SOLUTION INFO
Python · main.py
import sys
import heapq
def input():
return sys.stdin.readline().rstrip()
def find_parent(p):
if p_parents[p] != p:
p_parents[p] = find_parent(p_parents[p])
return p_parents[p]
def union_planet(p1, p2):
p1_parent = find_parent(p1)
p2_parent = find_parent(p2)
if p1_parent == p2_parent:
return False
if p1_parent < p2_parent:
p_parents[p2_parent] = p1_parent
else:
p_parents[p1_parent] = p2_parent
return True
N = int(input())
planets = []
p_parents = [i for i in range(N)]
edges = []
for i in range(N):
p_info = list(map(int, input().split()))
planets.append(p_info+[i])
for sort_n in range(3):
planets.sort(key=lambda x: x[sort_n])
for i in range(N - 1):
p1x, p1 = planets[i][sort_n], planets[i][3]
p2x, p2 = planets[i + 1][sort_n], planets[i + 1][3]
heapq.heappush(edges, [ abs(p1x - p2x), p1, p2 ])
answer = 0
cnt = 0
while edges:
cost, p1, p2 = heapq.heappop(edges)
if union_planet(p1, p2):
answer += cost
cnt += 1
if cnt == N - 1:
break
print(answer)
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