<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>

// OJ: https://leetcode.com/contest/weekly-contest-197/problems/best-position-for-a-service-centre/
// Author: github.com/lzl124631x
class Solution {
    double dist(vector<int> &a, vector<double> &b) {
        return sqrt(pow(a[0] - b[0], 2) + pow(a[1] - b[1], 2));
    }
    double all(vector<vector<int>> &A, vector<double> &p) {
        double ans = 0;
        for (auto &a : A) ans += dist(a, p);
        return ans;
    }
    const int dirs[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
public:
    double getMinDistSum(vector<vector<int>>& A) {
        double ans = DBL_MAX;
        vector<double> p(2, 0);
        double step = 100, eps = 1e-6;
        while (step > eps) {
            bool found = false;
            for (auto &dir : dirs) {
                vector<double> next = {p[0] + step * dir[0], p[1] + step * dir[1]};
                double d = all(A, next);
                if (d < ans) {
                    ans = d;
                    p = next;
                    found = true;
                    break;
                }
            }
            if (!found) step /= 2;
        }
        return ans;
    }
};
10500