<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>

#define MYPAIR pair<int,vector<int>>
class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        int total_cost=0;
        priority_queue< MYPAIR,vector<MYPAIR>, greater<MYPAIR> > edges; // cost,[p1 , p2] ,cost is sorted from small to large
        vector<set<int>> current_point;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                if(i != j)
                {
                    // add to queue
                    int dis = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
                    edges.push({dis,{i,j}}); // p1
                    // cout << i << " to "<< j <<",cost: "<< dis <<"\n";
                }
            }
        }
        
        // sort(edges.begin(),edges.end());
        // for(auto it:edges)
        // {
        //     cout << (it.second)[0] << " , "<<(it.second)[1] <<",cost: "<< it.first<<"\n";
        // }
        
        int pointCount = 0;
        while(pointCount < n-1)
        {
            int setOfp1 = -1,setOfp2 = -1;
            int p1 = (edges.top().second)[0],p2 = (edges.top().second)[1];
            for(int i=0;i < current_point.size() && (setOfp1 ==-1 || setOfp2 == -1);i++)
            {
                // traversal all sets
                if(current_point[i].find(p1) != current_point[i].end()) // find p1 in set
                {
                    setOfp1 = i;
                }
                if(current_point[i].find(p2) != current_point[i].end()) // find p2 in set
                {
                    setOfp2 = i;
                }
            }

            if(!((setOfp1 == setOfp2) && setOfp1 != -1))
            {
                // setOfp2 == setOfp1 but setOfp1,setOfp1 != -1 will be skip
                // add this edge
                total_cost += edges.top().first;
                pointCount++;
                // cout << p1 << ",set " << setOfp1 << " , "<< p2 <<",set " << setOfp2 <<" ,cost: "<< edges.top().first << "\n";
                if(setOfp1 == -1 && setOfp2 == -1)
                {
                    // neither of them are in any set
                    current_point.push_back({p1,p2});
                }
                else if(setOfp1 == -1)
                {
                    // p1 is new
                    current_point[setOfp2].insert(p1);
                }
                else if(setOfp2 == -1)
                {
                    // p2 is new
                    current_point[setOfp1].insert(p2);
                }
                else
                {
                    // p1 and p2 are in different set
                    // conbine setOfp1 and setOfp2
                    current_point[setOfp1].insert(current_point[setOfp2].begin(),current_point[setOfp2].end());
                    current_point.erase(current_point.begin() + setOfp2);
                }
            }
            edges.pop();
        }
        return total_cost;
        
    }
};
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