你想生产并售卖青铜器赚钱。你想不断选择部落作 为青铜器生产点每个生产点的初始投资费用为 2000 钱币建设时间需要 1 个季 度从下一个季度开始每季度每个生产点最多能制造出 100 件青铜器每件青铜 器的制造成本为 5 钱币。 你可以对青铜器统一定价 P一旦定了就不可改变且每个部落每季度的最 大需求量 Q 为： Q = CNP 其中N 为人口数 千P 为定价C 为常数 只要部落与部落之间有路每段

具体实现过程可以参考以下代码：

% 设置部落信息 tribes = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K'}; populations = [20, 30, 10, 25, 15, 35, 50, 40, 60, 30, 25]; roads = [0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0; 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0; 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0; 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0; 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0; 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0; 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0; 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1; 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0; 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1; 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0];

% 构建距离和配送成本矩阵 distances = zeros(length(tribes)); costs = zeros(length(tribes)); for i = 1:length(tribes) for j = i+1:length(tribes) if roads(i,j) == 1 distances(i,j) = 1; distances(j,i) = 1; if mod(i-1,3) ~= mod(j-1,3) % 不同国家之间的成本加1 costs(i,j) = 1; costs(j,i) = 1; end else distances(i,j) = Inf; distances(j,i) = Inf; costs(i,j) = Inf; costs(j,i) = Inf; end end end

% 步骤1：初始化投资 investments = zeros(length(tribes), 1); cash = 6000; k = 2; % 初始生产点个数

% 步骤2：求解线性规划问题 f = [distances(:); zeros(length(tribes), 1)]; Aeq = zeros(length(tribes), length(distances) + length(tribes)); beq = zeros(length(tribes), 1); lb = zeros(length(distances) + length(tribes), 1); ub = [Infones(length(distances), 1); 100ones(length(tribes), 1)]; for i = 1:length(tribes) for j = 1:length(tribes) if i == j Aeq(i, (i-1)*length(tribes)+j) = 1; else Aeq(i, (i-1)*length(tribes)+j) = -1; Aeq(i, length(distances)+i) = populations(j)/P; end end beq(i) = populations(i); end [x, fval] = intlinprog(f, 1:length(distances)+length(tribes), [], [], Aeq, beq, lb, ub);

% 步骤3：计算销售收入和运输成本 sales = zeros(k, 1); transport = zeros(length(tribes), k); for i = 1:length(tribes) for j = 1:k if x((i-1)*k+j) > 0 if mod(i-1,3) == mod(j-1,3) transport(i,j) = x((i-1)*k+j); else transport(i,j) = x((i-1)*k+j) + costs(i,j); end sales(j) = sales(j) + x((i-1)*k+j)*P; end end end cash = cash + sum(sales) - sum(transport(:));

% 步骤4：选择新增生产点的位置 new_tribe = 'L'; new_population = 50; new_demand = new_population/C; new_distances = distances; new_costs = costs; new_distances(:,end+1) = Inf; new_distances(end+1,:) = Inf; new_costs(:,end+1) = Inf; new_costs(end+1,:) = Inf;

% 计算连通分量 components = cell(1, k); for i = 1:k visited = false(length(tribes), 1); queue = []; component = []; queue(end+1) = i; while ~isempty(queue) node = queue(1); queue = queue(2:end); if ~visited(node) visited(node) = true; component(end+1) = node; for j = 1:length(tribes) if roads(node,j) == 1 && ~visited(j) queue(end+1) = j; end end end end components{i} = component; end

% 对于不连通的部分选择最低成本的新增生产点 unconnected = setdiff(1:length(tribes), cat(2, components{:})); [min_cost, min_index] = min(sum(new_distances(unconnected, k+1:k+length(components)), 1)); new_node = unconnected(min_index); new_distances(new_node, k+1:k+length(components)) = 1; new_distances(k+1:k+length(components), new_node) = 1; new_costs(new_node, k+1:k+length(components)) = 1; new_costs(k+1:k+length(components), new_node) = 1;

% 对于连通分量选择最佳新增生产点位置 for i = 1:length(components) component = components{i}; f = [new_distances(component, k+i); zeros(length(component), 1)]; Aeq = zeros(length(component), length(new_distances) + length(component)); beq = zeros(length(component), 1); lb = zeros(length(new_distances) + length(component), 1); ub = [Infones(length(new_distances), 1); 100ones(length(component), 1)]; for j = 1:length(component) Aeq(j, (component(j)-1)length(tribes)+k+i) = 1; beq(j) = new_demandlength(component); end [x, fval] = intlinprog(f, 1:length(new_distances)+length(component), [], [], Aeq, beq, lb, ub); for j = 1:length(component) new_distances(component(j), k+i) = x(j); new_costs(component(j), k+i) = mod(component(j)-1,3) == mod(k+i-1,3); end end

% 步骤5：求解线性规划问题 k = k+1; f = [new_distances(:); zeros(length(tribes), 1)]; Aeq = zeros(length(tribes), length(new_distances) + length(tribes)); beq = zeros(length(tribes), 1); lb = zeros(length(new_distances) + length(tribes), 1); ub = [Infones(length(new_distances), 1); 100ones(length(tribes), 1)]; for i = 1:length(tribes) for j = 1:length(tribes) if i == j Aeq(i, (i-1)*length(tribes)+j) = 1; else Aeq(i, (i-1)*length(tribes)+j) = -1; Aeq(i, length(new_distances)+i) = new_population/P; end end beq(i) = new_population; end [x, fval] = intlinprog(f, 1:length(new_distances)+length(tribes), [], [], Aeq, beq, lb, ub);

% 步骤6：重复步骤3-5，计算最终收益 for t = 1:3 % 步骤3：计算销售收入和运输成本 sales = zeros(k, 1); transport = zeros(length(tribes), k); for i = 1:length(tribes) for j = 1:k if x((i-1)*k+j) > 0 if mod(i-1,3) == mod(j-1,3) transport(i,j) = x((i-1)*k+j); else transport(i,j) = x((i-1)*k+j) + new_costs(i,j); end sales(j) = sales(j) + x((i-1)*k+j)*P; end end end cash = cash + sum(sales) - sum(transport(:));

% 步骤4：选择新增生产点的位置
new_tribe = ['L', num2str(t)];
new_population = 50 + t*10;
new_demand = new_population/C;
new_distances(:,end+1) = Inf;
new_distances(end+1,:) = Inf;
new_costs(:,end+1) = Inf;
new_costs(end+1,:) = Inf;

% 计算连通分