<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 DSU{
private:
public:
    vector<int> parents,clusters;
    DSU(int n){
        parents.resize(n);
        clusters.resize(n,1);
        for(int i=0;i<n;i++){
            parents[i]=i;
        }
    }
        int findUPar(int u){
            if(u==parents[u])return u;
            return parents[u]=findUPar(parents[u]);
        }
        bool unionBySize(int u,int v){
            int a=findUPar(u);
            int b=findUPar(v);
            if(a==b)return false;
            if(clusters[a]<clusters[b])swap(a,b);
            clusters[a]+=clusters[b];
            parents[b]=a;
            return true;
        }
};
class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        DSU* dsu = new DSU(n);
        vector<vector<int>> graph;
        for(int i=0; i<n ;i++){
            for(int j=i+1;j<n;j++){
                int d=abs(points[i][0]-points[j][0])+
                abs(points[i][1]-points[j][1]);
                graph.push_back({d,i,j});
            }
        }
        sort(graph.begin(),graph.end());
        int ans=0,cnt=0;
        for(auto &vec:graph){
            int wt=vec[0],u=vec[1],v=vec[2];
            if(dsu->unionBySize(u,v)){
                ans+=wt;
                cnt++;
            }
            if(cnt==n-1)break;
        }
        return ans;
    }
};
182035