<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 Edge {
public:
    int cost, x, y;

    Edge (int a, int b, int c) {
        cost = c;
        x = a;
        y = b;
    } 
};

// Overload the < operator.
bool operator<(const Edge& edge1, const Edge& edge2) {
    return edge1.cost > edge2.cost;
}

int dist (vector<int> &a, vector<int> &b) {
    return abs(b[0]-a[0]) + abs(b[1]-a[1]);
}

class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        int cost = 0;
        int edgeCount = 0;
        vector<int> visited(n, false);

        int firstPoint = 0;
        visited[0] = true;
        priority_queue<Edge, vector<Edge>> pq;

        for (int i=1;i<n;i++) {
            pq.push(Edge(firstPoint, i, dist(points[firstPoint], points[i])));
        }

        while (!pq.empty() && edgeCount < n-1) {
            Edge curr = pq.top();
            pq.pop();

            if (visited[curr.y] == false) {
                visited[curr.y] = true;
                cost += curr.cost;
                edgeCount++;

                for (int i=0;i<n;i++) {
                    pq.push(Edge(curr.y, i, dist(points[curr.y], points[i])));
                }
            }
        }

        return cost;
    }
};
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