Any border test cases? I am still getting WA
http://www.codechef.com/viewsolution/1954203
@anton_lunyov >> Please clarify this:
for (VI::iterator v = a[u].begin(); v != a[u].end(); ++v) {
// *v is the adjacent vertex itself
if (!mark[*v]) {
// if it was not visited before we move to it
dfs(*v);
}
}
In the loop part, the begin() and end() functions are called each time an iteration occurs, so isnāt it time consuming? Instead if we just store
x = a[u].begin()
and
y = a[u].end()
before entering the loop and then use x and y, do we save some time?
Can anybody explain what is the runtime of Union/Find approach. I used union by size approach.
Plz help me out in thisā¦i thought a lot on this problem and finally did itā¦though i got a WA i canāt figure out what was wrong with my codeā¦i wrote a c program which can be seen at
http://www.codechef.com/viewsolution/1900728
url:- CodeChef: Practical coding for everyone
I m getting WA cna someone help in finding a test case where my code fails??
thank u in advanceā¦
can any one tell me whats wrong with this codeā¦
its running on my pc but giving compilation error
http://ww2.codechef.com/viewsolution/2076377
Please help . Getting Wrong answers for even test case!!!
link to my code ā lU40Sl - Online C++0x Compiler & Debugging Tool - Ideone.com
well sorry for reviving such an old topic but there seems to be an error with my implementation,i dont know why i am getting a TLE here is my submission CodeChef: Practical coding for everyone i have used a dfs and a linked list approach could anyone illuminate my mistake.
Please star my projects and contribute if you are interested.
Can someone please point out the mistake in this code?
Code-(CodeChef: Practical coding for everyone)
Thanks
such a nice problem, thank you codechef.
Used union find but kept getting TLE. Code:gDZvZc - Online C++0x Compiler & Debugging Tool - Ideone.com
Would be really helpful if someone could point out the mistake in this.
very good question ā¦
You have loop for(int j=i; j<n; j++)
after each call to dfs. In the case of no edges in the graph you will have O(N^2)
solution. Obviously it should get TLE
@anton_lunyov >> So, if I consider the case of graph consisting of no edges separately, will it run within TL?
Replace c[p[i]]++
by c[find(i)]++
p[i]
is not always the actual root of the disjoint set containing i
.
Namely, after link operation.