<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>

using dpair = pair<double, int>;
class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();

        vector<vector<int>> graph(n, vector<int>(n, 0));

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                graph[i][j] = distance(points[i], points[j]);
            }
        }

        int totalCost = 0;
        unordered_set<int> visited;
        priority_queue<dpair, vector<dpair>, greater<dpair>> pq;

        pq.push({0 /* dist */, 0});

        while (!pq.empty()) {
            auto candidate = pq.top();
            pq.pop();

            if (visited.count(candidate.second)) {
                continue;
            }
            visited.insert(candidate.second);
            totalCost += candidate.first;

            for (int i = 0; i < n; i++) {
                if (i == candidate.second) {
                    continue;
                }

                pq.push({graph[candidate.second][i], i});
            }
        }

        return totalCost;
    }

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