<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 {
private:
    int getWeight(vector<int> p1, vector<int> p2) {
        return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1]);
    }

public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);

        //Prims minimum spanning tree algo

        // min heap {weight, {current node idx, parent node idx}}
        priority_queue<pair<int, pair<int, int>>,
                       vector<pair<int, pair<int, int>>>,
                       greater<pair<int, pair<int, int>>>> pq;
        
        int n = points.size();
        vector<bool> vis(n, false);

        int sum = 0;
        vector<pair<int, int>> mst;

        pq.push({0, {0, 0}}); //curr==parent for starting node

        while (!pq.empty()) {
            int curr = pq.top().second.first;
            int parent = pq.top().second.second;
            int wt = pq.top().first;
            pq.pop();

            if (vis[curr]) {
                continue;
            }

            vis[curr] = true;
            sum += wt;

            if (parent != curr)
                mst.push_back({parent, curr});

            // assume all nodes are connected to each other
            for (int k = 0; k < n; k++) {
                if (!vis[k]) {
                    pq.push({getWeight(points[curr], points[k]), {k, curr}});
                }
            }
        }

        for (auto it : mst) {
            cout << it.first << "->" << it.second << endl;
        }
        return sum;
    }
};
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