<p>You are given an array <code>points</code> representing integer coordinates of some points on a 2D-plane, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>The cost of connecting two points <code>[x<sub>i</sub>, y<sub>i</sub>]</code> and <code>[x<sub>j</sub>, y<sub>j</sub>]</code> is the <strong>manhattan distance</strong> between them: <code>|x<sub>i</sub> - x<sub>j</sub>| + |y<sub>i</sub> - y<sub>j</sub>|</code>, where <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

<p>Return <em>the minimum cost to make all points connected.</em> All points are connected if there is <strong>exactly one</strong> simple path between any two points.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/d.png" style="width: 214px; height: 268px;" />
<pre>
<strong>Input:</strong> points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
<strong>Output:</strong> 20
<strong>Explanation:</strong> 
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/c.png" style="width: 214px; height: 268px;" />
We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> points = [[3,12],[-2,5],[-4,1]]
<strong>Output:</strong> 18
</pre>

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

<ul>
	<li><code>1 &lt;= points.length &lt;= 1000</code></li>
	<li><code>-10<sup>6</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
	<li>All pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are distinct.</li>
</ul>

class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) 
    {
        int total_price = 0;
        int n = points.size ();
        std::unordered_set<int> visited;
        std::unordered_set<int> not_visited;
        std::vector<std::vector <int>> dist (n, std::vector<int> (n));
  
        for (int p1 = 0; p1 < n; ++p1)
        {
            not_visited.insert (p1);
            for (int p2 = 0; p2 < n; ++p2)
                dist[p1][p2] = abs (points [p1] [0] - points[p2] [0]) + abs (points[p1] [1] - points [p2] [1]);
        }

        visited.insert (0);
        not_visited.erase (0);

        while (! not_visited.empty ())
        {
            int min_dest = INT_MAX;
            int winner = -1;

            for (auto const& from: visited)
            {
                for (auto const& to : not_visited)
                {
                    if (dist [from][to] < min_dest)
                    {
                         min_dest = dist [from][to];
                         winner = to;
                    }
                }
            }

            total_price += min_dest;
            visited.insert (winner);
            not_visited.erase (winner);
        }

        return total_price;
    }
};
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