feat: 109. 冗余连接II新增python解法

problems/kamacoder/0109.冗余连接II.md

@@ -351,6 +351,92 @@ public class Main {
 ```
 ### Python
 
+```python
+from collections import defaultdict
+
+father = list()
+
+
+def find(u):
+    if u == father[u]:
+        return u
+    else:
+        father[u] = find(father[u])
+        return father[u]
+        
+        
+def is_same(u, v):
+    u = find(u)
+    v = find(v)
+    return u == v
+    
+    
+def join(u, v):
+    u = find(u)
+    v = find(v)
+    if u != v:
+        father[u] = v
+    
+    
+def is_tree_after_remove_edge(edges, edge, n):
+    # 初始化并查集
+    global father 
+    father = [i for i in range(n + 1)]
+    
+    for i in range(len(edges)):
+        if i == edge:
+            continue
+        s, t = edges[i]
+        if is_same(s, t): # 成環，即不是有向樹
+            return False
+        else: # 將s,t放入集合中
+            join(s, t)
+    return True
+    
+
+def get_remove_edge(edges):
+    # 初始化并查集
+    global father
+    father = [i for i in range(n + 1)]
+    
+    for s, t in edges:
+        if is_same(s, t):
+            print(s, t)
+            return
+        else:
+            join(s, t)
+        
+
+if __name__ == "__main__":
+    # 輸入
+    n = int(input())
+    edges = list()
+    in_degree = defaultdict(int)
+    
+    for i in range(n):
+        s, t = map(int, input().split())
+        in_degree[t] += 1
+        edges.append([s, t])
+        
+    # 尋找入度為2的邊，並紀錄其下標(index)
+    vec = list()
+    for i in range(n - 1, -1, -1):
+        if in_degree[edges[i][1]] == 2:
+            vec.append(i)
+            
+    # 輸出
+    if len(vec) > 0:
+        # 情況一：刪除輸出順序靠後的邊 
+        if is_tree_after_remove_edge(edges, vec[0], n):
+            print(edges[vec[0]][0], edges[vec[0]][1])
+        # 情況二：只能刪除特定的邊
+        else:
+            print(edges[vec[1]][0], edges[vec[1]][1])
+    else:
+        # 情況三： 原圖有環
+        get_remove_edge(edges)
+```
+
 ### Go
 
 ### Rust