ABC277 回想・解説(Python)
目次
前提
AtCoder Biginner Contest277のPythonの解答コードです。
- atcoder: https://atcoder.jp/contests/abc277
- github: https://github.com/hy-sksem/AtCoder/tree/master/ABC/abc277
A - ^-1
Xのインデックスを返す。
N, X = map(int, input().split())
P = list(map(int, input().split()))
for i, p in enumerate(P, start=1):
if p == X:
exit(print(i))
B - Playing Cards Validation
1文字目と2文字目のバリデーションをチェック。通るものをsetで管理。例外が発生したら、回答は"No"。
first = ("H", "D", "C", "S")
second = ("A", "2", "3", "4", "5", "6", "7", "8", "9", "T", "J", "Q", "K")
d = set()
N = int(input())
ans = "Yes"
for i in range(N):
s = input()
if s[0] in first and s[1] in second:
if s not in d:
d.add(s)
else:
ans = "No"
else:
ans = "No"
print(ans)
C - Ladder Takahashi
UnionFind
素直にUnionFindで実装しようとするとA,Bの制約が大きいのでTLEしてしまう。はしごのNはN<=2*105
なので、階数を別の数字に変換する。UFの初期化を2*105で済むので、実行時間に間に合う。変換するdictと変換した数字をもとに戻すdictを2つ用意して最終的には階数を返せるようにしておく。
from collections import defaultdict
class UnionFind:
"""Union Find木クラス
Attributes:
parent (list): 親要素を格納するリスト。親要素は自身のサイズを格納する。
Notes:
- parent[i] < 0 の場合、iは根。この時、-parent[i]はそのグループのサイズ
- parent[i] = -1 とすると、iは独立。
"""
def __init__(self, n):
"""
Args:
n (int): The number of objects. 要素数
"""
self.n = n
self.parents = [-1] * n
def find(self, x):
"""returns the root of x
Args:
x (int): The object to find the root of. 要素
Returns:
int: The root of x.
"""
if self.parents[x] < 0:
return x
else:
self.parents[x] = self.find(self.parents[x])
return self.parents[x]
def union(self, x, y):
"""unites the root of x and the root of y
Args:
x (int): The first object to union. 要素
y (int): The second object to union. 要素
"""
x = self.find(x)
y = self.find(y)
if x == y: # すでに同じ親
return
# union by size
if self.parents[x] > self.parents[y]:
x, y = y, x
self.parents[x] += self.parents[y]
self.parents[y] = x
def size(self, x):
"""returns the size of the group of x
Args:
x (int): The object to find the size of. 要素
Returns:
int: The size of the group of x.
"""
return -self.parents[self.find(x)]
def same(self, x, y):
"""returns True if the root of x is the same as the root of y
Args:
x (int): The first object to check. 要素
y (int): The second object to check. 要素
Returns:
bool: True if the root of x is the same as the root of y.
"""
return self.find(x) == self.find(y)
def members(self, x):
"""returns the members of the group of x
Args:
x (int): The object to find the members of. 要素
Returns:
list: The members of the group of x.
"""
root = self.find(x)
return [i for i in range(self.n) if self.find(i) == root]
def roots(self):
"""returns the roots
Returns:
list: The roots.
"""
return [i for i, x in enumerate(self.parents) if x < 0]
def group_count(self):
"""returns the number of groups
Returns:
int: The number of groups.
"""
return len(self.roots())
def all_group_members(self):
"""returns all groups
Returns:
list: All group members.
"""
return {r: self.members(r) for r in self.roots()}
N = int(input())
uf = UnionFind(2 * 10**5)
toInt = defaultdict(int)
toInt[1] = 1
toLadder = defaultdict(int)
toLadder[1] = 1
cnt = 2
for i in range(N):
a, b = map(int, input().split())
if toInt[a] == 0:
toInt[a] = cnt
toLadder[cnt] = a
cnt += 1
if toInt[b] == 0:
toInt[b] = cnt
toLadder[cnt] = b
cnt += 1
uf.union(toInt[a], toInt[b])
ans = 1
for i in uf.members(1):
ans = max(ans, toLadder[i])
print(ans)
BFS
はしごを辺、階数を頂点とみなして、BFSすると到達できる階数を走査できる。
from collections import defaultdict, deque
n = int(input())
d = defaultdict(list)
for _ in range(n):
a, b = map(int, input().split())
d[a].append(b)
d[b].append(a)
q = deque()
q.append(1)
S = {1}
while q:
v = q.popleft()
for i in d[v]:
if i not in S:
q.append(i)
S.add(i)
print(max(S))