<p>A delivery company wants to build a new service center in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new center in a position such that <strong>the sum of the euclidean distances to all customers is minimum</strong>.</p>

<p>Given an array <code>positions</code> where <code>positions[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> is the position of the <code>ith</code> customer on the map, return <em>the minimum sum of the euclidean distances</em> to all customers.</p>

<p>In other words, you need to choose the position of the service center <code>[x<sub>centre</sub>, y<sub>centre</sub>]</code> such that the following formula is minimized:</p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/06/25/q4_edited.jpg" />
<p>Answers within <code>10<sup>-5</sup></code> of the actual value will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/06/25/q4_e1.jpg" style="width: 377px; height: 362px;" />
<pre>
<strong>Input:</strong> positions = [[0,1],[1,0],[1,2],[2,1]]
<strong>Output:</strong> 4.00000
<strong>Explanation:</strong> As shown, you can see that choosing [x<sub>centre</sub>, y<sub>centre</sub>] = [1, 1] will make the distance to each customer = 1, the sum of all distances is 4 which is the minimum possible we can achieve.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/06/25/q4_e3.jpg" style="width: 419px; height: 419px;" />
<pre>
<strong>Input:</strong> positions = [[1,1],[3,3]]
<strong>Output:</strong> 2.82843
<strong>Explanation:</strong> The minimum possible sum of distances = sqrt(2) + sqrt(2) = 2.82843
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= positions.length &lt;= 50</code></li>
	<li><code>positions[i].length == 2</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 100</code></li>
</ul>

class Solution {
public:
    double dist(vector<vector<int>>& p, pair<double, double> x) {
        double res = 0;
        for (auto i : p) {
            double rsq = (i[0] - x.first) * (i[0] - x.first) + (i[1] - x.second) * (i[1] - x.second);
            res += sqrt(rsq);
        }
        return res;
    }
    double getMinDistSum(vector<vector<int>>& positions) {
        int runs = 64;
        int partition = 4;
        pair<double, double> BL = {DBL_MAX,DBL_MAX};
        pair<double, double> TR = {-1 * DBL_MAX, -1 * DBL_MAX};
        for (auto i : positions) {
            BL.first = min((double)(i[0]), BL.first);
            BL.second = min((double)(i[1]), BL.second);
            TR.first = max((double)(i[0]), TR.first);
            TR.second = max((double)(i[1]), TR.second);
        }

        for (int iter = 0; iter < runs; iter++) {
            pair<double, double> dim = {TR.first - BL.first, TR.second - BL.second};
            pair<double, double> center = BL;
            double md = DBL_MAX;
            for (int x = 0; x < partition; x++) {
                for (int y = 0; y < partition; y++) {
                    pair<double, double> test = {BL.first + (x + 0.5) * (dim.first / partition), BL.second + (y + 0.5) * (dim.second / partition)};
                    double td = dist(positions, test);
                    if (td < md) {
                        md = td;
                        center = test;
                    }
                }
            }

            BL = {center.first - 1.5 * (dim.first / partition), center.second - 1.5 * (dim.second / partition)};
            TR = {center.first + 1.5 * (dim.first / partition), center.second + 1.5 * (dim.second / partition)};

            // cout << center.first << " " << center.second << endl;
        }

        pair<double, double> res = {0.5 * (BL.first + TR.first), 0.5 * (BL.second + TR.second)};
        return dist(positions, res);
    }
};
80900