Разбор Educational Codeforces Round 37

7 лет назад, # |

+46

I had a different idea for E.

First, let's take the vertices whose degree is bigger than N/2. Any two such vertices must be in the same connected component, because their neighbor sets must overlap.

What about the other vertices? Well, their degrees are smaller than N/2. That means that N-deg(v)>deg(v) for each of them. but the sum of N-deg(v) is at most 2M, hence their sum of degrees is at most 2M as well. So we can apply a regular algorithm (dfs, or dsu) to those vertices.

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Skyscrapers

7 лет назад, # ^ |

You have to extend the set of vertices where degree is bigger than N/2 to its maximal, since vertices of degree no more then N/2 may also in the set.

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7 лет назад, # ^ |

Yes, I guess I wasn't clear.

For the vertices whose degree is less than N/2, you process ALL their edges — the ones leading to each other, as well as to the big set.

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Skyscrapers

7 лет назад, # ^ |

You're right.

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7 лет назад, # ^ |

← Rev. 2 →

Thank you very much for this explanation.

Now I got E

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1moretry

7 лет назад, # ^ |

Sorry, I don't get this part:

but the sum of N-deg(v) is at most 2M, hence their sum of degrees is at most 2M as well

Why "N-deg(v) is at most 2M", and what does "N-deg(v)" imply? and how we can derive "their sum of degrees is at most 2M" from that?

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7 лет назад, # ^ |

N — deg(v) counts the vertices w, such that (v,w) is NOT an edge in the graph.

There are M pairs (x,y) that are NOT edges, and 2M if we count both (x,y) and (y,x).

Thus if we add up (N — deg(v)) for every vertex v, we will get 2M. And if we add up (N — deg(v)) for only some vertices, we will get a number at most as big as 2M.

Now, if deg(v) < N — deg(v), then the sum of deg(v) in the set is also less than the sum of N — deg(v) in the set, which we have already established is at most 2M.

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Hemanth_1

7 лет назад, # ^ |

Nice idea

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computer_jock

18 месяцев назад, # ^ |

can you share your code for this logic. ?

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https://codeforces.net/contest/920/submission/34908754

18 месяцев назад, # ^ |

Wow, my code looked so different 5 years ago

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7 лет назад, # |

someone know why i get RE on Test 9 by using it=vis.lower_bound(*it) ,setvis for unvisited node and auto it=vis(iterator) 34911809

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7 лет назад, # |

← Rev. 2 →

someone know why i get Re on test9 by using it=vis.lower_bound(*it) ,

set<int>vis,it=vis(itertator) 34911809 ,I get AC after changing *it to x (int x=*it)

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VLamarca

7 лет назад, # ^ |

I belive its because you should not keep using the iterator it after you erased it from the set

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7 лет назад, # |

+14

Editorial for E looks unfinished.

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7 лет назад, # ^ |

Could someone explain it in a better way/more understandable way? Or Can someone provide the Implementation?

Thanks.

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apar03

7 лет назад, # ^ |

Here's the implementation: http://codeforces.net/contest/920/submission/34871219

Just keep all the unvisited nodes in a set and keep erasing vertices as you visit them in the bfs. You are now selectively only going to traverse through the unvisited nodes, which brings down the complexity to O(NlogN) [use of sets gives the logN factor.

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flyman3046

7 лет назад, # ^ |

Is it good to use unordered_set instead of set?

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apar03

7 лет назад, # ^ |

Yes, it should make the execution a bit faster

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flyman3046

7 лет назад, # ^ |

I tried HashSet in Java, which is unordered_set in c++. It actually TLE. But TreeSet (set in C++) passes the test.

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apar03

7 лет назад, # ^ |

← Rev. 2 →

In C++, using unordered_set reduces time. Here's my submission using unordered_set: http://codeforces.net/contest/920/submission/34926295

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7 лет назад, # ^ |

Worst-case complexity of unordered_set is O(n) because it uses hashing. Average case is O(1), so there must be a case that brings out it's worse complexity.

However, TreeSet/Set will always be O(logn)

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flyman3046

7 лет назад, # ^ |

My java submission 34926586 using HashSet TLE. But apar03's c++ submission 34926295 works fine. Maybe it is due to different implementations in different languages.

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7 лет назад, # ^ |

← Rev. 2 →

add 1~n node in "unvisited node"set , take out one of the node untile set is empty, for the node you take out ,denote it U, find the edge"you can go",which is [upper_bound s — find the smallest integer y from the set such that y > s.] in the tutorial and those nodes you find from node U will be one of the component.

keep the graph bu using STL:set it will be easy to implement.

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7 лет назад, # ^ |

I tried to implement and I'm getting Runtime error in CF whereas it is working fine on my computer. I got the error, it is because the set is dynamically getting changed during the recursion.

void dfs(int i,int p){
    merge(i,p);
    set<int>::iterator it;
    for(it=S.begin();it!=S.end();it++){
        if(M[ii(i,*it)]!=1){
            int x=*it;
            S.erase(x);
            dfs(x,i);
        }
    }
}

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tc-imba

7 лет назад, # ^ |

← Rev. 2 →

You made a mistake on C++ container iteration. When you want to erase something in an iteration, be sure to update the iterator, or it becomes an undefined behavior. For example, you should change your code to

void dfs(int i,int p){
    merge(i,p);
    set<int>::iterator it;
    for(it=S.begin();it!=S.end();){
        if(M[ii(i,*it)]!=1){
            int x=*it;
            it=S.erase(x); // it=S.erase(it) would be better
            dfs(x,i);
        }else ++it;
    }
}

All erase in iteration should be done so, see the cppreference of std::set::erase, or any other container's erase function

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7 лет назад, # ^ |

Yeah, thanks, man..! I got. I fixed it using the following.

void dfs(int i,int p){
    merge(i,p);
    set<int>::iterator it;
    for(it=S.begin();it!=S.end();){
        if(M[ii(i,*it)]!=1){
            int x=*it;
            S.erase(x);
            dfs(x,i);
            it=S.lower_bound(x);
        }
        else
        	it++;
    }
}

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7 лет назад, # ^ |

The advantage of my approach to yours is that it works for C++98 also, whereas yours works only for C++11 and above.

Anyway, Thank you.

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7 лет назад, # ^ |

← Rev. 3 →

I got it fully now.

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7 лет назад, # ^ |

Unfortunately, I'm not sure what to write there additionaly.

The algorithm is just typical DFS, but instead of iterating on adjacency list of current vertex and trying to find an unvisited vertex, we have it backwards: we iterate on the set of unvisited vertices and try to find the one that can be reached from current vertex.

Perhaps my implementation can help to understand it better.

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7 лет назад, # ^ |

I was confused about how we choose next vertex, so you could write about step "try to find the one that can be reached from current vertex" in details.

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7 лет назад, # ^ |

If the question is on a directed graph. I think we have to Iterate on all the edges of the set. Right? However, if we iterate on all the vertices in the unvisited set in this question also, I think the complexity remains the same because atmost after deg(v) (where v is a vertex in unvisited set and deg(v) is the degree of v in the given graph) v will be removed from the unvisited set. So overall complexity will be m*log(n).

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skpro19

7 лет назад, # |

-6

Can someone explain G?

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skpro19

7 лет назад, # ^ |

-11

What is p, x and k in the editorial?

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rkm0959

7 лет назад, # ^ |

+27

Did you even read the problem statement?

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7 лет назад, # |

Что за рюкзаковая динамика? В интернете не находится ничего адекватного по этой фразе.

Неужели трудно сделать разбор задачи чуть подробнее, видя что решили её единицы, при том что она на уровне сложности D.

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trillioner

7 лет назад, # ^ |

Поищи knapsack problem.

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Marco_L_T

7 лет назад, # |

For Problem E,can the problem be solved in the same complexity if the graph is directed?

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7 лет назад, # ^ |

if you cant add edge 1->2 it means you can add 2->1 ,so if the still add the edge 1->2,because the probelm is asking component so it doesnt matter. I think so...

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peltorator

7 лет назад, # |

In first string of problem G you have a little mistake. Not gcd(z, y) = 1 but gcd(z, p) = 1

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7 лет назад, # ^ |

Thank you, fixed.

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zhed

7 лет назад, # |

For F, you can do it with a BIT and a set. Store sums in the BIT, then for range updates, you can store the indices of all numbers that aren't 2 in the range in the set with lower_bound and traversing forward. If you set something to 2, erase it from your set.

The solution is still O(qlog n) but should be easier/faster to code.

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roll_no_1

6 лет назад, # ^ |

Here's my implementation of this idea with a segment tree and a set: 45337647. Hope it helps anyone looking to implement the problem this way as its much easier than the 2 segment trees idea in the editorial.

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ghoshsai5000

7 лет назад, # |

Can someone share their model solution to F ? I am not able to understand most of the codes.

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aijey

7 лет назад, # |

← Rev. 4 →

Hi guys, I'm having trouble with problem F.Here is my solution .I did everything that was written in editorial but my solution doesn't fit time limit in test case #69. Moreover, it works on other testcases ~1900 ms. Can somebody give me some pieces of advice?

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aijey

7 лет назад, # ^ |

I'd finally fixed it. In update procedure added one more verification. If somebody needs, here is my final solution

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Brankonymous

7 лет назад, # |

Can someone explain to me G-List Of Integers a bit more detailed? I don't understand the editorial solution

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7 лет назад, # ^ |

Well, the main difficulty is that you have to be familiar with inclusion-exclusion principle.

Let's use binary search to find the answer. Then we need to somehow calculate the number of good integers (good integers are coprime with p) from integer x to integer mid (the middle element in binary search). It's better to rewrite it as count(mid) - count(x), where count(x) is the number of good integers not exceeding x.

Now we somehow have to calculate this count(x). We need to find all good integers from [1, x]; these are all integers from [1, x] excluding those that are divisible by some prime divisor of p. To find the latter, we use inclusion-exclusion principle: let our sets A₁, A₂, ..., A_n be the sets of integers divisible by the first, the second, ..., the n-th prime divisor of p. And then we apply inclusion-exclusion formula as it is — since the maximum number of primes in factorization of p is 7, then we can just iterate through all possible combinations of these sets A₁ ... A_n.

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Brankonymous

7 лет назад, # ^ |

← Rev. 2 →

Thank you so much, I finally understand!

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T1duS

7 лет назад, # |

Can inclusion-exclusion be used to calculate euler's phi function? Won't it be a special case of it?

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7 лет назад, # ^ |

Well, actually if we set in A(p, y) mentioned in editorial p = y, then we will get exactly φ(p). So I believe Euler's function can be calculated using inclusion/exclusion, it's just not convenient to do it this way.

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7 лет назад, # |

← Rev. 2 →

I just wrote the same solution described in editorial and got TLE http://codeforces.net/contest/920/submission/34930869

Is the constant factor bad, or have I misunderstood something?

Nevermind, I'm doing range updates wrong.

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7 лет назад, # ^ |

← Rev. 3 →

The thing is that you can't just try updating each of the indices from [l, r] if you have REPLACE query.

You have to use the segment tree for your updates. It is difficult for me to explain. You have to make the queries in a similar manner to mass change queries in segment trees, but instead of pushing some values you need to do the following if you have to update the whole segment:

1) If the maximum element on this segment is 1 or 2, then just stop updating the segment, you don't need to do anything in it.

2) Otherwise, if the segment contains more than one element, divide it into two segments (just the way segment tree does it) and try updating these segments.

I think that our implementation may help. Take a look at how we have written upd function.

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