'''
状态压缩DP求解
'''
M, N = map(int, input().split())
grid = [0] * M
for line_idx in range(M):
arr = list(map(int, input().split()))
for i, val in enumerate(arr):
if val == 1:
grid[line_idx] |= (1 << i)
# dp(i, j) 表示前i行进行放置,最后一行的二进制状态是j的可能摆放次数
dp = [[0]*(1<<12) for _ in range(M)]
# 一行中是否有相邻的1
def is_valid(val):
cur_pos = -2
for i in range(10):
if val & (1 << i) != 0:
if i == cur_pos + 1:
return False
else:
cur_pos = i
return True
def is_match(val, mask):
for i in range(12):
if (val & (1<<i)) != 0 and (mask & (1<<i)) == 0:
return False
return True
# 先提前计算一行的状态能转换到的另外的一行状态
m = {val:[] for val in range(1<<N) }
for val in range(1 << N):
for new_val in range(1 << N):
if new_val & val == 0 and is_valid(new_val):
m[val].append(new_val)
for i in range(M):
for j in range(1<<N):
if is_valid(j) and is_match(j, grid[i]):
if i == 0:
dp[i][j] = 1
else:
for new_j in m[j]:
dp[i][j] += dp[i-1][new_j]
dp[i][j] %= 100000000
else:
dp[i][j] = 0
print( sum(dp[M-1]) % 100000000)
AcWing 状态压缩DP相关问题 327. 玉米田
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转载自blog.csdn.net/xiaohaowudi/article/details/107762305
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