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This time gfg did a good thing by assigning you to teach us.
Making us understand the problem - AGetting us to understand the approach - APresentation - AWriting in Code - F
Good explanation.NIT: Actually we can skip visited array using DP at its place.
Great explanation
Wrong time complexity. It should be O(V+E) as we are doing DFS and travelling all edges of the graph.
yes
Could someone explain me why?
This video really helped me a lot! I appreciate it!
Wow. I LOVED this video. It might come in GATE 2021
hey! is there a way to do this in an iterative form?
this was so simple, yet beautiful
Bang on Target, we are loving this.As always thanks for posting such excellent quality content, keep them rolling.
Any way you could add some test cases for the write up on the site? There isn't a problem anywhere that has them.
I enjoyed it . thank u
the hot dude without Indian accent was actually a nice surprise :D
Can you make a video on all possible paths from source to destination using dynamic programming?
cin>>n>>m; for(int i=1;iq; while(m--){ int a,b,c; cin>>a>>b>>c; if(a==b){ dis[a][b]=0; dis[b][a]=0; } dis[a][a]=0; dis[b][b]=0; dis[a][b]=min(dis[a][b],c); dis[b][a]=min(dis[a][b],c); } for(int k=1;k>b; if(dis[a][b]
Great explanation ☺
nice explanation !!!!!
Excellent explanation! Great job!
really useful
Why the need for visited?
how would this work if we were looking for a path from s to t? , that being where we already specified where we want to start and end
dfs?
What is the definition of longest path in DAG?Can you traverse same node multiple time while making sure chosen edge is not repeated?
Man. This kind of helps my issue, but I'm really trying to determine the longest path between two nodes in a weighted, directed graph. Not just the longest path in the graph in general. Hmmmmmmmmm.
bros a lefty... dam-
He is so gay - Arpit bala
This time gfg did a good thing by assigning you to teach us.
Making us understand the problem - A
Getting us to understand the approach - A
Presentation - A
Writing in Code - F
Good explanation.
NIT: Actually we can skip visited array using DP at its place.
Great explanation
Wrong time complexity. It should be O(V+E) as we are doing DFS and travelling all edges of the graph.
yes
Could someone explain me why?
This video really helped me a lot! I appreciate it!
Wow. I LOVED this video. It might come in GATE 2021
hey! is there a way to do this in an iterative form?
this was so simple, yet beautiful
Bang on Target, we are loving this.
As always thanks for posting such excellent quality content, keep them rolling.
Any way you could add some test cases for the write up on the site? There isn't a problem anywhere that has them.
I enjoyed it . thank u
the hot dude without Indian accent was actually a nice surprise :D
Can you make a video on all possible paths from source to destination using dynamic programming?
cin>>n>>m;
for(int i=1;iq;
while(m--){
int a,b,c;
cin>>a>>b>>c;
if(a==b){
dis[a][b]=0;
dis[b][a]=0;
}
dis[a][a]=0;
dis[b][b]=0;
dis[a][b]=min(dis[a][b],c);
dis[b][a]=min(dis[a][b],c);
}
for(int k=1;k>b;
if(dis[a][b]
Great explanation ☺
nice explanation !!!!!
Excellent explanation! Great job!
really useful
Why the need for visited?
how would this work if we were looking for a path from s to t? , that being where we already specified where we want to start and end
dfs?
What is the definition of longest path in DAG?
Can you traverse same node multiple time while making sure chosen edge is not repeated?
Man. This kind of helps my issue, but I'm really trying to determine the longest path between two nodes in a weighted, directed graph. Not just the longest path in the graph in general. Hmmmmmmmmm.
bros a lefty... dam-
He is so gay - Arpit bala
Great explanation