<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:
    typedef pair<int,int> P;
    typedef pair<int,pair<int,int>> PP;
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();

        map<P, vector<PP>> ump;

        for (int i = 0; i<n; i++) {
            vector<int> currPt = points[i];
            for (int j = i+1; j<n; j++) {
                vector<int> otherPt = points[j];

                int mdist = abs(currPt[0] - otherPt[0]) + abs(currPt[1] - otherPt[1]);

                ump[{currPt[0], currPt[1]}].push_back({mdist, {otherPt[0], otherPt[1]}});
                ump[{otherPt[0], otherPt[1]}].push_back({mdist, {currPt[0], currPt[1]}});
            }
        }

        priority_queue<PP, vector<PP>, greater<PP>> pq;

        set<P> inMST;

        pq.push({0,{points[0][0],points[0][1]}});

        int sum = 0;

        while (!pq.empty()) {
            PP topElem = pq.top();
            pq.pop();
            int currDist = topElem.first;
            P node = topElem.second;

            if (inMST.find(node) != inMST.end()) continue;

            sum += currDist;
            inMST.insert(node);

            for (auto neigh: ump[node]) {
                int nDist = neigh.first;
                P nNode = neigh.second;

                if (inMST.find(nNode) == inMST.end()) {
                    pq.push({nDist,nNode});
                }
            }
        }

        return sum;
    }
};
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