<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:
    def initPrim(self, points):
        self.distance = {}
        self.minheap = [[]]
        self.minheap[0] = [0, tuple(points[0])]
        self.edges = {}
        self.visited = {}
        self.numVisited = len(points)
        for i in range(len(points)):
            x = points[i]
            self.distance[tuple(x)] = float('inf')
            self.visited[tuple(x)] = False
            for j in range(i+1, len(points)):
                y = points[j]
                d = abs(x[0] - y[0]) + abs(x[1]-y[1])
                if tuple(x) in self.edges.keys():
                    self.edges[tuple(x)].append([d, tuple(y)])
                else:
                    self.edges[tuple(x)] = [[d, tuple(y)]]
                if tuple(y) in self.edges.keys():
                    self.edges[tuple(y)].append([d, tuple(x)])
                else:
                    self.edges[tuple(y)] = [[d, tuple(x)]]
        self.distance[tuple(points[0])] = 0
        return

    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        if len(points) ==1:
            return 0
        self.initPrim(points)
        ans = 0
        while self.minheap and self.numVisited != 0:
            dzero, x = heapq.heappop(self.minheap)
            if self.visited[x] == True:
                continue
            self.visited[x] = True
            self.numVisited -=1
            for e in self.edges[x]:
                d, y = e
                if d < self.distance[y] and self.visited[y] == False:
                    self.distance[y] = d
                    heapq.heappush(self.minheap, [self.distance[y], y])
        for i in self.distance:
            ans += self.distance[i]
        return ans
183764