旅行商问题分支界限法Python源码实现

简介

旅行商问题(Traveling Salesman Problem,TSP)是指给定一个城市的列表和每对城市之间的距离,找到一条最短路径,使得每个城市都被访问且只访问一次,最后返回起始城市。分支界限法是一种解决TSP问题的算法,它通过将问题划分为多个子问题,并使用界限函数来确定搜索空间,从而提高求解效率。

流程图

下面是整个分支界限法求解TSP问题的流程图:

journey
    Start --> Initialize the queue
    Initialize the visited list
    Initialize the minimum cost
    Initialize the root node
    Add the root node to the queue
    WHILE queue is not empty
        Pop node from queue
        IF node is the leaf node
            Update the minimum cost and path
        ELSE
            FOR each unvisited child node
                Calculate the cost of the child node
                Add the child node to the queue
                Mark the child node as visited
    Print the minimum cost and path
    End

代码实现

下面是使用Python语言实现分支界限法求解TSP问题的源码:

# Import library
import numpy as np

# Define the cost matrix
cost_matrix = np.array([[0, 10, 15, 20],
                        [10, 0, 35, 25],
                        [15, 35, 0, 30],
                        [20, 25, 30, 0]])

# Define the class for Node
class Node:
    def __init__(self, parent, level, path, bound, cost):
        self.parent = parent
        self.level = level
        self.path = path
        self.bound = bound
        self.cost = cost

# Define the function to calculate the bound
def bound(node, cost_matrix):
    n = len(cost_matrix)
    path = node.path
    for i in range(n):
        if i not in path:
            min_cost = min(cost_matrix[i])
            node.bound += min_cost
            for j in range(n):
                cost_matrix[j][i] = float("inf")
    return node.bound

# Define the function to solve TSP using branch and bound
def tsp_branch_bound(cost_matrix):
    n = len(cost_matrix)
    queue = []
    visited = []
    minimum_cost = float("inf")
    root = Node(None, 0, [0], 0, 0)
    queue.append(root)
    while queue:
        node = queue.pop(0)
        if node.level == n - 1:
            node.path.append(0)
            if node.cost + cost_matrix[node.path[-2]][0] < minimum_cost:
                minimum_cost = node.cost + cost_matrix[node.path[-2]][0]
                optimal_path = node.path
        else:
            for i in range(1, n):
                if i not in node.path:
                    child = Node(node, node.level + 1, node.path + [i], node.bound, node.cost + cost_matrix[node.path[-1]][i])
                    child.bound = bound(child, cost_matrix)
                    if child.bound < minimum_cost:
                        queue.append(child)
        queue.sort(key=lambda x: x.bound)
    print("Minimum cost:", minimum_cost)
    print("Optimal path:", optimal_path)

# Call the function
tsp_branch_bound(cost_matrix)

上述代码首先导入了numpy库,并定义了代价矩阵cost_matrix。然后定义了Node类,用于表示每个节点的信息。接下来定义了bound函数,用于计算每个节点的界限。最后定义了tsp_branch_bound函数,用于求解TSP问题。

在主函数中,首先初始化队列、访问列表、最小代价和根节点。然后将根节点添加到队列中。接下来进入循环,直到队列为空。在每次循环中,弹出一个节点,如果该节点是叶子节点,则更新最小代价和路径。否则,对于每个未访问的子节点,计算代价并将其添加到队列中,并将其标记为已访问。最后打印出最小代价和路径。

代码解释

  • cost_matrix:代价矩阵,表示每对城市之间的距离。
  • Node类:用于表示每个节点的信息,包括父节点、级别、路径、界限和代价。
  • bound函数:用于计算每个节点的