Python 图结构实现：计算马遍历 n*m 棋盘的所有途径总数

```python def is_valid_position(n, m, x, y): if x < 0 or x >= n or y < 0 or y >= m: return False return True def dfs(n, m, x, y, visited): directions = [(1, 2), (1, -2), (-1, 2), (-1, -2), (2, 1), (2, -1), (-2, 1), (-2, -1)] visited[x][y] = True count = 0 for dx, dy in directions: new_x, new_y = x + dx, y + dy if is_valid_position(n, m, new_x, new_y) and not visited[new_x][new_y]: count += dfs(n, m, new_x, new_y, visited) visited[x][y] = False if count == 0: return 1 if all(visited[i][j] for i in range(n) for j in range(m)) else 0 return count n, m, x, y = map(int, input().split()) visited = [[False for _ in range(m)] for _ in range(n)] visited[x][y] = True print(dfs(n, m, x, y, visited)) ``` 这段代码使用深度优先搜索（DFS）来遍历马在棋盘上的所有路径，当马无法再移动时，检查是否所有的点都已经被访问过，如果是则返回1，否则返回0。 ######[AI写代码神器 | 621点数解答 | 2024-11-18 19:34:21]