<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) {
        map<int,pair<int,int>> m;
        for(int i=0;i<points.size();i++)
        m[i]={points[i][0],points[i][1]};

        vector<int> adj[points.size()];
        for(int i=0;i<points.size();i++){
            for(int j=0;j<points.size();j++)
            adj[i].push_back(j);
        }
        priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> q;
        long long sum=0;
        q.push({0,0});
        vector<int> vis(points.size());
        // vis[0]=1;
        while(!q.empty()){
            int b=q.top().first;
            int a=q.top().second;
            q.pop();
            if(vis[a]==1)
            continue;
            vis[a]=1;
            sum+=b;
            for(int i=0;i<adj[a].size();i++){
                if(vis[adj[a][i]]==0){
                    long long dist=abs(m[i].first-m[a].first)+abs(m[i].second-m[a].second);
                    q.push({dist,adj[a][i]});
                }
            }

        }
        return sum;
    }
};
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