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https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-05 22:59:31 +08:00
@ -257,6 +257,54 @@ public class Main {
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### Python
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#### 深搜版
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```python
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position = [[1, 0], [0, 1], [-1, 0], [0, -1]]
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count = 0
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def dfs(grid, x, y):
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global count
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grid[x][y] = 0
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count += 1
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for i, j in position:
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next_x = x + i
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next_y = y + j
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if next_x < 0 or next_y < 0 or next_x >= len(grid) or next_y >= len(grid[0]):
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continue
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if grid[next_x][next_y] == 1:
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dfs(grid, next_x, next_y)
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n, m = map(int, input().split())
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# 邻接矩阵
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grid = []
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for i in range(n):
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grid.append(list(map(int, input().split())))
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# 清除边界上的连通分量
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for i in range(n):
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if grid[i][0] == 1:
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dfs(grid, i, 0)
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if grid[i][m - 1] == 1:
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dfs(grid, i, m - 1)
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for j in range(m):
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if grid[0][j] == 1:
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dfs(grid, 0, j)
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if grid[n - 1][j] == 1:
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dfs(grid, n - 1, j)
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count = 0 # 将count重置为0
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# 统计内部所有剩余的连通分量
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for i in range(n):
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for j in range(m):
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if grid[i][j] == 1:
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dfs(grid, i, j)
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print(count)
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```
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#### 广搜版
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```python
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from collections import deque
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@ -293,17 +341,22 @@ def bfs(r, c):
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for i in range(n):
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if g[i][0] == 1: bfs(i, 0)
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if g[i][m-1] == 1: bfs(i, m-1)
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if g[i][0] == 1:
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bfs(i, 0)
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if g[i][m-1] == 1:
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bfs(i, m-1)
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for i in range(m):
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if g[0][i] == 1: bfs(0, i)
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if g[n-1][i] == 1: bfs(n-1, i)
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if g[0][i] == 1:
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bfs(0, i)
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if g[n-1][i] == 1:
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bfs(n-1, i)
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count = 0
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for i in range(n):
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for j in range(m):
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if g[i][j] == 1: bfs(i, j)
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if g[i][j] == 1:
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bfs(i, j)
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print(count)
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```
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