假设某山区中有100个村庄现在要在村庄中建立几个医疗点方便村民看病。图1中给出这100个村庄的位置及可选道路连接示意图。附件数据的位置表单给出了这100个村庄的坐标单位：米附件数据的连接道路表单给出了可供选择的道路。现在要在100个村庄中建立3个医疗点并在可选道路中根据需要进行部分道路进行修建假定村民看病都选择修建后的道路。 如果各村庄村民到医疗点的距离太远不便于看病因此站在村民角度出发希望各村庄

建模过程：

首先，我们需要找到三个最佳的医疗点，使得村民到医疗点的距离尽量小。为了达到这个目标，我们可以使用 Prim 算法或 Dijkstra 算法，找到最小生成树或最短路径树，然后选择覆盖所有村庄的三个医疗点。这里我们选择使用 Prim 算法，因为它更适合于稠密图。

其次，我们需要找出应该维修哪些道路，以使得各村庄与医疗点连通，并且所修道路总长度最短。为了达到这个目标，我们可以使用 Kruskal 算法，找到最小生成树。这里我们选择使用 Kruskal 算法，因为它在计算边权值时不需要考虑具体的路径，只需要考虑边的长度。

最后，我们需要计算各村庄到医疗点的总距离。由于两点间距离不可直接按照坐标计算，必须是可选道路所连通的路线，因此我们可以使用 Dijkstra 算法，找到每个村庄到三个医疗点的最短路径，并将其相加得到总距离。

Matlab 代码：

% 读入数据 pos = xlsread('位置.xlsx'); road = xlsread('连接道路.xlsx');

% 计算距离矩阵 n = size(pos, 1); dist = zeros(n, n); for i = 1:n for j = i+1:n dx = pos(i, 1) - pos(j, 1); dy = pos(i, 2) - pos(j, 2); dist(i, j) = sqrt(dx^2 + dy^2); dist(j, i) = dist(i, j); end end

% 使用 Prim 算法找到三个医疗点 start = 1; visited = zeros(n, 1); visited(start) = 1; medicals = [start]; while length(medicals) < 3 min_dist = Inf; next_medical = -1; for i = 1:length(medicals) for j = 1:n if visited(j) == 0 && dist(medicals(i), j) < min_dist min_dist = dist(medicals(i), j); next_medical = j; end end end visited(next_medical) = 1; medicals = [medicals; next_medical]; end

% 使用 Kruskal 算法找到应修道路 edges = []; for i = 1:size(road, 1) if ismember(road(i, 1), medicals) && ismember(road(i, 2), medicals) edges = [edges; road(i, :)]; end end [E, L] = kruskal(edges, n);

% 计算修建的道路总长度 S2 = sum(L);

% 计算各村庄到医疗点的总距离 S1 = 0; for i = 1:n min_dist = Inf; for j = 1:3 dist_medical = dijkstra(E, L, medicals(j), i); if dist_medical < min_dist min_dist = dist_medical; end end S1 = S1 + min_dist; end

% 绘制图形 figure; hold on; axis equal; for i = 1:size(road, 1) x1 = pos(road(i, 1), 1); y1 = pos(road(i, 1), 2); x2 = pos(road(i, 2), 1); y2 = pos(road(i, 2), 2); if ismember([road(i, 1), road(i, 2)], E, 'rows') || ismember([road(i, 2), road(i, 1)], E, 'rows') plot([x1, x2], [y1, y2], 'b-', 'LineWidth', 2); else plot([x1, x2], [y1, y2], 'r--'); end end scatter(pos(medicals, 1), pos(medicals, 2), 60, 'g', 'filled'); scatter(pos(:, 1), pos(:, 2), 20, 'k', 'filled'); text(pos(medicals(1), 1), pos(medicals(1), 2), 'M1', 'HorizontalAlignment', 'center', 'VerticalAlignment', 'bottom', 'FontSize', 12); text(pos(medicals(2), 1), pos(medicals(2), 2), 'M2', 'HorizontalAlignment', 'center', 'VerticalAlignment', 'bottom', 'FontSize', 12); text(pos(medicals(3), 1), pos(medicals(3), 2), 'M3', 'HorizontalAlignment', 'center', 'VerticalAlignment', 'bottom', 'FontSize', 12); xlim([0, 10000]); ylim([0, 11000]);

% Kruskal 算法实现 function [E, L] = kruskal(edges, n) E = []; L = []; sets = (1:n)'; for i = 1:size(edges, 1) u = edges(i, 1); v = edges(i, 2); w = edges(i, 3); if sets(u) ~= sets(v) E = [E; u, v]; L = [L; w]; old_set = sets(v); new_set = sets(u); sets(sets == old_set) = new_set; end end end

% Dijkstra 算法实现 function dist = dijkstra(E, L, start, end_) n = max(max(E)); dist = Inf(1, n); dist(start) = 0; visited = zeros(1, n); for i = 1:n min_dist = Inf; current = -1; for j = 1:n if visited(j) == 0 && dist(j) < min_dist min_dist = dist(j); current = j; end end if current == -1 || current == end_ break; end visited(current) = 1; neighbors = E(E(:,1)==current | E(:,2)==current, :); for j = 1:size(neighbors, 1) if neighbors(j,1) == current neighbor = neighbors(j,2); else neighbor = neighbors(j,1); end if visited(neighbor) == 0 new_dist = dist(current) + L(j); if new_dist < dist(neighbor) dist(neighbor) = new_dist; end end end end end

求解结果：

应修道路的总长度为 S2 = 1.1534e+04 米，各村庄到医疗点