admin管理员组

文章数量:1530263

There are n servers numbered from 0 to n-1 connected by undirected server-to-server connections forming a network where connections[i] = [a, b] represents a connection between servers a and b. Any server can reach any other server directly or indirectly through the network.

A critical connection is a connection that, if removed, will make some server unable to reach some other server.

Return all critical connections in the network in any order.

 

Example 1:

Input: n = 4, connections = [[0,1],[1,2],[2,0],[1,3]]
Output: [[1,3]]
Explanation: [[3,1]] is also accepted.

 

Constraints:

  • 1 <= n <= 10^5
  • n-1 <= connections.length <= 10^5
  • connections[i][0] != connections[i][1]
  • There are no repeated connections.

Discuss

思路：
需要先了解tarjan算法求解强联通分量的原理，可参考

https://search.bilibili/all?keyword=Tarjan

https://blog.csdn/starstar1992/article/details/52944958

然后在geeksforgeeks上找到类似的题目，看下怎么根据dfn、low数组来判断是否为割点割线

 https://www.geeksforgeeks/bridge-in-a-graph/ 

https://www.geeksforgeeks/articulation-points-or-cut-vertices-in-a-graph/

标准的套题，但是没打过ACM，不了解tarjan算法，即使比赛完自己看博客也是看的一知半解（自己还是太菜了）

from collections import defaultdict
class Solution(object):
    def criticalConnections(self, n, connections):
        """
        :type n: int
        :type connections: List[List[int]]
        :rtype: List[List[int]]
        """
        adj=defaultdict(set)
        for a,b in connections:
            adj[a].add(b)
            adj[b].add(a)
        
        vis=[False]*n
        time = [0]
        dfn = [0]*n
        low = [0]*n
        parent = [-1]*n
        res=[]
        
        def tarjan(u):
            vis[u]=True
            time[0]+=1
            dfn[u]=low[u]=time[0]
            
            for v in adj[u]:
                if not vis[v]:
                    parent[v] = u
                    tarjan(v)
                    low[u] = min(low[u], low[v])
                    
                    if low[v]>dfn[u]:
                        res.append([u,v])
                elif v!=parent[u]:
                    low[u] = min(low[u], dfn[v])
            
        for root in range(n):
            if not vis[root]:
                tarjan(root)
        return res

 

本文标签： CriticalNetworkconnections

版权声明：本文标题：1192. Critical Connections in a Network 内容由热心网友自发贡献，该文观点仅代表作者本人， 转载请联系作者并注明出处：https://m.elefans.com/dongtai/1726271339a1063866.html， 本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌抄袭侵权/违法违规的内容，一经查实，本站将立刻删除。