結果

問題 No.1254 補強への架け橋
コンテスト
ユーザー yuusanlondon
提出日時 2020-10-09 22:49:35
言語 Python3
(3.13.1 + numpy 2.2.1 + scipy 1.14.1)
結果
RE  
実行時間 -
コード長 3,020 bytes
コンパイル時間 92 ms
コンパイル使用メモリ 12,672 KB
実行使用メモリ 52,600 KB
最終ジャッジ日時 2024-07-20 13:17:15
合計ジャッジ時間 34,387 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 121 RE * 2
権限があれば一括ダウンロードができます

ソースコード

diff # 

# Python program to find bridges in a given undirected graph 
#Complexity : O(V+E) 
   
from collections import defaultdict 
   
#This class represents an undirected graph using adjacency list representation 
class Graph: 
   
    def __init__(self,vertices): 
        self.V= vertices #No. of vertices 
        self.graph = defaultdict(list) # default dictionary to store graph 
        self.Time = 0
   
    # function to add an edge to graph 
    def addEdge(self,u,v,i): 
        self.graph[u].append((v,i)) 
        self.graph[v].append((u,i)) 
   
    '''A recursive function that finds and prints bridges 
    using DFS traversal 
    u --> The vertex to be visited next 
    visited[] --> keeps tract of visited vertices 
    disc[] --> Stores discovery times of visited vertices 
    parent[] --> Stores parent vertices in DFS tree'''
    def bridgeUtil(self,u, visited, parent, low, disc): 
      
        global bridges
        # Mark the current node as visited and print it 
        visited[u]= True
  
        # Initialize discovery time and low value 
        disc[u] = self.Time 
        low[u] = self.Time 
        self.Time += 1
  
        #Recur for all the vertices adjacent to this vertex 
        for (v,i) in self.graph[u]: 
            # If v is not visited yet, then make it a child of u 
            # in DFS tree and recur for it 
            if visited[v] == False : 
                parent[v] = u 
                self.bridgeUtil(v, visited, parent, low, disc) 
  
                # Check if the subtree rooted with v has a connection to 
                # one of the ancestors of u 
                low[u] = min(low[u], low[v]) 
  
  
                ''' If the lowest vertex reachable from subtree 
                under v is below u in DFS tree, then u-v is 
                a bridge'''
                if low[v] > disc[u]: 
                    bridges[i]=1
      
                      
            elif v != parent[u]: # Update low value of u for parent function calls. 
                low[u] = min(low[u], disc[v]) 
  
  
    # DFS based function to find all bridges. It uses recursive 
    # function bridgeUtil() 
    def bridge(self): 
        global bridges
        # Mark all the vertices as not visited and Initialize parent and visited,  
        # and ap(articulation point) arrays 
        visited = [False] * (self.V) 
        disc = [float("Inf")] * (self.V) 
        low = [float("Inf")] * (self.V) 
        parent = [-1] * (self.V) 
  
        # Call the recursive helper function to find bridges 
        # in DFS tree rooted with vertex 'i' 
        for i in range(self.V): 
            if visited[i] == False: 
                self.bridgeUtil(i, visited, parent, low, disc) 
          
n=int(input())
# Create a graph given in the above diagram 
g1 = Graph(n)
bridges=[0]*n
for i in range(n):
  a,b=map(int,input().split())
  g1.addEdge(a-1,b-1, i)
g1.bridge()
ans=[]
for i in range(n):
  if bridges[i]==0:
    ans.append(str(i+1))
print(len(ans))
print(' '.join(ans))
0