我自己制作了一个Matlab脚本，用于解决应用分支定界算法的旅行商问题。
1）在我的例子中，我有4个城市（A，B，C，D）。我直接定义了距离矩阵。
2）在另一个文件（generate_distance_matrix）中，我编写了一些用于加载excel文件的代码（Test）。在那个excel文件中，我随机选择一些坐标（x1，x2）：
如何将这两个脚本组合起来？我如何以迭代的方式调整脚本？因为目前我只为4个城市编写了代码（TSP_test）。
我的脚本的工作步骤是：减少初始成本矩阵（行和列缩减）
评估下一个节点。例如（A => B）
设置M（A，B）= 0
将行A和列B设置为inf
设置M（B，A）= inf
评估成本值
对下一个节点重复这些步骤。
我希望有人能帮我解决问题。
如果我的剧本中的内容不明确，请询问。
在这里，您可以看到分支定界算法的脚本：
% Script for solving the TSP by using branch and bound
% Step 1: Write the initial cost matrix and reduce it (Row reduction and
% Column reduction
% Step 2: From the reduced matrix from step 1, M(A,B) = 0
% Set row A and column B to inf
% Set M(B,A) to inf
clear all
clc
% Rows from the inital cost matrix
% There are 4 cities (city A...city D)
A = [inf, 4, 12, 7];
B = [5, inf, inf, 18];
C = [11, inf, inf, 6];
D = [10, 2, 3, inf];
% Initial cost matrix representing the distances between the cities
% for example: the distance between city A and B is 4
M_initial_cost_matrix = [A; B; C; D];
% Reduce the initial cost matrix
row_reduction = min(M_initial_cost_matrix, [], 2);
row_reduction(isinf(row_reduction)) = 0;
M_row_reduced = M_initial_cost_matrix - row_reduction;
column_reduction = min(M_row_reduced);
M_reduced_1 = M_row_reduced - column_reduction;
M_reduced_1(isnan(M_reduced_1)) = inf;
% Evaluate the cost value
cost_1 = sum(row_reduction) + sum(column_reduction)
% Distance between city A and B (node 2)
M_reduced = M_reduced_1
M_reduced(1, :) = inf;
M_reduced(:, 2) = inf;
M_reduced(2, 1) = inf;
row_reduction = min(M_reduced, [], 2);
row_reduction(isinf(row_reduction)) = 0;
M_row_reduced = M_reduced - row_reduction;
column_reduction = min(M_row_reduced);
column_reduction(isinf(column_reduction)) = 0;
M_reduced_2 = M_row_reduced - column_reduction;
M_reduced_2(isnan(M_reduced_2)) = inf
cost_2 = cost_1 + sum(row_reduction) + sum(column_reduction) + M_reduced_1(1, 2)
% Distance between city A and C (node 3)
M_reduced = M_reduced_1
M_reduced(1, :) = inf;
M_reduced(:, 3) = inf;
M_reduced(3, 1) = inf;
row_reduction = min(M_reduced, [], 2);
row_reduction(isinf(row_reduction)) = 0;
M_row_reduced = M_reduced - row_reduction;
column_reduction = min(M_row_reduced);
column_reduction(isinf(column_reduction)) = 0;
M_reduced_3 = M_row_reduced - column_reduction;
M_reduced_3(isnan(M_reduced_3)) = inf;
cost_3 = cost_1 + sum(row_reduction) + sum(column_reduction) + M_reduced_1(1, 3)
% Distance between city A and D (node 4)
M_reduced = M_reduced_1
M_reduced(1, :) = inf;
M_reduced(:, 4) = inf;
M_reduced(4, 1) = inf;
row_reduction = min(M_reduced, [], 2);
row_reduction(isinf(row_reduction)) = 0;
M_row_reduced = M_reduced - row_reduction;
column_reduction = min(M_row_reduced);
column_reduction(isinf(column_reduction)) = 0;
M_reduced_4 = M_row_reduced - column_reduction;
M_reduced_4(isnan(M_reduced_4)) = inf;
cost_4 = cost_1 + sum(row_reduction) + sum(column_reduction) + M_reduced_1(1, 4)
% Lowest cost
lowest_cost = min([cost_2, cost_3, cost_4])
% Distance between city A, C, B (node 5)
M_reduced = M_reduced_3
M_reduced(3, :) = inf;
M_reduced(:, 2) = inf;
M_reduced(2, 3) = inf;
row_reduction = min(M_reduced, [], 2);
row_reduction(isinf(row_reduction)) = 0;
M_row_reduced = M_reduced - row_reduction;
column_reduction = min(M_row_reduced);
column_reduction(isinf(column_reduction)) = 0;
M_reduced_5 = M_row_reduced - column_reduction;
M_reduced_5(isnan(M_reduced_5)) = inf;
cost_5 = cost_3 + sum(row_reduction) + sum(column_reduction) + M_reduced_3(3, 2)
% Distance between city A, C, D (node 6)
M_reduced = M_reduced_3
M_reduced(3, :) = inf;
M_reduced(:, 4) = inf;
M_reduced(4, 3) = inf;
row_reduction = min(M_reduced, [], 2);
row_reduction(isinf(row_reduction)) = 0;
M_row_reduced = M_reduced - row_reduction;
column_reduction = min(M_row_reduced);
column_reduction(isinf(column_reduction)) = 0;
M_reduced_6 = M_row_reduced - column_reduction;
M_reduced_6(isnan(M_reduced_6)) = inf;
cost_6 = cost_3 + sum(row_reduction) + sum(column_reduction) + M_reduced_3(3, 4)
% Distance between city A, C, D, B (node 7)
M_reduced = M_reduced_3
M_reduced(4, :) = inf;
M_reduced(:, 2) = inf;
M_reduced(2, 4) = inf;
row_reduction = min(M_reduced, [], 2);
row_reduction(isinf(row_reduction)) = 0;
M_row_reduced = M_reduced - row_reduction;
column_reduction = min(M_row_reduced);
column_reduction(isinf(column_reduction)) = 0;
M_reduced_7 = M_row_reduced - column_reduction;
M_reduced_7(isnan(M_reduced_7)) = inf;
cost_7 = cost_3 + sum(row_reduction) + sum(column_reduction) + M_reduced_3(4, 2)
% Optimal path is: A, C, D, B, A with a cost of 25
这里是用于从名为“test”的excel文件生成距离矩阵的脚本。
% column 1: ID for city
% column 2: X coordinate
% column 3: Y coordinate
data = xlsread('Test.xlsx',1)
Dist = dist(data(:,2:3)')
Dist(Dist==0) = inf
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