結果

問題 No.2173 Nightcord
ユーザー chineristAC
提出日時 2022-12-25 03:09:12
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,080 ms / 2,525 ms
コード長 3,399 bytes
コンパイル時間 265 ms
コンパイル使用メモリ 82,448 KB
実行使用メモリ 84,076 KB
最終ジャッジ日時 2024-11-17 14:31:35
合計ジャッジ時間 16,105 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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 * 3
other AC * 54
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 sys,random,bisect
from collections import deque,defaultdict,Counter
from heapq import heapify,heappop,heappush
from itertools import cycle, permutations
from math import log,gcd

input = lambda :sys.stdin.readline().rstrip()
mi = lambda :map(int,input().split())
li = lambda :list(mi())

def cross3(a, b, c):
    return (b[0]-a[0])*(c[1]-a[1]) - (b[1]-a[1])*(c[0]-a[0])

# ps = [(x, y), ...]: ソートされた座標list
def convex_hull(ps):
    qs = []
    N = len(ps)
    for p in ps:
        # 一直線上で高々2点にする場合は ">=" にする
        while len(qs) > 1 and cross3(qs[-1], qs[-2], p) > 0:
            qs.pop()
        qs.append(p)
    t = len(qs)
    for i in range(N-2, -1, -1):
        p = ps[i]
        while len(qs) > t and cross3(qs[-1], qs[-2], p) > 0:
            qs.pop()
        qs.append(p)
    return qs

# O(N)
def inside_convex_polygon0(p0, qs):
    L = len(qs)
    D = [cross3(qs[i-1], p0, qs[i]) for i in range(L)]
    return all(e >= 0 for e in D) or all(e <= 0 for e in D)

# O(log N)
def inside_convex_polygon(p0, qs):
    L = len(qs)
    left = 1; right = L
    q0 = qs[0]
    while left+1 < right:
        mid = (left + right) >> 1
        if cross3(q0, p0, qs[mid]) <= 0:
            left = mid
        else:
            right = mid
    if left == L-1:
        left -= 1
    qi = qs[left]; qj = qs[left+1]
    v0 = cross3(q0, qi, qj)
    v1 = cross3(q0, p0, qj)
    v2 = cross3(q0, qi, p0)
    if v0 < 0:
        v1 = -v1; v2 = -v2
    return 0 <= v1 and 0 <= v2 and v1 + v2 <= v0

# 線分同士の交点判定
def dot3(O, A, B):
    ox, oy = O; ax, ay = A; bx, by = B
    return (ax - ox) * (bx - ox) + (ay - oy) * (by - oy)
def cross3(O, A, B):
    ox, oy = O; ax, ay = A; bx, by = B
    return (ax - ox) * (by - oy) - (bx - ox) * (ay - oy)
def dist2(A, B):
    ax, ay = A; bx, by = B
    return (ax - bx) ** 2 + (ay - by) ** 2
def is_intersection(P0, P1, Q0, Q1):
    C0 = cross3(P0, P1, Q0)
    C1 = cross3(P0, P1, Q1)
    D0 = cross3(Q0, Q1, P0)
    D1 = cross3(Q0, Q1, P1)
    if C0 == C1 == 0:
        E0 = dot3(P0, P1, Q0)
        E1 = dot3(P0, P1, Q1)
        if not E0 < E1:
            E0, E1 = E1, E0
        return E0 <= dist2(P0, P1) and 0 <= E1
    return C0 * C1 <= 0 and D0 * D1 <= 0





def solve_3(N,K,star):

    for p in range(2):
        for x,y in star[p]:
            S = set()
            for xx,yy in star[p^1]:
                xx,yy = xx-x,yy-y
                g = gcd(xx,yy)
                xx,yy = xx//g,yy//g
                if (-xx,-yy) in S:
                    return "Yes"
                S.add((xx,yy))

    return "No"

def solve_4(N,K,star):
    

    star[0].sort()
    star[1].sort()

    ch = [convex_hull(star[p]) for p in range(2)]
    for p in range(2):
        if len(star[p^1]) > 2:
            for i in range(len(star[p])):
                x,y = star[p][i]
                if inside_convex_polygon((x,y),ch[p^1]):
                    return "Yes"
        if len(star[p]) == 2:
            a,b = star[p][0],star[p][1]
            for i in range(len(ch[p^1])-1):
                c,d = ch[p^1][i],ch[p^1][i+1]
                if is_intersection(a,b,c,d):
                    return "Yes"
    
    return "No"



N,K = mi()
star = [[] for p in range(2)]
for _ in range(N):
    x,y,c = mi()
    
    star[c-1].append((x,y))

if K == 3:
    print(solve_3(N,K,star))
else:
    print(solve_4(N,K,star))
0