leetcode 1584. Min Cost to Connect All Points（python）

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描述

You are given an array points representing integer coordinates of some points on a 2D-plane, where points[i] = [xi, yi].

Return the minimum cost to make all points connected. All points are connected if there is exactly one simple path between any two points.

Example 1:

Input: points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
Output: 20
Explanation: 
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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.
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Example 2:

Input: points = [[3,12],[-2,5],[-4,1]]
Output: 18
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Note:

1 <= points.length <= 1000
-10^6 <= xi, yi <= 10^6
All pairs (xi, yi) are distinct.
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解析

根据题意，给定一个数组 points 表示 2D 平面上某些点的整数坐标，其中 points[i] = [xi, yi]。连接两个点 [xi, yi] 和 [xj, yj] 的成本是它们之间的曼哈顿距离：|xi - xj| + |yi - yj|，其中 |val| 表示 val 的绝对值。返回使所有点连接的最小成本。

这道题其实考察的是 Kruskal 算法，因为要让平面中两个点之间的距离最短，且任意两点之间只有一条路径，最后的连接值尽可能小，能够满足任意两点之间有且仅有一条权重路径的只有最小生成树，而最小生成树有一个经典解法，就是 Kruskal 算法，这是一个贪心算法。

解答

class Solution(object):
    def minCostConnectPoints(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        dist = lambda x, y: abs(points[x][0] - points[y][0]) + abs(points[x][1] - points[y][1])

        n = len(points)
        dsu = DisjointSetUnion(n)
        edges = list()

        for i in range(n):
            for j in range(i + 1, n):
                edges.append((dist(i, j), i, j))
        
        edges.sort()
        
        ret, num = 0, 1
        for length, x, y in edges:
            if dsu.unionSet(x, y):
                ret += length
                num += 1
                if num == n:
                    break
        
        return ret

class DisjointSetUnion:
    def __init__(self, n):
        self.n = n
        self.rank = [1] * n
        self.f = list(range(n))
    
    def find(self, x):
        if self.f[x] == x:
            return x
        self.f[x] = self.find(self.f[x])
        return self.f[x]
    
    def unionSet(self, x, y):
        fx, fy = self.find(x), self.find(y)
        if fx == fy:
            return False

        if self.rank[fx] < self.rank[fy]:
            fx, fy = fy, fx
        
        self.rank[fx] += self.rank[fy]
        self.f[fy] = fx
        return True      
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运行结果

Runtime: 3026 ms, faster than 59.86% of Python online submissions for Min Cost to Connect All Points.
Memory Usage: 80.9 MB, less than 68.03% of Python online submissions for Min Cost to Connect All Points.
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原题链接

leetcode.com/problems/mi…

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