<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 {
    struct cmp {
        bool operator()(pair<int, int>& a, pair<int, int>& b) {
            return a.second > b.second;
        }
    };
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        if (points.size() <= 1) {
            return 0;
        }  

        // Construct a graph using adjacent list ;
        int n = points.size();
        vector<vector<pair<int, int>>> graph(n, vector<pair<int, int>>());
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int d = getDistance(points[i], points[j]);
                graph[i].push_back({j, d});
                graph[j].push_back({i, d});
            }
        }

        //Using prim algorithm to find the MST
        priority_queue<pair<int, int>, vector<pair<int, int>>, cmp> pq;
        vector<int> dist(n, INT_MAX);
        unordered_set<int> visited;

        int result = 0;
        dist[0] = 0;
        pq.push({0, 0});
        while (!pq.empty()) {
            int cur = pq.top().first;
            int curDist = pq.top().second;
            pq.pop();

            if (visited.count(cur)) continue;
            visited.insert(cur);

            result += curDist;
            if (visited.size() == n) {
                return result;
            }

            for (auto next : graph[cur]) {
                int nextP = next.first;
                int nextDist = next.second;
                if (visited.count(nextP))  {
                    continue;
                }
                if (nextDist < dist[nextP]) {
                    dist[nextP] = nextDist;
                    pq.push({nextP, nextDist});
                }
            }
        }

        return result;
    }
    
private:
    int getDistance(vector<int> p1, vector<int> p2) {
        return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1]); 
    }

};
249095