803

Author: ProHiryu

DFS/traditionalDFS/803.md

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+## Bricks Falling When Hit
+
+#### Description
+
+[link](https://leetcode.com/problems/bricks-falling-when-hit/)
+
+---
+
+#### Solution
+
+- See Code
+
+---
+
+#### Code
+
+> 最坏情况：O(n^2)
+
+```python
+class Solution:
+    def hitBricks(self, grid: List[List[int]], hits: List[List[int]]) -> List[int]:
+        m, n = len(grid), len(grid[0])
+        
+        # Connect unconnected bricks and Sum them all
+        def dfs(i, j):
+            if not (0<=i<m and 0<=j<n) or grid[i][j]!=1:
+                return 0
+            ret = 1
+            grid[i][j] = 2
+            ret += sum(dfs(x, y) for x, y in [(i-1, j), (i+1, j), (i, j-1), (i, j+1)])
+            return ret
+        
+        # Check whether (i, j) is connected to Not Falling Bricks
+        def is_connected(i, j):
+            return i==0 or any([0<=x<m and 0<=y<n and grid[x][y]==2 for x, y in [(i-1, j), (i+1, j), (i, j-1), (i, j+1)]])
+        
+        # Mark whether there is a brick at the each hit
+        for i, j in hits:
+            grid[i][j] -= 1
+                
+        # Get grid after all hits
+        for i in range(n):
+            dfs(0, i)
+        
+        # Reversely add the block of each hits and get count of newly add bricks
+        ret = [0]*len(hits)
+        for k in reversed(range(len(hits))):
+            i, j = hits[k]
+            grid[i][j] += 1
+            if grid[i][j]==1 and is_connected(i, j):
+                ret[k] = dfs(i, j)-1
+            
+        return ret
+```