結果

問題 No.274 The Wall
ユーザー prin_kemkemprin_kemkem
提出日時 2023-05-11 21:47:06
言語 PyPy3
(7.3.15)
結果
TLE  
実行時間 -
コード長 3,247 bytes
コンパイル時間 333 ms
コンパイル使用メモリ 82,048 KB
実行使用メモリ 795,228 KB
最終ジャッジ日時 2024-11-27 17:51:16
合計ジャッジ時間 12,345 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 21 TLE * 1
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 # 

from collections import defaultdict, deque, Counter
import copy
from itertools import combinations, permutations, product, accumulate, groupby
from heapq import heapify, heappop, heappush
import math
import bisect
from pprint import pprint
import sys
# sys.setrecursionlimit(700000)
input = lambda: sys.stdin.readline().rstrip('\n')
inf = float('inf')
mod1 = 10**9+7
mod2 = 998244353
def ceil_div(x, y): return -(-x//y)

#################################################

def sccd(adj, radj):
    n = len(adj)
    been = [False]*n
    P = []
    for s in range(n):
        if been[s]: continue
        stack = [s]
        while stack:
            now = stack.pop()
            if now >= 0:
                if been[now]: continue
                been[now] = True
                stack.append(~now)
                for nxt in adj[now]:
                    if been[nxt]: continue
                    stack.append(nxt)
            else:
                P.append(~now)
    scc_idx = [None]*n
    idx = 0
    for s in reversed(P):
        if not been[s]: continue
        been[s] = False
        stack = [s]
        while stack:
            now = stack.pop()
            scc_idx[now] = idx
            for nxt in radj[now]:
                if not been[nxt]: continue
                been[nxt] = False
                stack.append(nxt)
        idx += 1
    return scc_idx

class Two_SAT:
    def __init__(self, n) -> None:
        self.n = n
        self.adj = [[] for _ in range(2*n)]
        self.radj = [[] for _ in range(2*n)]
    def add(self, x, y): # 否定したいときは~xを入れる
        if x >= 0 and y >= 0:
            self.adj[self.n+x].append(y)
            self.radj[y].append(self.n+x)
            self.adj[self.n+y].append(x)
            self.radj[x].append(self.n+y)
        elif x >= 0:
            y = ~y
            self.adj[self.n+x].append(self.n+y)
            self.radj[self.n+y].append(self.n+x)
            self.adj[y].append(x)
            self.radj[x].append(y)
        elif y >= 0:
            x = ~x
            self.adj[x].append(y)
            self.radj[y].append(x)
            self.adj[self.n+y].append(self.n+x)
            self.radj[self.n+x].append(self.n+y)
        else:
            x = ~x; y = ~y
            self.adj[x].append(self.n+y)
            self.radj[self.n+y].append(x)
            self.adj[y].append(self.n+x)
            self.radj[self.n+x].append(y)
    def solve(self):
        scc_idx = sccd(self.adj, self.radj)
        for x in range(self.n):
            i, j = scc_idx[x], scc_idx[self.n+x]
            if i == j:
                return None
        return 0

def is_intersect(li, ri, lj, rj):
    if li > lj: li, ri, lj, rj = ri, rj, li, lj
    return lj <= ri

N, M = map(int, input().split())
blocks = [tuple(map(int, input().split())) for _ in range(N)]
sat = Two_SAT(N)
for i in range(N):
    li, ri = blocks[i]
    for j in range(i+1, N):
        lj, rj = blocks[j]
        if is_intersect(li, ri, lj, rj): sat.add(~i, ~j)
        if is_intersect(li, ri, M-1-rj, M-1-lj): sat.add(~i, j)
        if is_intersect(M-1-ri, M-1-li, lj, rj): sat.add(i, ~j)
        if is_intersect(M-1-ri, M-1-li, M-1-rj, M-1-lj): sat.add(i, j)
del blocks
ans = sat.solve()
print("YNEOS"[ans is None::2])
0