📅  最后修改于: 2023-12-03 15:27:26.397000             🧑  作者: Mango
本题是《算法竞赛入门经典》一书中的题目,要求实现一个算法,检查一个给定的 N*N 的矩阵是否满足以下条件:
具体实现要求,请见下方的代码部分。
def check_mx(mx, n):
directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
vis = [[False] * n for _ in range(n)]
def dfs(x, y):
vis[x][y] = True
for direction in directions:
nx, ny = x + direction[0], y + direction[1]
if 0 <= nx < n and 0 <= ny < n and mx[nx][ny] == 1 and not vis[nx][ny]:
dfs(nx, ny)
for i in range(n):
for j in range(n):
if mx[i][j] == 1 and not vis[i][j]:
dfs(i, j)
cnt = 0
for ii in range(n):
for jj in range(n):
if mx[ii][jj] == 1 and vis[ii][jj]:
cnt += 1
if cnt % 2 == 1:
return False
return True
我们可以通过以下代码测试上述算法:
if __name__ == '__main__':
mx1 = [[0, 1, 0], [1, 0, 1], [0, 1, 0]]
assert check_mx(mx1, 3) == True
mx2 = [[0, 1, 0], [1, 1, 1], [0, 1, 0]]
assert check_mx(mx2, 3) == False
mx3 = [[1, 1], [1, 0]]
assert check_mx(mx3, 2) == False
mx4 = [[1, 1], [1, 1]]
assert check_mx(mx4, 2) == True
print("All test cases pass")
通过以上实现和测试,我们可以检验一个给定的矩阵是否满足上述三个条件。