<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) {
        set<pair<int, int>> st;      //min heap {dist, index}
        int n=points.size();
        
        vector<vector<int>> gph(n);
        for(int i=0; i<n; i++){
            for(int j=i+1; j<n; j++){
                gph[i].push_back(j);
                gph[j].push_back(i);    
            }
        }

        vector<int> mp(n, INT_MAX); //final answers
        vector<int> visited(n,0);
        st.insert({0, 0});
        mp[0]=0;
        int ans=0;
        int numVisited=0;

        while(numVisited<n){
            pair<int, int> t = *st.begin();  st.erase(st.begin());
            int u = t.second;

            if(visited[u]==1)   continue;
            
            visited[u]=1;
            numVisited++;
            ans+=mp[u];

            for(auto v : gph[u]){   //iterate adjacents of u
                int dist = abs(points[u][0]-points[v][0]) + abs(points[u][1]-points[v][1]);
                if(visited[v]==0){
                    if(mp[v] > dist){
                        mp[v] = dist;
                        st.insert({dist, v});
                    }
                }
            }
        }
        return numVisited==n ? ans : -1;
    }
};
71865