文章目录
Prim Algorithm
Kruskal Algorithm
LeetCode1584. 连接所有点的最小费用
题目描述
Prim解法 - 这里是稠密图 用O(n^2) 解法
class Solution:
def minCostConnectPoints(self, points: List[List[int]]) -> int:
cost, n = 0, len(points)
vis = [0] * n
distance = [float('inf')] * n
distance[0] = 0
for _ in range(n):
node = -1
mid_cost = float('inf')
for i in range(n):
if not vis[i] and distance[i] < mid_cost:
mid_cost = distance[i]
node = i
cost += mid_cost
vis[node] = 1
for j in range(n):
if not vis[j]:
a = abs(points[node][0] - points[j][0])
b = abs(points[node][1] - points[j][1])
distance[j] = min(distance[j], a+b)
return cost
Kruskal解法
class Solution:
def minCostConnectPoints(self, points: List[List[int]]) -> int:
# if len(points) < 2: return 0
data_dict = {
} # (start, end): distance
for i in range(len(points)):
for j in range(i+1, len(points)):
a = abs(points[i][0] - points[j][0])
b = abs(points[i][1] - points[j][1])
data_dict[(i, j)] = a+b
data = sorted(data_dict.items(), key=lambda item: item[1])
tree = [_ for _ in range(len(points))]
def find_root(x):
if tree[x] == x: return x
else: return find_root(tree[x])
res = 0
for one, cost in data:
a, b = one
a_root = find_root(a)
b_root = find_root(b)
if a_root != b_root:
res += cost
tree[a_root] = b_root
return res