多旅行商问题matlab代码
时间: 2023-05-23 09:06:25 浏览: 95
A:以下是一个使用动态规划解决多旅行商问题的 MATLAB 代码示例:
```matlab
% 输入城市坐标
cities = [0 0; 1 1; 2 1; 1 2; 2 2];
% 求解旅行商路径
path = solveTSP(cities);
% 输出路径
disp(path);
% 动态规划算法求解 TSP
function path = solveTSP(cities)
n = size(cities, 1);
dist = distMat(cities);
dp = zeros(2^n, n);
path = zeros(1, n);
% 初始化,从任意一点开始,记为第 1 个点
for i = 2 : n
dp(bitset(0, i-1), i) = dist(1, i);
end
% 动态规划求解
for s = 3 : n
C = combnk(2:n, s-1);
for i = 1 : size(C, 1)
S = C(i, :);
M = zeros(1, n);
for j = 1 : length(S)
state = bitset(0, S(j)-1);
for k = 1 : length(S)
if j ~= k
state = bitset(state, S(k)-1);
end
end
for k = 1 : n
if bitget(state, k-1) == 1
M(k) = M(k) + dist(k, S(j)) + dp(bitset(state, j-1), S(j));
end
end
end
for j = 1 : length(S)
state = bitset(0, S(j)-1);
for k = 1 : length(S)
if j ~= k
state = bitset(state, S(k)-1);
end
end
for k = 1 : n
if bitget(state, k-1) == 1
dp(bitset(state, j-1), k) = min(dp(bitset(state, j-1), k), M(k) - dist(k, S(j)) - dp(bitset(state, j-1), S(j)) + dist(S(j), k));
end
end
end
end
end
% 回到起点
state = 2^n-1;
[~, idx] = min(dp(state, :));
path(n) = idx;
for i = n-1 : -1: 1
state = bitset(state, path(i+1)-1, 0);
[~, idx] = min(dp(state, path(i+1)));
path(i) = idx;
end
end
% 计算城市距离矩阵
function dist = distMat(cities)
n = size(cities, 1);
dist = zeros(n, n);
for i = 1 : n
for j = i+1 : n
dist(i, j) = norm(cities(i, :) - cities(j, :));
dist(j, i) = dist(i, j);
end
end
end
```
此代码实现了一个使用了动态规划算法的多旅行商问题求解器。该算法通过计算所有子问题的最短路径长度,然后组合得出最终的旅行商路径解。
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