[22.02.25] 2667 - BFS 풀이법 추가

Author: par3k

Diff for: Python_BOJ_2022/2667.py

@@ -4,25 +4,56 @@
 visited = [[0] * n for _ in range(n)]
 dx, dy = [-1, 1, 0, 0], [0, 0, -1, 1]
 
+# DFS - 재귀를 이용한 풀이
+# cnt = 0
+# ans = []
+
+# def recursive(x, y):
+#     global cnt
+#     visited[x][y] = 1
+#     if graph[x][y] == 1:
+#         cnt += 1
+#     for i in range(4):
+#         nx, ny = x + dx[i], y + dy[i]
+#         if 0 <= nx < n and 0 <= ny <n:
+#             if not visited[nx][ny] and graph[nx][ny] == 1:
+#                 recursive(nx, ny)
+
+
+# for i in range(n):
+#     for j in range(n):
+#         if graph[i][j] == 1 and not visited[i][j]:
+#             recursive(i, j)
+#             ans.append(cnt)
+#             cnt = 0
+
+# BFS 풀이
+from collections import deque
+
 cnt = 0
 ans = []
 
-def recursive(x, y):
+def bfs(x, y):
     global cnt
+    queue = deque()
+    queue.append((x, y))
     visited[x][y] = 1
     if graph[x][y] == 1:
         cnt += 1
-    for i in range(4):
-        nx, ny = x + dx[i], y + dy[i]
-        if 0 <= nx < n and 0 <= ny <n:
-            if not visited[nx][ny] and graph[nx][ny] == 1:
-                recursive(nx, ny)
-
+    while queue:
+        x, y = queue.popleft()
+        for i in range(4):
+            nx, ny = x + dx[i], y + dy[i]
+            if 0 <= nx < n and 0 <= ny < n:
+                if graph[nx][ny] == 1 and not visited[nx][ny]:
+                    visited[nx][ny] = 1
+                    cnt += 1
+                    queue.append((nx, ny))
 
 for i in range(n):
     for j in range(n):
         if graph[i][j] == 1 and not visited[i][j]:
-            recursive(i, j)
+            bfs(i, j)
             ans.append(cnt)
             cnt = 0