Problem Link Anyone please help me to understand the problem . For this example :
3 3 3 1 2 1 2 3
If we set
label[ 2 ] = 1 , label[ 3 ] = 2 , label[ 1 ] = 3 ; Answer should be : 2 3 1 but answer is 3 2 1 , but why ?
Here is two implementation of this problem
Time took by Sparse Table is 0.336 s ,where Segment Tree took 0.324 s. Memory took by Sparse Table is 8732 KB ,where Segment Tree took 3640 KB.
Problem is static,thats why Sparse Table should be fast!But here i couldn't see this..Where is the problem?Could anyone please find that out ?
Thanks in advance :)
I got Runtime Error!I have been trying to find out this fault for 4hours.But,i cant get my fault.Could anyone please help me where is my error? Code is here.If i got wrong answer,then i will check that out.But why i am getting Runtime Error? Problem Link is here probelm link
I am seeing this error for the first time in any judge.Can anybody have any idea what is it?oh!I have missed a point!CodeForces staus page today shows many judgement error! Thanks in advance!
Could anyone please explain me clearly why we have to do
Yeah!I know that is for lowest discovery time determination.But my question is that all the time we put minimum time in the current node comparing with its adjacent nodes,but not increase the time.My question is two or more nodes cant visit at a time.But lowest time show that,it can do that.But how? Could anyone please clear me out?Actually,i read more than 5/6 resources,but i couldn't get that.There all says about subtree,ancestor,blah blah,but my memory couldn't make a sense for that,what they actually want to say.
Thanks in advance.
Following is the psuedocode of finding out articulation point.
time = 0 function DFS(adj, disc, low, visited, parent, AP, vertex, V) visited[vertex] = true disc[vertex] = low[vertex] = time+1 child = 0 for i = 0 to V if adj[vertex][i] == true if visited[i] == false child = child + 1 parent[i] = vertex DFS(adj, disc, low, visited, parent, AP, i, n, time+1) low[vertex] = minimum(low[vertex], low[i]) if parent[vertex] == nil and child > 1 AP[vertex] = true if parent[vertex] != nil and low[i] >= disc[vertex] AP[vertex] = true else if parent[vertex] != i low[vertex] = minimum(low[vertex], disc[i])
I have tried my level best to understand this problem.Could anyone else please explain me how i solve this problem?
Note: I know Binary Indexed tree and want to solve this problem by this algorithm.Google search couldn't make a sense for myself to solve this problem.Everywhere,i saw,sort query according to it's value 'k'.And suggestion was that sort your given array also..But,i failed to get that!Pure explanation needed!
Thanks in advance!
Z[k] would be longest substring starting at k which is also prefix of the string. Now R-i + 1 gives the length of remaining characters in the Z box. Since we copy the value of Z[k] to optimize the calculation of Z values we need to check whether these values are inside the boundary of the Z box
Why not we calculate the remaining part including k index?K index is in the substring,so why skip this index?
I want to solve this problem by DSU.. But,unfortunately,i don't get that..i have already checked Tutorial,someone else code..but,till now,i'm in darkness It's an Bipartite graph base problem.But,what is happening here ? Anyone plz help me. Thanks in advance
To practise more on Number Theory realted problem,is there any oj except spoj,uva where i'll get a catagorized problem? And want to know that which topics on Number theory should i konw for competitive programming?Let's help me by giving helpfull tutorial link.
Thanks!Thanks in advance ! Happy coding!