結果
| 問題 | No.2375 watasou and hibit's baseball | 
| コンテスト | |
| ユーザー |  | 
| 提出日時 | 2023-07-07 22:43:59 | 
| 言語 | PyPy3 (7.3.15) | 
| 結果 | 
                                AC
                                 
                             | 
| 実行時間 | 1,018 ms / 2,000 ms | 
| コード長 | 2,104 bytes | 
| コンパイル時間 | 236 ms | 
| コンパイル使用メモリ | 82,520 KB | 
| 実行使用メモリ | 129,640 KB | 
| 最終ジャッジ日時 | 2024-07-21 19:00:24 | 
| 合計ジャッジ時間 | 21,878 ms | 
| ジャッジサーバーID (参考情報) | judge3 / judge4 | 
(要ログイン)
| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 3 | 
| other | AC * 36 | 
ソースコード
from math import gcd
from bisect import bisect_left, bisect_right
from heapq import heappop, heappush
from collections import defaultdict, deque, Counter
import sys
sys.setrecursionlimit(5*10**5)
input = sys.stdin.readline
# n <= 1<<60 以下くらい
def popcnt(n):
    c = (n & 0x5555555555555555) + ((n >> 1) & 0x5555555555555555)
    c = (c & 0x3333333333333333) + ((c >> 2) & 0x3333333333333333)
    c = (c & 0x0f0f0f0f0f0f0f0f) + ((c >> 4) & 0x0f0f0f0f0f0f0f0f)
    c = (c & 0x00ff00ff00ff00ff) + ((c >> 8) & 0x00ff00ff00ff00ff)
    c = (c & 0x0000ffff0000ffff) + ((c >> 16) & 0x0000ffff0000ffff)
    c = (c & 0x00000000ffffffff) + ((c >> 32) & 0x00000000ffffffff)
    return c
n, A, B = map(int, input().split())
ball = [list(map(int, input().split())) for i in range(n)]
dp = [[[0]*(n+1) for i in range(n+1)] for i in range(1 << n)]
for i in range(n):
    dp[1 << i][i][-1] = 1
def dist(i, j):
    xi, yi, ki = ball[i]
    xj, yj, kj = ball[j]
    return abs(xi-xj) + abs(yi-yj)
def speed(i,j):
    xi, yi, ki = ball[i]
    xj, yj, kj = ball[j]
    return abs(ki-kj)
def judge1(i, j, cnt):
    ok = 0
    if cnt == 2 and A <= dist(i, j):
        ok = 1
    if B <= speed(i,j):
        ok = 1
    return ok
def judge2(i, j, k):
    ok = 0
    if A <= dist(i,j) + dist(i,k):
        ok = 1
    return ok
ans = 1
for bit in range(1 << n):
    cnt = popcnt(bit)
    if cnt <= 1: continue
    for crr in range(n):
        if not (bit >> crr) & 1:
            continue
        for pre1 in range(n):
            if not (bit >> pre1) & 1:
                continue
            ok = judge1(crr, pre1, cnt)
            if cnt == 2 and ok:
                dp[bit][crr][pre1] |= dp[bit-(1<<crr)][pre1][-1]
                ans = max(ans, dp[bit][crr][pre1]*2)
            if cnt == 2:  continue
            for pre2 in range(n+1):
                if not (bit >> pre2) & 1:
                    continue
                if judge2(crr, pre1, pre2) or ok:
                    dp[bit][crr][pre1] |= dp[bit - (1<<crr)][pre1][pre2]
                    ans = max(ans, dp[bit][crr][pre1]*cnt)
print(ans)
            
            
            
        