MATLAB 优化医疗点和配送点位置

% 读取数据 pos = readtable('位置.xlsx'); link = readtable('连接道路.xlsx');

% 构建距离矩阵 n = size(pos,1); d = zeros(n); for i = 1:n for j = i+1:n idx = find((link.2==i & link.3==j) | (link.2==j & link.3==i)); if ~isempty(idx) d(i,j) = link.4(idx); d(j,i) = d(i,j); end end end

% 枚举所有可能的医疗点位置 min_S1 = inf; for i = 1:n-2 for j = i+1:n-1 for k = j+1:n % 计算各村庄到这三个医疗点的距离 dist = [d(:,i) d(:,j) d(:,k)]; [min_dist,idx] = min(dist,[],2); % 计算总距离S1 S1 = sum(min_dist); % 更新最小值 if S1 < min_S1 min_S1 = S1; ans_idx = [i j k]; ans_dist = dist; end end end end

% 枚举所有可能的配送点位置 min_S2 = inf; for i = 1:n-3 for j = i+1:n-2 for k = j+1:n-1 for l = k+1:n % 计算各医疗点到这四个配送点的距离 dist = [d(ans_idx(1),:) ; d(ans_idx(2),:) ; d(ans_idx(3),:) ; d(i,:) ; d(j,:) ; d(k,:) ; d(l,:)]; min_dist = min(dist,[],1); % 计算总距离S2 S2 = sum(min_dist); % 更新最小值 if S2 < min_S2 min_S2 = S2; ans_idx2 = [i j k l]; ans_dist2 = dist; end end end end end

% 输出结果 fprintf('问题2的答案为：\n'); fprintf('医疗点位置分别为：%d %d %d\n',ans_idx); fprintf('配送点位置分别为：%d %d %d %d\n',ans_idx2); fprintf('总距离S1为：%.1f\n',min_S1); fprintf('总距离S2为：%.1f\n',min_S2);

% 作图 figure; hold on; plot(pos.2,pos.3,'ko','MarkerSize',4); plot(pos.2(ans_idx),pos.3(ans_idx),'ro','MarkerSize',8,'LineWidth',2); plot(pos.2(ans_idx2),pos.3(ans_idx2),'bs','MarkerSize',8,'LineWidth',2); for i = 1:n [~,min_idx] = min(ans_dist(i,:)); switch min_idx case 1 plot([pos(i) pos.2(ans_idx(1))],[pos.3(i) pos.3(ans_idx(1))],'r-'); case 2 plot([pos.2(i) pos.2(ans_idx(2))],[pos.3(i) pos.3(ans_idx(2))],'r-'); case 3 plot([pos.2(i) pos.2(ans_idx(3))],[pos.3(i) pos.3(ans_idx(3))],'r-'); end [~,min_idx] = min(ans_dist2(:,i)); switch min_idx case 1 plot([pos(i) pos.2(ans_idx2(1))],[pos.3(i) pos.3(ans_idx2(1))],'b-'); case 2 plot([pos(i) pos.2(ans_idx2(2))],[pos.3(i) pos.3(ans_idx2(2))],'b-'); case 3 plot([pos(i) pos.2(ans_idx2(3))],[pos.3(i) pos.3(ans_idx2(3))],'b-'); case 4 plot([pos(i) pos.2(ans_idx2(4))],[pos.3(i) pos.3(ans_idx2(4))],'b-'); end end axis equal; title('问题2的结果');