<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 {  // O(E * logE)  // good for sparse graph
public:
	int minCostConnectPoints(vector<vector<int>>& A) {
		int n = A.size(), res = 0;
		if (n == 1) return 0;
		vector<int> u(n);
		vector<vector<int>> G(n, vector<int>(n, 1e7));
		priority_queue<vector<int>, vector<vector<int>>, greater<>> qu;
		for (int i = 0; i < n; ++i)
			for (int j = i + 1; j < n; ++j)
			{
				int d = abs(A[j][0] - A[i][0]) + abs(A[j][1] - A[i][1]);
				G[i][j] = G[j][i] = d;
			}
		qu.push({ 0, 0 });
		while (n)
		{
			auto &t = qu.top();
			int d = t[0], i = t[1];
			qu.pop();
			if (u[i]) continue;
			u[i] = 1;
			res += d;
			for (int j = 1; j < A.size(); ++j) if (!u[j])
			{
				int w = G[i][j];
				// if (w < v[j])
				{
					// v[j] = w;
					qu.push({ w, j });
				}
			}
			--n;
		}
		return res;
	}
};
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