% VRP - 基于IVY算法的TWVRP最短距离求解
% 数据准备
% 假设有一组客户点的坐标和对应的时间窗信息
% 假设数据已经存储在 coordinates、timeWindows 和 demands 变量中
% 参数设置
numCustomers = size(coordinates, 1); % 客户点数量
vehicleCapacity = 100; % 车辆容量
numVehicles = 5; % 车辆数量
% 构建距离矩阵
distanceMatrix = zeros(numCustomers+1, numCustomers+1); % +1是为了包含仓库点
for i = 1:numCustomers+1
for j = 1:numCustomers+1
distanceMatrix(i, j) = pdist([coordinates(i, 😃; coordinates(j, 😃], ‘euclidean’);
end
end
% IVY算法求解
bestDistance = Inf;
bestSolution = [];
for iter = 1:100 % 迭代次数
% 随机生成初始解
solution = cell(numVehicles, 1);
for k = 1:numVehicles
% 随机选择一个客户点作为起始点
startNode = randi(numCustomers) + 1; % +1是为了排除仓库点
% 初始化路径
path = [1, startNode, 1]; % 1代表仓库点
% 计算剩余容量
remainingCapacity = vehicleCapacity - demands(startNode);
% 随机构建路径
while remainingCapacity > 0
validNodes = setdiff(2:numCustomers+1, path); % 排除已经访问过的点
validDemands = demands(validNodes);
feasibleNodes = validNodes(validDemands <= remainingCapacity);
if isempty(feasibleNodes)
break;
end
nextNode = randsample(feasibleNodes, 1);
path = [path, nextNode];
remainingCapacity = remainingCapacity - demands(nextNode);
end
solution{k} = path;
end
% 评估解的距离
totalDistance = 0;
for k = 1:numVehiclespath = solution{k};for i = 1:length(path)-1totalDistance = totalDistance + distanceMatrix(path(i), path(i+1));end
end% 更新最优解
if totalDistance < bestDistancebestDistance = totalDistance;bestSolution = solution;
end
end
% 输出最优解距离和路径
disp(‘Best Distance:’);
disp(bestDistance);
disp(‘Best Solution:’);
for k = 1:numVehicles
disp(bestSolution{k});
end
