I know the common way to do a topological sort is using DFS with recursion. But how would you do it using stack<int>
instead of recursion? I need to obtain the reversed post-order but I'm kinda stuck:
The graph is a vector<vector<int> >
adjacency list
The following is the DFS which I want to use for topological sort
bool visited[MAX]={0};
stack<int> dfs, postOrder;
vector<int> newVec;
vector<int>::iterator it;
for(int i=0;i<MAX;i++){
if(visited[i]==false){
dfs.push(i);
}
while(!dfs.empty()){
int node=dfs.top();
dfs.pop();
visited[node]=true;
newVec=graph[node]; //vector of neighboors
for(it=newVec.begin();it!=newVec.end();it++){
int son=*it;
if(visited[son]==false){
dfs.push(son);
}
}
}
}
In DFS, we start from a vertex, we first print it and then recursively call DFS for its adjacent vertices. In topological sorting, we use a temporary stack. We don't print the vertex immediately, we first recursively call topological sorting for all its adjacent vertices, then push it to a stack.
Topological Sorting can be done by both DFS as well as BFS,this post however is concerned with the BFS approach of topological sorting popularly know as Khan's Algorithm.
Topological sort simply involves running DFS on an entire graph and adding each node to the global ordering of nodes, but only after all of a node's children are visited. This ensures that parent nodes will be ordered before their child nodes, and honors the forward direction of edges in the ordering.
Yes it's always true. The reverse of a DFS on a DAG gives the topological sort.
In order to construct the postOrder
list you need to know the time when your algorithm has finished processing the last child of node k
.
One way to figure out when you have popped the last child off the stack is to put special marks on the stack to indicate spots where the children of a particular node are starting. You could change the type of your dfs
stack to vector<pair<bool,int>>
. When the bool
is set to true
, it indicates a parent; false
indicates a child.
When you pop a "child pair" (i.e. one with the first member of the pair set to false
) off the stack, you run the code that you currently have, i.e. push all their children onto the stack with your for
loop. Before entering the for
loop, however, you should push make_pair(true, node)
onto the stack to mark the beginning of all children of this node
.
When you pop a "parent pair" off the stack, you push the parent index onto the postOrder
, and move on:
vector<vector<int> > graph;
vector<bool> visited(max);
stack<pair<bool,int>> dfs;
stack<int> postOrder;
for(int i=0 ; i < max ; i++){
if(!visited[i]){
dfs.push(make_pair(false,i));
}
while(!dfs.empty()){
pair<bool,int> node=dfs.top();
dfs.pop();
if (node.first) {
postOrder.push(node.second);
continue;
}
if (visited[node.second]) {
continue;
}
visited[node.second]=true;
dfs.push(make_pair(true, node.second));
const auto& newVec=graph[node.second]; //vector of neighboors
for(const auto son: newVec){
if(!visited[son]){
dfs.push(make_pair(false, son));
}
}
}
}
Demo on ideone.
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