输入:待排列的全部圆的半径R集合,圆的总数n 输出:最小圆排列距离Vmin,最优排列方案Path 算法: 1. 令Vmin为一个理论上绝对足够长的距离, Path=[] 2. 令ANode=zeros(n+1,n), ANode(1,:)=1:n, ANode(n+1,:)=1 (最后一

行记录X中已确定部分数量)

function [Vmin, Path] = minCircleArrange(R, n)
% 输入：R - 待排列的圆的半径集合
%       n - 圆的数量
% 输出：Vmin - 最小圆排列距离
%       Path - 最优排列方案

% 定义全局变量，用于记录当前最优解和最优解对应的排列方案
global Vmin Path
Vmin = inf;
Path = [];

% 初始化 ANode, AVal 和 ABound
ANode = zeros(n+1, n);
ANode(1, :) = 1:n;
ANode(n+1, :) = 1;
AVal = inf(1, n);
ABound = inf(1, n);
AVal(1) = 0;
ABound(1) = 0;

% 递归计算最小圆排列距离
minCircleArrangeRec(R, n, ANode, AVal, ABound, 1);

end

function minCircleArrangeRec(R, n, ANode, AVal, ABound, j)
% 递归计算最小圆排列距离

global Vmin Path

% 如果当前代价函数下界大于等于当前最优解，则直接返回
if ABound(j) >= Vmin
    return
end

% 如果当前已确定的圆的数量等于 n，则计算当前代价函数值，并更新最优解和最优解对应的排列方案
if ANode(n+1, j) == n
    d = calcDistance(R, ANode(1:n, j));
    if d < Vmin
        Vmin = d;
        Path = ANode(1:n, j);
    end
    return
end

% 选择 AVal 中最小值对应的节点进行扩展
[~, loc] = min(AVal);
X = ANode(1:n, loc);
k = ANode(n+1, loc);

% 从 ANode, AVal 和 ABound 中删除该节点
ANode(:, loc) = [];
AVal(loc) = [];
ABound(loc) = [];

% 对 X(k+1) 进行循环
for i = setdiff(1:n, X(1:k))
    % 如果 X(1:k+1) 为非叶结点，则计算代价函数下界
    if k+1 < n
        ABoundNew = calcBound(R, X(1:k+1));
        if ABoundNew >= Vmin
            continue
        end
    end
    
    % 将 [X; i] 加入 ANode，更新 AVal 和 ABound
    ANodeNew = [X; i];
    AValNew = AValFun(R, ANodeNew);
    ABoundNew = calcBound(R, ANodeNew);
    ANodeNew(n+1, :) = k+1;
    ANode = [ANode, ANodeNew];
    AVal = [AVal, AValNew];
    ABound = [ABound, ABoundNew];
    
    % 递归计算下一层节点
    minCircleArrangeRec(R, n, ANode, AVal, ABound, size(ANode, 2));
    
    % 从 ANode, AVal 和 ABound 中删除新增节点
    ANode(:, end) = [];
    AVal(end) = [];
    ABound(end) = [];
end

end

function dist = calcDistance(R, X)
% 计算圆排列距离

n = length(X);
dist = 0;
for i = 1:n-1
    for j = i+1:n
        dist = dist + 2*sqrt(R(X(i)) * R(X(j)));
    end
end

end

function val = AValFun(R, X)
% 计算节点对应的代价函数值

val = calcDistance(R, X);

end

function bound = calcBound(R, X)
% 计算节点对应的代价函数下界

n = length(X);
if n == 1
    bound = 0;
else
    bound = calcDistance(R, X(1:n-1));
    r = max(R(X(1:n-1)));
    for i = 1:n-1
        bound = bound - 2*sqrt(r*R(X(i)));
    end
    r = max(R(X));
    for i = 1:n-1
        bound = bound + 2*sqrt(r*R(X(i)));
    end
end

end

注意，由于该算法采用递归的方式进行搜索，可能会出现 Stack Overflow 的问题。如果出现这种情况，可以考虑使用迭代加深搜索或者其他避免 Stack Overflow 的方法。