Seeing authors solution , I think wouldn’t it be easier if we just use C++ bitset ? which supports the operation . I am getting WA though here
What is wrong with my code: CodeChef: Practical coding for everyone ?? i expected that it will give TLE, but it gives WA.
seriously thought of strassen’s algorithm but dropped the idea thinking it wont work…damn !!
My unfinished code is given below:
sa=int(input())
flist=[int(input(), 2) for t in range(sa)]
total=0 for i in range(sa-1):
for j in range(i+1, sa):
if flist[i]&flist[j]!=0:
total+=1
The code finds mutual friends; however, if A and B are already friends and share common friends, total is increased by 1,which is unwanted. I am at a loss at how to implement this condition efficiently; furthermore, the editorial does not explain how to do this either. What is the best way to include this restriction?
Thanks,
minimario
Complexity can be reduced to nn(n-1)/128.
The denominator comes from 64 and 2. Use long instead of int. Using symmetry, we compare each element only with lower indexed users. That’s 1+2+…+n-1=n(n-1)/2. We do this n times. This takes 6.25*10^7 operations, well within time constraints.
Not really… using long and checking for only lower indices, the operations can be reduced to (n)(n)(n-1)/((64)*(2))=6.25X10^7 operations…That falls well within the time limit.
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nice tutorial 
but can someone help me how to solve with strassens matrix multiplication. Can i get a code for implementing it??
I solved it using binary search.
Firstly I stored all the pairs which were not friends together as a pair in a vector.
Then I checked if both of these person have any common friend , by binary search .
If yes , ans += 2 .
My code : CodeChef: Practical coding for everyone
Elegant and Simple .
quality & timing of editorials have seriously improved on codechef…nic work
why tle ? 
please help…
will this work? all pair shortest paths of cost 2
Here are solutions both in JAVA and C++.
Both of them use BitSet concept. I’ll try and write this in terms for even a layman to understand. If you’re a beginner, i’m assuming you know what the ‘&’ bitwise operator returns for 2 binary values. If you don’t then just google it. It’s pretty easy
C++:
- build up the bitset. ‘bitset gr[no. of vertices]’
- now run a loop for each vertex. Check for the vertices for this vertex which do not share a common edge (bit will not be set e.i 1 here)
- simply put if((i!=j)&&((gr[i]&gr[j]).any()>0)) then ans++; [.any() method checks if the number gr[i]&gr[j] has any bit set as 1. This would obviously physically mean that they share one or more than 1 common edge]
- Voila!. Once you’ve interated through all loops, you have your answer.
My C++ submission: CodeChef: Practical coding for everyone
JAVA:
Same concept. Its gets much easier here.JAVA has an awesome ‘Bitset.intersects(BitSet)’ method. Intersect means that they have atleast one common bit set.
1.Create an ArrayList of Bitsets and build the graph up.
2.Now same. run a loop for each vertex to check for non adjacent vertices. After that, simply check when (i!=j && !gr.get(i).get(j)) → if(gr.get(i).intersects(gr.get(j))), ans++;
CAUTION: the second .get(j) is a bitset method. Not the arrayList one. It checks if the bit is set at position j
DONE!
My JAVA submission: CodeChef: Practical coding for everyone
Meanwhile, i’m picking up some of my old WA and TLE submission and try and make a few optimisations here 
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here is my solution-i2Hw32 - Online Java Compiler & Debugging Tool - Ideone.com. can any tell why it is showing wrong answer!!
can we solve it using Disjoint sets concept
consider 1 and 2 to be friend 2 and 3 to be friend and 3 and 4 to be friend then 1 and 4 does not have any mutual friends as friend of 1 is 2 and friend of 4 is 3.
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I have to say i love the idea of 2 persons interacting with each in the editorial. Amazing stuff!!
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is O((n^3) / 32) fast enough to get AC? how much operations do your servers do per second? where can I see the specs of the cpu?
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Runtime error!
N is at most 1000.I don’t think so!!!
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