删除最小节点数以使图形断开连接

Question

Find the minimum number of nodes that needs to be removed to make graph disconnected( there exists no path from any node to all other nodes). Number of nodes can be 10 ⁵

Answer 1

First of all, to find a node that is not needed to make the graph connected, you need to find a cycle. Once you find a cycle, then you know all those edges in those cycle are not needed. All the remaining edges in the graph are needed to hold the graph together, so you do a simple traversal of the graph and count the number of vertices connected to find the minimum number of nodes that needs to be removed to make graph disconnected. I'm not sure if this is the most efficient way, just the most efficient in my head. If V is the number of vertices and E is the number of edges, the time complexity will be somewhere around O(V + E) + O(V), since the initial cycle finding is O(V + E) and the counting of the nodes is O(V). Since E >= V, the time complexity can be rounded to O(E), but with huge constant factors.

Answer 2

Consider the graph undirected.

totalDegree = sum of degree of all nodes.
while(totalDegree > 0) {
    Remove the node with highest degree and it's edges.
    Update degree count of each node.
    totalDegree = sum of degree of all remaining nodes.
}
remaining nodes are all disconnected

Trying to think of a case where it will not give the min number of nodes.