写一个matlab求解设定起点,给出起点坐标,带容量限制的路径规划代码
时间: 2024-03-01 11:56:24 浏览: 23
下面是一个简单的Matlab代码,使用整数线性规划方法求解带容量限制的路径规划问题,其中起点为用户指定的点,终点为(10,10),节点数量为5,容量限制为每个节点的容量不能超过2,路径长度为各段长度之和:
```matlab
% 定义起点、终点和各节点的坐标
start = input('请输入起点坐标:');
endp = [10, 10];
nodes = [3, 1; 4, 4; 6, 2; 8, 5; 9, 8];
% 定义容量限制和路径长度
capacity = 2;
dist = pdist2([start; nodes; endp], [start; nodes; endp]);
% 定义决策变量
n = length(nodes) + 2;
x = intvar(n, n, 'binary');
% 定义目标函数
obj = sum(sum(x .* dist));
% 定义约束条件
constraints = [];
for i = 1:n
% 起点和终点的限制
if i == 1
constraints = [constraints, sum(x(i, :)) == 1];
elseif i == n
constraints = [constraints, sum(x(:, i)) == 1];
else
% 容量限制
constraints = [constraints, sum(x(i, :)) <= capacity];
constraints = [constraints, sum(x(:, i)) <= capacity];
% 入度和出度限制
constraints = [constraints, sum(x(i, :)) == sum(x(:, i))];
end
% 路径连通性
for j = 1:n
if j ~= i
constraints = [constraints, x(i, j) + x(j, i) <= 1];
end
end
end
% 求解
options = sdpsettings('verbose', 0);
sol = optimize(constraints, obj, options);
% 输出结果
if sol.problem == 0
path = [];
node = 1;
while node ~= n
path = [path, node];
node = find(value(x(node, :)) == 1);
end
path = [path, n];
disp(path);
else
disp('No feasible solution');
end
```
这段代码与上一个代码基本相同,只是在定义起点时使用了input函数让用户输入起点的坐标。