写出分治法求最小点对距离过程matlab

以下是使用分治法求解最小点对距离的MATLAB代码：

function [d, p1, p2] = minDist(points) % 递归基：如果只剩下一个或两个点，直接计算距离并返回 n = size(points, 1); if n == 1 d = inf; p1 = 1; p2 = 1; elseif n == 2 d = norm(points(1,:) - points(2,:)); p1 = 1; p2 = 2; else % 分治：将点集分成左右两个子集 mid = floor(n / 2); [d1, p1_l, p1_r] = minDist(points(1:mid,:)); [d2, p2_l, p2_r] = minDist(points(mid+1:end,:)); if d1 < d2 d = d1; p1 = p1_l; p2 = p1_r; else d = d2; p1 = p2_l + mid; p2 = p2_r + mid; end

% 计算跨越中线的最小距离
x_mid = (points(mid,1) + points(mid+1,1)) / 2;
idx = (points(:,1) >= x_mid - d) & (points(:,1) <= x_mid + d);
strip = points(idx,:);
strip = sortrows(strip, 2);
m = size(strip, 1);
for i = 1:m-1
    for j = i+1:min(i+7,m)
        dist = norm(strip(i,:) - strip(j,:));
        if dist < d
            d = dist;
            p1 = find(ismember(points, strip(i,:), 'rows'));
            p2 = find(ismember(points, strip(j,:), 'rows'));
        end
    end
end

end

end

其中，points是一个n*2的矩阵，每一行表示一个点的坐标；d表示最小的点对距离；p1和p2表示最小距离的两个点在points中的索引。