<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:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n=points.size();
        vector<vector<pair<int,int>>> neighbours(n);
        int min_weight=INT_MAX;
        pair<pair<int,int>,pair<int,int>> start;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(i==j) continue;
                int weight=abs(points[i][0]-points[j][0])+abs(points[i][1]-points[j][1]);
                neighbours[i].push_back({j,weight});
            }
        }

        vector<int> visited(n,0);

        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>> pq;
        for(auto neighbour: neighbours[0]){
            pq.push({neighbour.second,0,neighbour.first});
        }
        visited[0]=1;

        // vector<int> dist(n,INT_MAX);

        int ans=0;

        while(!pq.empty()){
            auto p = pq.top(); pq.pop();
            if(visited[p[2]]==1) continue;
            visited[p[2]]=1;
            ans+=p[0];
            for(auto neighbour: neighbours[p[2]]){
                if(visited[neighbour.first]==0) pq.push({neighbour.second,p[2],neighbour.first});
            }
        }
        return ans;



    }
};
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