1584. Min Cost to Connect All Points刷题笔记

用prim算法暴力求解，可以出来但是超时了

cost = 0
        clique = [0]
        
        while len(clique)<len(points):
            distance = float('inf')
            toadd = 0            
            
            for i in range(1,len(points)):
                if i in clique:
                    continue


                for j in clique:
                    temp_dis = abs(points[i][0]-points[j][0])+abs(points[i][1]-points[j][1])
                    if temp_dis < distance:
                        toadd = i
                        distance = temp_dis
            cost += distance
            clique.append(toadd)
        return cost

Prim加上用堆排序的算法

class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        manhattan = lambda p1, p2: abs(p1[0]-p2[0]) + abs(p1[1]-p2[1])
        n, c = len(points), collections.defaultdict(list)
        for i in range(n):
            for j in range(i+1, n):
                d = manhattan(points[i], points[j])
                c[i].append((d, j))
                c[j].append((d, i))
        cnt, ans, visited, heap = 1, 0, [0] * n, c[0]
        visited[0] = 1
        heapq.heapify(heap)
        while heap:
            d, j = heapq.heappop(heap)
            if not visited[j]:
                visited[j], cnt, ans = 1, cnt+1, ans+d
                for record in c[j]: heapq.heappush(heap, record)
            if cnt >= n: break        
        return ans