#860418 (PyPy3) No.1605 Matrix Shape

• 問題: No.1605 Matrix Shape
• ユーザー: buey_t
• 提出日時: 2023-04-17 11:59:33
• 言語: PyPy3 (7.3.15)
• 結果: AC
• 実行時間: 603 ms / 2,000 ms
• コード長: 2,858 bytes
• コンパイル時間: 195 ms
• コンパイル使用メモリ: 82,412 KB
• 実行使用メモリ: 117,408 KB
• 最終ジャッジ日時: 2024-10-12 13:32:01
• 合計ジャッジ時間: 12,334 ms
• ジャッジサーバーID （参考情報）: judge2 / judge

ファイルパターン	結果
sample	AC * 3
other	AC * 34

ソースコード

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from math import sqrt,sin,cos,tan,ceil,radians,floor,gcd,exp,log,log10,log2,factorial,fsum
from heapq import heapify, heappop, heappush
from bisect import bisect_left, bisect_right
from copy import deepcopy
import copy
import random
from collections import deque,Counter,defaultdict
from itertools import permutations,combinations
from decimal import Decimal,ROUND_HALF_UP
#tmp = Decimal(mid).quantize(Decimal('0'), rounding=ROUND_HALF_UP)
from functools import lru_cache, reduce
#@lru_cache(maxsize=None)
from operator import add,sub,mul,xor,and_,or_,itemgetter
INF = 10**18
mod1 = 10**9+7
mod2 = 998244353
#DecimalならPython
#再帰ならPython!!!!!!!!!!!!!!!!!!!!!!!!!!
class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
        #親だけ
        self.edge = [0]*n
    def find(self, 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):
        x = self.find(x)
        y = self.find(y)
        if self.parents[x] > self.parents[y]:
            x, y = y, x
        
        self.edge[x] += 1
        
        if x == y:
            return
        self.parents[x] += self.parents[y]
        self.edge[x] += self.edge[y]
        self.parents[y] = x
    def size(self, x):
        return -self.parents[self.find(x)]
    def same(self, x, y):
        return self.find(x) == self.find(y)
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]
    def group_count(self):
        return len(self.roots())
    def all_group_members(self):
        group_members = defaultdict(list)
        for member in range(self.n):
            group_members[self.find(member)].append(member)
        return group_members
    def __str__(self):
        return '\n'.join(f'{r}: {m}' for r, m in self.all_group_members().items())
'''
グラフっぽく考えられそう
最初と終わりさえ決めれば間は関係ない
全部を通って終わりまでたどりつけるか？という問題
sccして1個なら全部行ける
1個じゃなくて、パスがあったらその組はいける
入次数0から始める
そこから全部通れるか
サイクルはSCCで対処可能
'''
N = int(input())
lim = 2*(10**5)+1
uf = UnionFind(lim)
memo = defaultdict(int)
for i in range(N):
    h,w = map(int, input().split())
    memo[h-1] += 1
    memo[w-1] -= 1
    uf.union(h,w)
M = len(memo.keys())
if uf.group_count() != lim-M+1:
    print(0)
    exit()
ls = list(memo.values())
ls.sort()
if ls[0] == ls[-1] == 0:
    print(M)
elif -ls[0] == ls[-1] == 1 and ls.count(0) == len(ls)-2:
    print(1)
else:
    print(0)
 
 
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