結果
問題 | No.2411 Reverse Directions |
ユーザー | FromBooska |
提出日時 | 2023-08-15 19:55:03 |
言語 | PyPy3 (7.3.15) |
結果 |
WA
|
実行時間 | - |
コード長 | 3,411 bytes |
コンパイル時間 | 277 ms |
コンパイル使用メモリ | 82,304 KB |
実行使用メモリ | 99,840 KB |
最終ジャッジ日時 | 2024-11-24 04:26:26 |
合計ジャッジ時間 | 5,938 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 39 ms
54,528 KB |
testcase_01 | AC | 40 ms
54,144 KB |
testcase_02 | AC | 40 ms
54,272 KB |
testcase_03 | AC | 38 ms
54,400 KB |
testcase_04 | AC | 77 ms
99,840 KB |
testcase_05 | AC | 39 ms
54,632 KB |
testcase_06 | AC | 38 ms
54,144 KB |
testcase_07 | AC | 98 ms
78,956 KB |
testcase_08 | AC | 38 ms
54,144 KB |
testcase_09 | AC | 39 ms
54,016 KB |
testcase_10 | WA | - |
testcase_11 | WA | - |
testcase_12 | WA | - |
testcase_13 | AC | 72 ms
76,544 KB |
testcase_14 | WA | - |
testcase_15 | WA | - |
testcase_16 | AC | 64 ms
70,784 KB |
testcase_17 | WA | - |
testcase_18 | WA | - |
testcase_19 | AC | 40 ms
54,656 KB |
testcase_20 | WA | - |
testcase_21 | AC | 81 ms
77,116 KB |
testcase_22 | WA | - |
testcase_23 | AC | 80 ms
76,928 KB |
testcase_24 | WA | - |
testcase_25 | AC | 60 ms
71,424 KB |
testcase_26 | AC | 39 ms
54,400 KB |
testcase_27 | AC | 65 ms
73,472 KB |
testcase_28 | WA | - |
testcase_29 | WA | - |
testcase_30 | WA | - |
testcase_31 | AC | 84 ms
77,364 KB |
ソースコード
# BFS, distance # L-1歩とR歩では同じ座標にいる、そうしないとゴールに行けない # L to Rでは同じところを行ったり来たりすればいい、それ以上の動き必要なしだから # つまりL-1歩で3個の縦または横のフリーゾーンの真ん中にいる必要 # (R+1-L)%2==0が必要 # (K-goal_distance-(R+1-L))%2==0も必要 # WA、distance<=L-1までの場所で行ったり来たりすることもできるな H, W, K, L, R = map(int, input().split()) S = [] for i in range(H): s = input() S.append(s) from collections import deque d = [[+1, 0], [-1, 0], [0, +1], [0, -1]] sh, sw = 0, 0 que = deque() que.append((sh, sw)) INF = 10**4 distance = [[INF]*W for h in range(H)] distance[sh][sw] = 0 while que: ch, cw = que.popleft() for dh, dw in d: if 0 <= ch+dh < H and 0 <= cw+dw < W: if S[ch+dh][cw+dw] != '#': if distance[ch+dh][cw+dw] > distance[ch][cw]+1: distance[ch+dh][cw+dw] = distance[ch][cw]+1 que.append([ch+dh, cw+dw]) #print(distance) def check(ch, cw): if L-1 - distance[ch][cw] < 0 or (L-1 - distance[ch][cw])%2==1: return -1 # 行ったり来たりチェック L1_direction = -1 #縦0、横1 back_forth_test = False if 1 <= ch < H-1 and S[ch-1][cw] == S[ch][cw] == S[ch+1][cw] == '.': back_forth_test = True L1_direction = 0 if 1 <= cw < W-1 and S[ch][cw-1] == S[ch][cw] == S[ch][cw+1] == '.': back_forth_test = True L1_direction = 1 if back_forth_test == False: return -1 return L1_direction goal_distance = distance[H-1][W-1] if goal_distance == INF: print('No') exit() LR = R+1-L if LR%2 == 1: print('No') exit() if K-goal_distance-LR < 0 or (K-goal_distance-LR)%2 == 1: print('No') exit() for h in range(H): for w in range(W): dire = check(h, w) if dire == -1: continue direction1 = ['U', 'D', 'L', 'R'] route1 = [] current = (h, w) for i in range(L-1, -1, -1): ch, cw = current for j in range(4): dh, dw = d[j] if 0 <= ch+dh < H and 0 <= cw+dw < W: if distance[ch+dh][cw+dw] == i-1: current = (ch+dh, cw+dw) route1.append(direction1[j]) break route2 = [] if dire == 0: for k in range(LR//2): route2.append('U') route2.append('D') elif dire == 1: for k in range(LR//2): route2.append('R') route2.append('L') direction3 = ['D', 'U', 'R', 'L'] route3 = [] current = (h, w) for i in range(L-1, goal_distance): #print('current', current) ch, cw = current for j in range(4): dh, dw = d[j] if 0 <= ch+dh < H and 0 <= cw+dw < W: if distance[ch+dh][cw+dw] == i+1: current = (ch+dh, cw+dw) route3.append(direction3[j]) break ans = ''.join(route1[::-1]) + ''.join(route2) + ''.join(route3) print('Yes') print(ans) exit() print('No')