的解中找到最优解。

import queue

class Node:
    def __init__(self, x, y, step, s):
        self.x = x
        self.y = y
        self.step = step
        self.s = s

    def __lt__(self, other):
        return self.step < other.step
def bfs():
    dx = [0, 0, 1, -1]
    dy = [1, -1, 0, 0]
    q = queue.PriorityQueue()
    q.put(Node(x1, y1, 0, str(x1) + ',' + str(y1)))
    while not q.empty():
        now = q.get()
        if now.x == x2 and now.y == y2:
            print("线路长：", now.step)
            print(now.s)
            return
        for i in range(4):
            nx = now.x + dx[i]
            ny = now.y + dy[i]
            if nx < 1 or nx > n or ny < 1 or ny > m or a[nx][ny] == 1:
                continue
            a[nx][ny] = 1
            q.put(Node(nx, ny, now.step + 1, now.s + '-->' + str(nx) + ',' + str(ny)))
    print("没有找到路径")

n, m = map(int, input("请输入方阵的行数和列数：").split())
a = [[0] * (m + 1) for _ in range(n + 1)]
print("请输入方阵，不能通过的地方输入1，能通过的地方输入0：")
for i in range(1, n + 1):
    a[i] = [0] + list(map(int, input().split()))
x1, y1 = map(int, input("请输入起点坐标：").split())
x2, y2 = map(int, input("请输入终点坐标：").split())
a[x1][y1] = 1
bfs()