如何在有向图中找到凹坑？

Question

Question : we have a directed graph , how to find a hole (pit) in a graph on Ɵ(n) time complexity .

a pit on graph : if an vertex with n-1 degree for (input) and degree 0 for (output) , we have a pit on graph .

Answer 1

Am I missing something? Don't search the graph following graph edges. Just iterate over all n vertices in the graph and test each one for the number of incoming and outgoing edges. I assume for each vertex you can store a count of incoming and outgoing edges. This would scale O(n), if you have the edge counts.

@REPLY: We'd have to know how your graph is implemented to get more specific. But I mean something like:

foreach( node in graph )
     if (( node.numberInputEdges == numNodes -1 ) && 
         ( node.numberOutputEdges == 0 ))
         print ( "this node is a pit" )

Answer 2

I don't think you need to search graph at all. you can just count indegree and outdegree of every node. you just have to look for which node- indegree of a node is (n-1) and outdegree is '0'.

Considering you know "how many edges" are there.

int outdegree[n]={0}; // Storing outdegree of each node
int indegree[n]={0}; /// Storing indegree of each node


 while(m--)  // m is number of edges
 {
    scanf("%d %d",&a,&b);  // this means there is an edge from 'a' to 'b'. a-->b
    outdegree[a]++;
    indegree[b]++;
 }
 int sink;  
 for(i=1;i<=n;i++)
 {
    if((outdegree[i]==0 )&& (indegree[i]==(n-1)))
        sink=i;
 }
 cout<<"Sink/Pit is: "<<sink<<endl;

Answer 3

If you have a adjacency list implementation , just scan it,and you will find in O(n) time,if there is a vertex with outdegree 0. If you have a adjacency matrix, you can use DFS.

In DFS , just just increment the counter indegree whenver we find a outgoing node.

A pseudo code :

for each adjacent node :
   indegree->neighbour ++;
   if (neighbour not visited ) dfs(neighbour)   

Finaly scan indegree array to get which node has indegree (n-1) ,bcoz every node will have an edge to the Pit .

Answer 4

There can only be one "pit" in a graph (because of the deg_in = N-1). So search for all nodes with deg_out = 0. If two, stop. If one, check the deg_in. This is O(N).