結果

問題 No.1917 LCMST
ユーザー とりゐとりゐ
提出日時 2022-03-01 06:53:34
言語 PyPy3
(7.3.15)
結果
MLE  
実行時間 -
コード長 7,434 bytes
コンパイル時間 318 ms
コンパイル使用メモリ 82,176 KB
実行使用メモリ 851,328 KB
最終ジャッジ日時 2024-06-28 08:02:24
合計ジャッジ時間 9,500 ms
ジャッジサーバーID
(参考情報)
judge5 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 62 ms
72,320 KB
testcase_01 AC 63 ms
67,072 KB
testcase_02 AC 63 ms
67,072 KB
testcase_03 AC 213 ms
82,108 KB
testcase_04 AC 223 ms
82,132 KB
testcase_05 AC 233 ms
82,812 KB
testcase_06 AC 219 ms
82,568 KB
testcase_07 AC 247 ms
82,436 KB
testcase_08 AC 65 ms
67,072 KB
testcase_09 MLE -
testcase_10 -- -
testcase_11 -- -
testcase_12 -- -
testcase_13 -- -
testcase_14 -- -
testcase_15 -- -
testcase_16 -- -
testcase_17 -- -
testcase_18 -- -
testcase_19 -- -
testcase_20 -- -
testcase_21 -- -
testcase_22 -- -
testcase_23 -- -
testcase_24 -- -
testcase_25 -- -
testcase_26 -- -
testcase_27 -- -
testcase_28 -- -
testcase_29 -- -
testcase_30 -- -
testcase_31 -- -
testcase_32 -- -
testcase_33 -- -
testcase_34 -- -
testcase_35 -- -
testcase_36 -- -
testcase_37 -- -
testcase_38 -- -
testcase_39 -- -
testcase_40 -- -
testcase_41 -- -
testcase_42 -- -
testcase_43 -- -
testcase_44 -- -
権限があれば一括ダウンロードができます

ソースコード

diff #

import math


class FibonacciHeap:

    # internal node class
    class Node:
        def __init__(self, key, value):
            self.key = key
            self.value = value
            self.parent = self.child = self.left = self.right = None
            self.degree = 0
            self.mark = False

    # function to iterate through a doubly linked list
    def iterate(self, head):
        node = stop = head
        flag = False
        while True:
            if node == stop and flag is True:
                break
            elif node == stop:
                flag = True
            yield node
            node = node.right

    # pointer to the head and minimum node in the root list
    root_list, min_node = None, None

    # maintain total node count in full fibonacci heap
    total_nodes = 0

    # return min node in O(1) time
    def find_min(self):
        return self.min_node

    # extract (delete) the min node from the heap in O(log n) time
    # amortized cost analysis can be found here (http://bit.ly/1ow1Clm)
    def extract_min(self):
        z = self.min_node
        if z is not None:
            if z.child is not None:
                # attach child nodes to root list
                children = [x for x in self.iterate(z.child)]
                for i in range(0, len(children)):
                    self.merge_with_root_list(children[i])
                    children[i].parent = None
            self.remove_from_root_list(z)
            # set new min node in heap
            if z == z.right:
                self.min_node = self.root_list = None
            else:
                self.min_node = z.right
                self.consolidate()
            self.total_nodes -= 1
        return z

    # insert new node into the unordered root list in O(1) time
    # returns the node so that it can be used for decrease_key later
    def insert(self, key, value=None):
        n = self.Node(key, value)
        n.left = n.right = n
        self.merge_with_root_list(n)
        if self.min_node is None or n.key < self.min_node.key:
            self.min_node = n
        self.total_nodes += 1
        return n

    # modify the key of some node in the heap in O(1) time
    def decrease_key(self, x, k):
        if k > x.key:
            return None
        x.key = k
        y = x.parent
        if y is not None and x.key < y.key:
            self.cut(x, y)
            self.cascading_cut(y)
        if x.key < self.min_node.key:
            self.min_node = x

    # merge two fibonacci heaps in O(1) time by concatenating the root lists
    # the root of the new root list becomes equal to the first list and the second
    # list is simply appended to the end (then the proper min node is determined)
    def merge(self, h2):
        H = FibonacciHeap()
        H.root_list, H.min_node = self.root_list, self.min_node
        # fix pointers when merging the two heaps
        last = h2.root_list.left
        h2.root_list.left = H.root_list.left
        H.root_list.left.right = h2.root_list
        H.root_list.left = last
        H.root_list.left.right = H.root_list
        # update min node if needed
        if h2.min_node.key < H.min_node.key:
            H.min_node = h2.min_node
        # update total nodes
        H.total_nodes = self.total_nodes + h2.total_nodes
        return H

    # if a child node becomes smaller than its parent node we
    # cut this child node off and bring it up to the root list
    def cut(self, x, y):
        self.remove_from_child_list(y, x)
        y.degree -= 1
        self.merge_with_root_list(x)
        x.parent = None
        x.mark = False

    # cascading cut of parent node to obtain good time bounds
    def cascading_cut(self, y):
        z = y.parent
        if z is not None:
            if y.mark is False:
                y.mark = True
            else:
                self.cut(y, z)
                self.cascading_cut(z)

    # combine root nodes of equal degree to consolidate the heap
    # by creating a list of unordered binomial trees
    def consolidate(self):
        A = [None] * int(math.log(self.total_nodes) * 2)
        nodes = [w for w in self.iterate(self.root_list)]
        for w in range(0, len(nodes)):
            x = nodes[w]
            d = x.degree
            while A[d] != None:
                y = A[d]
                if x.key > y.key:
                    temp = x
                    x, y = y, temp
                self.heap_link(y, x)
                A[d] = None
                d += 1
            A[d] = x
        # find new min node - no need to reconstruct new root list below
        # because root list was iteratively changing as we were moving
        # nodes around in the above loop
        for i in range(0, len(A)):
            if A[i] is not None:
                if A[i].key < self.min_node.key:
                    self.min_node = A[i]

    # actual linking of one node to another in the root list
    # while also updating the child linked list
    def heap_link(self, y, x):
        self.remove_from_root_list(y)
        y.left = y.right = y
        self.merge_with_child_list(x, y)
        x.degree += 1
        y.parent = x
        y.mark = False

    # merge a node with the doubly linked root list
    def merge_with_root_list(self, node):
        if self.root_list is None:
            self.root_list = node
        else:
            node.right = self.root_list.right
            node.left = self.root_list
            self.root_list.right.left = node
            self.root_list.right = node

    # merge a node with the doubly linked child list of a root node
    def merge_with_child_list(self, parent, node):
        if parent.child is None:
            parent.child = node
        else:
            node.right = parent.child.right
            node.left = parent.child
            parent.child.right.left = node
            parent.child.right = node

    # remove a node from the doubly linked root list
    def remove_from_root_list(self, node):
        if node == self.root_list:
            self.root_list = node.right
        node.left.right = node.right
        node.right.left = node.left

    # remove a node from the doubly linked child list
    def remove_from_child_list(self, parent, node):
        if parent.child == parent.child.right:
            parent.child = None
        elif parent.child == node:
            parent.child = node.right
            node.right.parent = parent
        node.left.right = node.right
        node.right.left = node.left

from heapq import heappush, heappop, heapify
def MST(n,edge):
  used=[0]*n
  f=FibonacciHeap()
  for c,v in edge[0]:
    f.insert((c,v))
  used[0]=1
  ans=0
  tmp=1
  while tmp<n:
    m=f.extract_min()
    cost,v=m.key
    if used[v]:
      continue
    used[v]=1
    tmp+=1
    ans+=cost
    for c,w in edge[v]:
      if used[w]:
          continue
      f.insert((c,w))
  return ans

n=int(input())
a=list(map(int,input().split()))
m=10**5
d=[-1]*(m+1)
cnt=[0]*(m+1)

ans=0
for i in a:
  if cnt[i]==0:
    cnt[i]=1
  else:
    ans+=i


id=[-1]*(m+1)
size=0
for i in range(m+1):
  if cnt[i]:
    id[i]=size
    size+=1
edge=[[] for i in range(size)]
for i in range(1,m+1):
  for j in range(i,m+1,i):
    if cnt[j]==1:
      if d[i]==-1:
        d[i]=j
      else:
        id1=id[j]
        id2=id[d[i]]
        edge[id1].append((d[i]*j//i,id2))
        edge[id2].append((d[i]*j//i,id1))

ans+=MST(size,edge)
print(ans)
0