给定一个大小为(NxM)的网格，将使用整数填充。
网格中单元格的填充应采用以下方式：
- 令A，B和C为三个像元，并且B和C与A共享一侧。
- 单元格B和C的值必须是不同的。
- 令L为网格中不同整数的数量。
- 每个像元应包含从1到L的值。
任务是找到L的最小值和所有结果网格。
例子：
Input: N = 1, M = 2
Output:
L = 2
grid = {1, 2}
Input: 2 3
Output:
L = 3
grid = {{1, 2, 3},
{1, 2, 3}}
Explanation: Integers in the neighbors
of cell (2, 2) are 1, 2 and 3.
All numbers are pairwise distinct.
方法：
假定与另一个单元共享一侧的两个单元必须是不同的。对于每个此类单元格，网格中最多可能有8个单元格，其值必须不同。
它将出现4种颜色问题：填充区域所需的最大颜色将为4。
- 对于N <4或M <4
所需的整数数量可能从1到4不等。
检查8个单元格，然后填充当前单元格。
如果8个像元中不同整数的数量小于L，则用剩余的所有整数填充当前像元，否则用L + 1个整数填充当前像元。 - 对于N> = 4和M> = 4
根据4种颜色问题，所需的整数数目必须为4。
使用4×4矩阵填充NxM矩阵。1 2 3 4 1 2 3 4 3 4 1 2 3 4 1 2
下面是上述方法的实现：
执行：
# Python 3 implementation of
# above approach
# Function to display the matrix
def display_matrix(A):
for i in A:
print(*i)
# Function for calculation
def cal_main(A, L, x, i, j):
s = set()
# Checking 8 cells and
# then fill the current cell.
if (i - 2) >= 0:
s.add(A[i - 2][j])
if (i + 2) < N:
s.add(A[i + 2][j])
if (j - 2) >= 0:
s.add(A[i][j - 2])
if (j + 2) < M:
s.add(A[i][j + 2])
if (i - 1) >= 0 and (j - 1) >= 0:
s.add(A[i - 1][j - 1])
if (i - 1) >= 0 and (j + 1) < M:
s.add(A[i - 1][j + 1])
if (i + 1) < N and (j - 1) >= 0:
s.add(A[i + 1][j - 1])
if (i + 1) < N and (j + 1) < M:
s.add(A[i + 1][j + 1])
# Set to contain distinct value
# of integers in 8 cells.
s = s.difference({0})
if len(s) < L:
# Set contain remaining integers
w = x.difference(s)
# fill the current cell
# with maximum remaining integer
A[i][j] = max(w)
else:
# fill the current cells with L + 1 integer.
A[i][j] = L + 1
L += 1
# Increase the value of L
x.add(L)
return A, L, x
# Function to find the number
# of distinct integers
def solve(N, M):
# initialise the list (NxM) with 0.
A = []
for i in range(N):
K = []
for j in range(M):
K.append(0)
A.append(K)
# Set to contain distinct
# value of integers from 1-L
x = set()
L = 0
# Number of integer required
# may vary from 1 to 4.
if N < 4 or M < 4:
if N > M: # if N is greater
for i in range(N):
for j in range(M):
cal_main(A, L, x, i, j)
else:
# if M is greater
for j in range(M):
for i in range(N):
cal_main(A, L, x, i, j)
else:
# Number of integer required
# must be 4
L = 4
# 4×4 matrix to fill the NxM matrix.
m4 = [[1, 2, 3, 4],
[1, 2, 3, 4],
[3, 4, 1, 2],
[3, 4, 1, 2]]
for i in range(4):
for j in range(4):
A[i][j] = m4[i][j]
for i in range(4, N):
for j in range(4):
A[i][j] = m4[i % 4][j]
for j in range(4, M):
for i in range(N):
A[i][j] = A[i][j % 4]
print(L)
display_matrix(A)
# Driver Code
if __name__ == "__main__":
# sample input
# Number of rows and columns
N, M = 10, 5
solve(N, M)
输出：
4
1 2 3 4 1 2 3 4 1 2
1 2 3 4 1 2 3 4 1 2
3 4 1 2 3 4 1 2 3 4
3 4 1 2 3 4 1 2 3 4
1 2 3 4 1 2 3 4 1 2
1 2 3 4 1 2 3 4 1 2
3 4 1 2 3 4 1 2 3 4
3 4 1 2 3 4 1 2 3 4
1 2 3 4 1 2 3 4 1 2
1 2 3 4 1 2 3 4 1 2