我正在开发一个 2 人棋盘游戏（例如connect 4），参数棋盘大小为h，w。我想使用hw大小的位板检查获胜条件。
在国际象棋等棋盘大小固定的游戏中，位棋盘通常用某种 64 位整数表示。当h和w不是常数并且可能非常大（假设 30*30）时，位板是个好主意吗？如果是这样，C/C++ 中的任何数据类型是否可以处理大型位板以保持其性能？
由于我目前正在研究 python，因此也非常感谢使用这种语言的解决方案！:)
提前致谢
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我之前写了这段代码只是为了玩游戏概念。没有智力行为涉及。只是随机动作来演示游戏。我想这对您来说并不重要，因为您只是在寻找快速检查获胜条件。这个实现很快，因为我尽了最大努力避免 for 循环并且只使用内置的 python/numpy 函数（有一些技巧）。
import numpy as np
row_size = 6
col_size = 7
symbols = {1:'A', -1:'B', 0:' '}
def was_winning_move(S, P, current_row_idx,current_col_idx):
#****** Column Win ******
current_col = S[:,current_col_idx]
P_idx= np.where(current_col== P)[0]
#if the difference between indexes are one, that means they are consecutive.
#we need at least 4 consecutive index. So 3 Ture value
is_idx_consecutive = sum(np.diff(P_idx)==1)>=3
if is_idx_consecutive:
return True
#****** Column Win ******
current_row = S[current_row_idx,:]
P_idx= np.where(current_row== P)[0]
is_idx_consecutive = sum(np.diff(P_idx)==1)>=3
if is_idx_consecutive:
return True
#****** Diag Win ******
offeset_from_diag = current_col_idx - current_row_idx
current_diag = S.diagonal(offeset_from_diag)
P_idx= np.where(current_diag== P)[0]
is_idx_consecutive = sum(np.diff(P_idx)==1)>=3
if is_idx_consecutive:
return True
#****** off-Diag Win ******
#here 1) reverse rows, 2)find new index, 3)find offest and proceed as diag
reversed_rows = S[::-1,:] #1
new_row_idx = row_size - 1 - current_row_idx #2
offeset_from_diag = current_col_idx - new_row_idx #3
current_off_diag = reversed_rows.diagonal(offeset_from_diag)
P_idx= np.where(current_off_diag== P)[0]
is_idx_consecutive = sum(np.diff(P_idx)==1)>=3
if is_idx_consecutive:
return True
return False
def move_at_random(S,P):
selected_col_idx = np.random.permutation(range(col_size))[0]
#print selected_col_idx
#we should fill in matrix from bottom to top. So find the last filled row in col and fill the upper row
last_filled_row = np.where(S[:,selected_col_idx] != 0)[0]
#it is possible that there is no filled array. like the begining of the game
#in this case we start with last row e.g row : -1
if last_filled_row.size != 0:
current_row_idx = last_filled_row[0] - 1
else:
current_row_idx = -1
#print 'col[{0}], row[{1}]'.format(selected_col,current_row)
S[current_row_idx, selected_col_idx] = P
return (S,current_row_idx,selected_col_idx)
def move_still_possible(S):
return not (S[S==0].size == 0)
def print_game_state(S):
B = np.copy(S).astype(object)
for n in [-1, 0, 1]:
B[B==n] = symbols[n]
print B
def play_game():
#initiate game state
game_state = np.zeros((6,7),dtype=int)
player = 1
mvcntr = 1
no_winner_yet = True
while no_winner_yet and move_still_possible(game_state):
#get player symbol
name = symbols[player]
game_state, current_row, current_col = move_at_random(game_state, player)
#print '******',player,(current_row, current_col)
#print current game state
print_game_state(game_state)
#check if the move was a winning move
if was_winning_move(game_state,player,current_row, current_col):
print 'player %s wins after %d moves' % (name, mvcntr)
no_winner_yet = False
# switch player and increase move counter
player *= -1
mvcntr += 1
if no_winner_yet:
print 'game ended in a draw'
player = 0
return game_state,player,mvcntr
if __name__ == '__main__':
S, P, mvcntr = play_game()
如果您有任何问题，请告诉我
更新：说明：
在每次移动时，查看穿过当前单元格的列、行、对角线和次对角线，并找到具有当前符号的连续单元格。避免扫描整个电路板。
在每个方向上提取单元格：
柱子：
current_col = S[:,current_col_idx]
排：
current_row = S[current_row_idx,:]
对角线：找到所需对角线与主对角线的偏移量：
diag_offset = current_col_idx - current_row_idx
current_diag = S.diagonal(offset)
非对角线：
反转矩阵的行：
S_reversed_rows = S[::-1,:]
查找新矩阵中的行索引
new_row_idx = row_size - 1 - current_row_idx
current_offdiag = S.diagonal(offset)
于 2015-05-19T08:41:34.913 回答