[BOJ] 20057 마법사 상어와 토네이도 (G3)

Author: poodlepoodle

최어진/9주차/260223.py

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+# 백준 20057번: 마법사 상어와 토네이도 (2회차)
+
+import sys
+
+input = sys.stdin.readline
+
+# N <= 499
+N = int(input())
+sands = [list(map(int, input().rstrip().split())) for _ in range(N)]
+
+# 시간복잡도 자체는...
+#   1. 격자의 모든 칸에 대해서 모래 날리기 -> O(N^2) ~= 10^5
+#   2. 각 칸마다 날라간 모래 계산 ~= 10
+#   결론: 시간복잡도 때문에 문제 생길 것 같지는 않음
+
+moves = [
+    [0, -1], # <
+    [1, 0], # v
+    [0, 1], # >
+    [-1, 0], # ^
+]
+
+splits = [
+    [
+        [-1, 0, 0.01],
+        [1, 0, 0.01],
+        [-1, -1, 0.07],
+        [1, -1, 0.07],
+        [-2, -1, 0.02],
+        [2, -1, 0.02],
+        [-1, -2, 0.1],
+        [1, -2, 0.1],
+        [0, -3, 0.05],
+    ],
+    [
+        [0, -1, 0.01],
+        [0, 1, 0.01],
+        [1, -1, 0.07],
+        [1, 1, 0.07],
+        [1, -2, 0.02],
+        [1, 2, 0.02],
+        [2, -1, 0.1],
+        [2, 1, 0.1],
+        [3, 0, 0.05],
+    ],
+    [
+        [-1, 0, 0.01],
+        [1, 0, 0.01],
+        [-1, 1, 0.07],
+        [1, 1, 0.07],
+        [-2, 1, 0.02],
+        [2, 1, 0.02],
+        [-1, 2, 0.1],
+        [1, 2, 0.1],
+        [0, 3, 0.05],
+    ],
+    [
+        [0, -1, 0.01],
+        [0, 1, 0.01],
+        [-1, -1, 0.07],
+        [-1, 1, 0.07],
+        [-1, -2, 0.02],
+        [-1, 2, 0.02],
+        [-2, -1, 0.1],
+        [-2, 1, 0.1],
+        [-3, 0, 0.05],
+    ],
+]
+
+remains = [
+    [0, -2],
+    [2, 0],
+    [0, 2],
+    [-2, 0]
+]
+
+r, c = N // 2, N // 2
+direction = 0
+answer = 0
+
+def is_in_board(r, c):
+    return 0 <= r < N and 0 <= c < N
+
+def print_sands(current_r, current_c):
+    global answer
+    for r in range(N):
+        for c in range(N):
+            if r == current_r and c == current_c: print('#', end=' ')
+            elif not sands[r][c]: print('.', end=' ')
+            else: print(sands[r][c], end=' ')
+        print()
+    print(f'answer: {answer}')
+    print()
+
+def split(r, c, direction):
+    global answer
+    # print(f'split(r: {r}, c: {c}), direction: {direction}')
+
+    # y칸의 좌표 저장
+    dr, dc = moves[direction]
+    yr, yc = r + dr, c + dc
+    y_sands = sands[yr][yc]
+    if not y_sands: return
+    
+    for dr, dc, portion in splits[direction]:
+        # 이동할 모래의 양 계산, 소수점 아래 버림
+        z_sands = int(y_sands * portion)
+        # 모래의 이동
+        sands[yr][yc] -= z_sands
+        if is_in_board(r + dr, c + dc):
+            sands[r + dr][c + dc] += z_sands
+        else:
+            answer += z_sands
+
+    dr, dc = remains[direction]
+    ar, ac = r + dr, c + dc
+    if is_in_board(ar, ac):
+        sands[ar][ac] += sands[yr][yc]
+    else:
+        answer += sands[yr][yc]
+    sands[yr][yc] = 0
+
+for i in range(1, N):
+    for _ in range(i):
+        split(r, c, direction)
+        # print_sands(r, c)
+        dr, dc = moves[direction]
+        r, c = r + dr, c + dc
+    direction = (direction + 1) % 4
+    for _ in range(i):
+        split(r, c, direction)
+        # print_sands(r, c)
+        dr, dc = moves[direction]
+        r, c = r + dr, c + dc
+    direction = (direction + 1) % 4
+for _ in range(i):
+    split(r, c, direction)
+    # print_sands(r, c)
+    dr, dc = moves[direction]
+    r, c = r + dr, c + dc
+direction = (direction + 1) % 4
+
+print(answer)