<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:
    // Djikstra's algorithm
    // Node weights are manhattan distance to weights
    // Visit all nodes. All nodes are neighbours of all nodes.
    // Track visited. Unlike djikstra, we should not care about the dists table

    typedef pair<int,int> tup;

    inline int dist(vector<vector<int>>& points, int fromIndx, int toIndx) {
        vector<int> from = points.at(fromIndx);
        vector<int> to = points.at(toIndx);

        return abs(from.at(0) - to.at(0)) + abs(from.at(1) - to.at(1));
    }

    int minCostConnectPoints(vector<vector<int>>& points) {
        if (points.size() <= 1) return 0;

        // Track visited nodes. Once this reaches n points, we have visited every possible node.
        unordered_set<int> vis;

        // Min heap (cost, fromIndx)
        priority_queue<tup, vector<tup>, greater<tup>> pq;

        // Running cost.
        int totalCost = 0;

        // Push first point.
        pq.push({0, 0});
        while (!pq.empty()) {
            auto [cost, fromIndx] = pq.top();
            pq.pop(); 

            // Check visited.
            if (vis.contains(fromIndx)) {
                continue;
            }
            vis.insert(fromIndx);

            // Increment running cost. Consider this node as added to MST, since we are visiting now.
            totalCost += cost;

            // Visit all neighbours.
            for (int toIndx = 0; toIndx < points.size(); ++toIndx) {
                if (fromIndx == toIndx) {
                    // Ignore i == itself
                    // This is pedantic. The check from before should also check this.
                    continue;
                }

                int newCost = dist(points, fromIndx, toIndx);
                pq.push({newCost, toIndx});
            }
        }

        return totalCost;
    }
};
359265