Codeforces Round #597 (Div. 2)

5 years ago, # ^ |

+13

huh.. very complicated, i will go and change the css to red.

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NeiH

5 years ago, # ^ |

+17

It's very trouble :<

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freesteed

5 years ago, # |

Expect for the constest!

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5 years ago, # ^ |

want for the contest!

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huangzirui

5 years ago, # |

I wish I can get expert in the contest again. :)

Good luck to everyone!

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hjain1157

5 years ago, # |

+21

I m back Bois.. Had a busy October but now I am back to top the charts again.

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IstiakAhmedSifat

5 years ago, # |

+10

The contestants had to face some difficulty in the previous contest due to several server failure. Though it was overcome in few seconds but its disappointing to reload the page over and over again during the contest hour. Div-2 rating changes also had to roll back because of some problems and we had to wait for several hours to get the rating changes in the last contest. Hope the authority will take care of these matters in this round.

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DeliciousFlatChest

5 years ago, # |

Auto comment: topic has been updated by DeliciousFlatChest (previous revision, new revision, compare).

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Falla7

5 years ago, # |

Best of luck in your first contest!

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yogi

5 years ago, # |

Why first and second question has same score distribution....does that mean they have same difficulty level??

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amoo_johnny

5 years ago, # |

← Rev. 2 →

-8

Hope we have a fair race :)

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Ketchum

5 years ago, # |

-8

I hope that I can turn it into a blue name.

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ValeriyG

5 years ago, # |

← Rev. 2 →

-35

Another DISGUSTING reading contest.

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v-O_O-v

5 years ago, # ^ |

+17

Reading is also a part of CP. So you should adjust yourself to read long questions. Else make the question look smaller to you. How? You figure it out ;).

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ValeriyG

5 years ago, # ^ |

← Rev. 2 →

-15

I don't like to read tonns of text after hard work day. It tires the brain completely.

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v-O_O-v

5 years ago, # ^ |

+14

Do you think everyone is only doing CP? Are they not tired?. Let's not make any excuses and just read the question there is no other way.

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Zadymiarz12345

5 years ago, # ^ |

Well, I guess that's just your excuse for not being able to solve problems on this contest. The statements were actually well written and short. Go practice instead of complaining.

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Origenes

5 years ago, # ^ |

+22

Well, I don't know if the statements were well-written but problem E's statement is definitely not short.

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Zadymiarz12345

5 years ago, # ^ |

You're right, they were well written except for E. When I saw that statement I didn't even try to read it and went for problem F.

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grg124

5 years ago, # |

← Rev. 3 →

strong pre test cases for problems at code forces I ever seen!!

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It_Wasnt_Me

5 years ago, # ^ |

Strongest pretest 13 I have ever seen

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It_Wasnt_Me

5 years ago, # |

Wrong answer on pretest 13 on problem D, No idea why. :(

Any one have idea what is going on in pretest 13 ?

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5 years ago, # |

← Rev. 2 →

TC 9 of problem C?

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SkySurfer

5 years ago, # ^ |

Were u taking modulo when calculating the fibonacci ????

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5 years ago, # ^ |

That test case got me too. I spent the last hour trying to fix whatever was wrong my fibonacci-number calculator.

I was modding by 10^9+7 everywhere I could see, but for some reason, my 89-th and onward Fib numbers were wrong. So of course the Fib products were also wrong too. Please send help

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5 years ago, # ^ |

I think. You should replace ret*=fibNum(x); as ret = (ret * 1ll * fibNum(x)) % M and also temp = (temp1 * 1ll * temp2) % M;

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5 years ago, # ^ |

Not sure what the || (or? or something else...) is meant to do or what difference ret*=a vs ret=ret*a would make, care to explain?

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shahnirmit2503

5 years ago, # ^ |

I think he means 1LL. LL makes it to long long

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5 years ago, # ^ |

that is 1LL.

which leads to long long int.

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5 years ago, # ^ |

Your solution will get AC after doing that.

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5 years ago, # ^ |

Thank you so much for explaining that starboy_jb. I will never forget to cast again.

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5 years ago, # ^ |

If you need help tell the approach of your solution rather than asking what was the test case. After some time you will get to know it right? :)

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5 years ago, # ^ |

← Rev. 2 →

I found out that ways to arrange characters of length i which are consecutive, can be found out by this formula,

Pseudocode

if(i is even) dp[i] = dp[i-1]*2 +1;
else dp[i] = dp[i-1]*2 - 1;

In my solution, i was doing like this:

 dp[0] = dp[1] = 0LL;
 dp[2] = 2LL;
 dp[3] = 3LL;
for(int i=4;i<=sz(s);i++){
        dp[i] = (dp[i-1]*2LL)%MOD;
        if(i%2) dp[i] = (dp[i]+1LL)%MOD;
        else dp[i] = (dp[i]-1LL + MOD)%MOD;
        //cout<<i<<" "<<dp[i]<<"\n";
    }

Is the problem with Modulo that i am doing? Please someone help.

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5 years ago, # ^ |

← Rev. 2 →

First of all, your first "if" statement has identical branches.(now ok)

Secondly, dp[1] should be 1

Thirdly, by the formula you have given dp[4] = dp[3] * 2 + 1 = 7, but actually dp[4] = 5. For example :

intput:

uuuu

1.uuuu 2.wuu 3.uwu 4.uuw 5.ww

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5 years ago, # ^ |

Here is my output of first 10 numbers. Check once again, i think you are misreading something.

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5 years ago, # ^ |

dp[5] = 8 not 11

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5 years ago, # ^ |

Then how I was able to pass 8 test cases? lol

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5 years ago, # ^ |

till 8 TC,

the continuous occurrence of n and u is not greater than 4.

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5 years ago, # ^ |

Thanks anyways, I was calculating repeating cases again and again.

1.uuuuu  2.wuuu 3. wwu 4.wuw
         5.uwuu 6.uww
         7.uuwu 8.wwu
         9.uuuw 10.wuw 11.uww

where 8,10 and 11 are repeating. Wasted 1 hour thinking about something wrong with computations.

Thank you :)

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5 years ago, # ^ |

I calculated according to your first two lines of your comment (the if else part), you said dp[i] = dp[i-1] * 2 + 1 if i is even, maybe you have mistaken even with odd there, anyway, your formula is incorrect I guess.

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cocoaNcheese

5 years ago, # |

+17

how to solve D?

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dummkopf

5 years ago, # ^ |

Kruskal or Prim to find smallest tree. However the graph must be modified a little bit. Add a new node (let's say it's the power supply for the power stations) and the edge from this source to each of n nodes is equal to the cost to build a station there.

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phantom11

5 years ago, # ^ |

Dont we need to make sure that there is atleast one power station? Wouldnt your approach fail when the power stations cost are really high, but the points are close by?

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shenxy13

5 years ago, # ^ |

However, your minimum spanning tree will also need to include the new node, which represents a sort of "super power supply" and ensures that at least one node is connected to the new node, and thus has a power station.

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dummkopf

5 years ago, # ^ |

Hmm? a tree is a connected graph, thus the newly added node (power supply) must belong to tree too. If the power supply is connected to node i, then you put a station at node i. Since the supply is always connected to some city because of connectedness, there is always at least one power station.

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mraron

5 years ago, # ^ |

No because we are finding a spanning tree which is connected if the original graph is connected too thus there'll be a path from the power station to every city.

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barun511

5 years ago, # |

+25

What the hell was pretest 13 in D :(((

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sourabh_jangid

5 years ago, # ^ |

← Rev. 2 →

Did you consider the case when $$$(x_i, y_i) = (x_j, y_j)$$$ even if $$$i!= j$$$. I think this is the mistake, I am not sure though, because I too got wrong answer on test case $$$13$$$. I didn't consider this case, so maybe this is the mistake I made.

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pk842

5 years ago, # ^ |

← Rev. 2 →

IS it matters? Because dijkstra will take care of this fact I guess. CAn you please give the example by which you debug your code?

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barun511

5 years ago, # ^ |

I considered this case. I was not finding MST though, which I think was the expected soln

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not_original

5 years ago, # |

approach for problem C?

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algo3rythm

5 years ago, # ^ |

I was getting some kind of fibonacci sequence

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SeismicToss

5 years ago, # ^ |

say we have len length of string of either n or u, in how many ways we can make pair in the string

so if we make x pairs of 'n' ,we are left with len-2*x 'n' characters and we have x+1 blocks where we can fill them , so distribute len-2*x objects among x+1 persons such that each get zero or more with (len-x,x) ways

and iterate x from 0 to len/2

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5 years ago, # ^ |

Any chain of U's can be broken up in fib(length) ways:

Proof: if there are n of them: UU...U we can condition on whether to turn the first U into a W, or not: If we do, then it becomes W(U...U) (with n-2 U's) If we don't, then it becomes U(U...U) (with n-1 U's)

Now each k-chain of U's or N's in the string has Fib(k) possible strings, so we multiply Fib(k) over all the k's

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algo3rythm

5 years ago, # ^ |

← Rev. 2 →

Nice

But how did you calculate the Fibonacci sequence?

I was doing mod 1e9+7 during the calculation and I'm getting wrong answer on pretest 6

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not_original

5 years ago, # ^ |

Any chain of U's can be broken up in fib(length) ways

How did you come up with this intuition?

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5 years ago, # ^ |

For my case...I understood this by observation, I calculated the results for the first few lengths of chain and then realized the pattern :3

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5 years ago, # ^ |

← Rev. 3 →

Consider u as 1 and w as 2 , then how many possible combinations exist such that sum is n.

Let n = 4

uuuu = 1 + 1 + 1 + 1

wuu = 2 + 1 + 1

uwu = 1 + 2 + 1

uuw = 1 + 1 + 2

ww = 2 + 2

There exists 5 ways which is nothing but fib[4].

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5 years ago, # ^ |

I saw it in a book (Proofs That Really Count) in the first chapter. It explained that fibonacci numbers can be thought of as the number of ways to tile a 1xn grid using (unlimited) dominoes of size 1x2 and 1x1. There you condition on whether you place a rectangle or square domino in the leftmost place.

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SeismicToss

5 years ago, # |

+15

what is test case 13 in D

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barun511

5 years ago, # |

+10

I thought this was an amazing round :D What lovely well framed questions

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5 years ago, # |

Does problem D using Prime Algorithm? I did it but got WA on test case 4. What is TC 4 :(

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EternalAlexander

5 years ago, # ^ |

+25

I think you mean the Prim Algorithm. As far as I know, yes.

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5 years ago, # ^ |

+14

Yes, Prim Algorithm. Sadly, i still didn't solve it. After solving A, B, C for around 50 minutes. I came to problem D, the statement is so long and demotivate me a lot. After relaxing 10 mins, I decided to read the statement because I had 1 hour remaining time and the problem didn't hard as much as I think. If I started to solve it sooner, I would have more chances to debug. It's experiment for me. Very nice contest =)))

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EternalAlexander

5 years ago, # ^ |

+43

Yeah, I understand. I was puzzled too when I saw the super long statement. Just calm down and don't be afraid of those things. I believe you can do much better in the future. Good luck!

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meme

5 years ago, # ^ |

+13

Slightly off-topic, but I love comments like this for the example they set for the CF/CP community. :')

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5 years ago, # ^ |

I solved D with Prim's algo.

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5 years ago, # ^ |

← Rev. 2 →

Dijkstra?I did using Dijkstra....But I guess the algorithm is similar XD

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5 years ago, # ^ |

It's almost the same algorithm.

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5 years ago, # ^ |

← Rev. 2 →

-11

-is-this-fft- Is this prim?

Code

vector<int> indices;
    vector<pair<int, int> > edges;
    long long ans = 0;

    int tot  = 0;
    while(!q.empty()) {
        pair<long long, pair<int, int> > tp = q.top();
        q.pop();
        ans += tp.first;
        if(mark[tp.second.first]) continue;
        mark[tp.second.first] = 1;
        if(tp.second.second == -1) indices.push_back(tp.second.first);
        else edges.push_back(tp.second);
        for(int j = 0; j < n; j++) {
            if(mark[j]) continue;
            int i = tp.second.first;
            long long m = abs(x[j] - x[i]) + abs(y[j] - y[i]);
            long long cost2 = m * (k[j] + k[i]);
            if(cost2 < cost[j]) {
                q.push(make_pair(cost2, make_pair(j, i)));
            }
        }
        tot++;
        if(tot == n) break;
    }

Edit: I need to re-order these two statements for AC. lol

ans += tp.first;
if(mark[tp.second.first]) continue;

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5 years ago, # ^ |

It seems so

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5 years ago, # ^ |

-8

hmm.. 13 is a devil's number i must say.

test cases should be 1..12 and then 14..30

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5 years ago, # ^ |

Kruskal is fair enough

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sillentKiller

5 years ago, # ^ |

Check out if you have used "long long int" instead of "int". I also got WA at test case 4, and this helped me :)

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5 years ago, # ^ |

I used long long instead of int, but It didn't solve. Btw, what is the difference between long long and long long int? are they same?

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sillentKiller

5 years ago, # ^ |

← Rev. 2 →

Yes, they are identical.

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12mango

5 years ago, # ^ |

My situation is exactly the same as yours.Long long didn't save me.(•́ω•̀ ٥)

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12mango

5 years ago, # ^ |

Hey bro~I find a lil issue in my code,I defined a variable as int but it should be long long.Thus the program got overflow.Maybe you should check the type of each variable carefully.

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5 years ago, # ^ |

Thanks bro, I definitely solve it again. I have just woken up.

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5 years ago, # ^ |

I lack of dummy city.

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pk842

5 years ago, # |

What is pretest 13 of D? Wasted almost half hour :(

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rishi_07

5 years ago, # ^ |

← Rev. 2 →

Try this:

4 
1 1
1 2
1 3
1 4
100 101 102 2
2 2 2 100

Ans: 110

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skarlord

5 years ago, # ^ |

Try this test

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dummkopf

5 years ago, # |

UwU

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5 years ago, # |

+19

dpforces

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Misericorde

5 years ago, # ^ |

So true.

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harshit15

5 years ago, # |

How to solve F?

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5 years ago, # ^ |

Dp with digits (in this case is binary digits).

You have a+b = a xor b + a&b. Then a + b = a xor b if and only if a&b = 0.

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Origenes

5 years ago, # ^ |

+10

I pre-solved it with digit dp. Let $$$f(pos, smaller_1, larger_1, smaller_2, larger_2)$$$ be the number of way to fill the first $$$pos$$$ binary bits such that [ the first number's prefix is already smaller than $$$r$$$ ], [ the first number's prefix is already larger than $$$l$$$ ], [ the second...], [ the second...] and their prefixes have no set bits in the same position so far. The transition is not hard to come up with. The answer is the sum of all entries in $$$f(0)$$$.

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phantom11

5 years ago, # ^ |

I am trying to understand your solution and transition. Could you please explain the transitions and the continue statements? I think this is a much generic approach. Thanks

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Origenes

5 years ago, # ^ |

PMed you :)

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__Apocalypse__

5 years ago, # ^ |

Could you please forward the message to me too. I am also confused as to what would be the transitions. Thanks in advance.

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TShT

5 years ago, # ^ |

Could you please forward the message to me too ))

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ItzABC

5 years ago, # |

I finally realized that it was problem E until I submitted my code(since mirror sites don't show the indices of problems
I fooled myself XDDDDDD

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5 years ago, # |

+16

Nice round. F is a fairly common problem btw. But it is still nice :)

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DeliciousFlatChest

5 years ago, # |

Auto comment: topic has been updated by DeliciousFlatChest (previous revision, new revision, compare).

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coderanant

5 years ago, # |

Nice Contest and problems.

But A was quite similar to this problem.

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CP_Sucks

5 years ago, # ^ |

orz

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po1ycarp

5 years ago, # |

← Rev. 3 →

Find out why this gets WA8 (takes less than 5 seconds), i couldn't during the whole contest :DDDDDDDD

Solution

Question #2: which lines cause the trouble?

Answer

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djm03178

5 years ago, # ^ |

Answer to #2. Because the range of index starts from 0, not 1.

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Partharora1010

5 years ago, # |

Ques.D) Shichikuji and Power Grid: In this question which concept and data structure are used. I think so Kruskal for the shortest path and DP for choosing the powerhouse or connection.

Suggest or provide some explanation, I am a beginner for this type of question.

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coderanant

5 years ago, # ^ |

← Rev. 2 →

I used priority queue to solve.

First, push the cost of setting plants at each city into priority queue along with city number.
Then, start where cost of setting plant is smallest.
Then push the cost making a connection from this city to every unelectrified city (along with city details) and push them into priority queue. Extract minimum cost (setting plant or making a connection) and go to step 3 (n-1 times).

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Partharora1010

5 years ago, # ^ |

okay cool, a different approach

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fried-chicken

5 years ago, # ^ |

+12

it sounds like you just invented Prim algorithm lol

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5 years ago, # ^ |

My solution: Using Kruskal:

Create all the possible edges and their weights(using formula in the statement). Sort the edges according to their weights.

Suppose you are trying to connect the connected component U and the connected component V using the edges (x, y). Basically you have 2 options:

Build a power station in a city in U and a power station in a city in V (Just pick the city with less cost). You don't have to build the edge (x, y) anymore.
If you decide to build the edge (x, y) to connect U and V, then you have to build a power station in a city in the union of U and V. Just pick the smallest one.

Just pick the optimal options for each edge (x, y) and add up the cost.

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yougaindra

5 years ago, # |

In A I guessed from test cases that a and b needs to be coprime, and the submission worked, But I don't get why is this correct answer. For example a = 3, b = 7 there are infinitely many numbers which are coprime to 3 as well as 7, then aren't there infinite black numbers?

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Tsugiru

5 years ago, # ^ |

Check this: https://en.m.wikipedia.org/wiki/Coin_problem

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GLAYS

5 years ago, # ^ |

← Rev. 2 →

A number $$$X$$$ will have color white iff $$$X$$$ can be written as a combination of $$$(a,b)$$$

==> There exists $$$k_1,k_2>=0$$$ such that $$$X=k_1*a+k_2*b$$$.

So the numbers $$$X$$$ having color black will be the ones such that that equation has no solutions. Since that's the Diophantine Equation(to which there's either an infinite number of solutions, or there are none), having no solutions means $$$X$$$ is not a multiple of $$$gcd(a,b)$$$.

So if $$$gcd(a,b)=1$$$ then there's no such X.

Else there will be many; infinitely many.

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5 years ago, # |

+23

D can be solved using a trick: create a virtual vertex, let's denote it as n + 1, then for every vertex connect it to (n + 1)th vertex with an edge having a weight equal to the vextex's c value. Then the problem becomes finding the MST of a given graph.

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5 years ago, # ^ |

You don't need a virtual vertex for Prim's algo.

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5 years ago, # ^ |

I used Kruskal

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5 years ago, # ^ |

Woah !! Awesome trick.

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Roach00

5 years ago, # ^ |

can you please tell how you deduced that mst will work here, I mean it looks like baccktracking with unioin find to me.

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anime

5 years ago, # |

+26

The competition was awesome !!

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Bilal.Coder

10 months ago, # ^ |

← Rev. 2 →

Assalomu aleykum!

Beksultan aka kandaisyz? Men I.Razzakov liseyinde 8-klassmyn. ICT olymp. 2023 Summer Camp 7-klassymda bargam, Alihan Mingbayev menen. Siz sabak bergensiz. Siz menen bailanyshyp kenesh suraiyn degem. Oblastnoygo otkom, daiardanyp jatam azyr. Kantip, kanday chalywsam onugom? degendey suroolor. Oshogo bailanyshyp koiunuzchu suranych?!

Instagram : https://www.instagram.com/bilal.oo99/ Codeforces : https://mirror.codeforces.com/profile/Bilal.Coder Telegram : https://t.me/BilalCoder

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5 years ago, # |

Maybe I'm the only one who solve B using dp and C using 2-dimension dp (I didn't think of Fibonacci or any stuffs related to it)

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5 years ago, # ^ |

Do you use EternalFire ?

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5 years ago, # ^ |

ofcourse it's a certificate I got last year after participating a programming contest and now it's my mousepad

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asipu_pawan

5 years ago, # |

← Rev. 2 →

In problem C, I was using square free Fibonacci series. But, I got WA on pretest 9. Can someone help me?? My solution

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5 years ago, # ^ |

Square free Fibonacci series is wrong. You should use normal Fibonacci series.

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asipu_pawan

5 years ago, # ^ |

Ok, I got my mistake. Thanks

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5 years ago, # ^ |

This might overflow since both are int:

ans= (ans * k)%mod;

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asipu_pawan

5 years ago, # ^ |

I have used #define for that

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liouzhou_101

5 years ago, # |

+40

Anyone notices that there seems to be no one (at least very very few people) that fails system test. Amazing contest!

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kuksag

5 years ago, # |

← Rev. 2 →

+23

Damn, these 4 ms help me become a candidate master!

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yogi

5 years ago, # |

I converted the problem C into "how many ways you can get a sum s using coins {1,2}"..then somehow didn't able to implement it in time...Its kind of sad

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5 years ago, # ^ |

That is a very nice solution.

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yogi

5 years ago, # ^ |

but I later figured it out that it will not work since,in counting number of ways a sum can be generated from fixed number of coins, order doesn't matters,but in our case order of selection will matter...eg. sum=3 can be generated from coin-sets {1,2} in two different ways {1,1,1} ans {1,2}(note that {1,2} and {2,1} are same) but the answer should be 3.It will convert to staircase problem counting number of ways to reach stair n if you can climb 1 or 2 stair at a time....correct me if I am wrong

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5 years ago, # ^ |

Yes, I don't think you are wrong

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Decayed

5 years ago, # |

← Rev. 2 →

Can someone help me with solution of D.. Here's my submission 64039124
My idea was to check if a node can be connected with any other node such that the cost for wiring is minimum and less than or equal to the cost to build a power station on that node. If such a node is available then construct an edge between them. Otherwise build a power station on that node.
Lastly, if a component has no such node which has power station in it , then build a power station that requires minimum cost in one of the node of the component..

It is generating way too much minimum cost for test case 13. I couldnt find why

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skarlord

5 years ago, # ^ |

Try this:

Ans is 15 and not 17

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Decayed

5 years ago, # ^ |

← Rev. 2 →

This solution also generates 15
This is the output
15
1
1
2
2 3
3 1

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mello

5 years ago, # ^ |

The correct answer is 18

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Decayed

5 years ago, # ^ |

Thanks :)

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Araz.G

5 years ago, # |

A very balanced contest with strong cases , thank you :)

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Ivan_Frost

5 years ago, # |

Great round! Thank you)

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anakib1

5 years ago, # |

I had following idea to D. Lets make minimum spanning tree of graph with edges $$$c(i, j) = (k_i+k_j) * d(i, j)$$$ for all $$$i , j$$$. Firstly lets add the cheapest plant. Then sort edges in non-increasing order and we will try to delete every edge. If we are considering edge $$$(x,y)$$$ let $$$c_1$$$ will be component with vert $$$x$$$ of tree without edge $$$(x, y)$$$ and $$$c_2$$$ will be component with vert $$$y$$$ of tree without edge $$$(x, y)$$$. Only one of this components has installed plant now. Let it be $$$c_2$$$ . Then find cost of the cheapest plant in $$$c_2$$$ and check if its better to add plant instead of edge from spanning tree. If its better to add plant, we delete edge $$$(x, y)$$$ and add plant.

But, my solution got WA13. I will be grateful if you point out my mistake.

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5 years ago, # ^ |

Not entirely sure but could be that you are not using the "cost" of power plant while sorting, you are only using wiring cost while sorting? Both types of costs need to be consider.

edges.pb(tii((k[i] + k[j]) * dist(i, j, coord), i, j));

SORT(edges);

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Ali_Tavakoli

5 years ago, # |

+15

Thanks DeliciousFlatChest !

That was a good contest !

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blinkshot

5 years ago, # |

+37

However, she died when she was only in fifth grade so she is not smart enough for this.

Wait, what codeforces?

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Ali_Tavakoli

5 years ago, # ^ |

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Len

5 years ago, # ^ |

Maybe it's reference to this character.

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5 years ago, # ^ |

Honestly, didn't read this part until you pointed out, XD...

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-LHVS-

5 years ago, # |

Has anyone solved D using dynamic programming?

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BumbleBee

5 years ago, # |

Another solution for D is the O(n^2) version of Prim's algorithm which works better on a dense graph. It doesn't need extra dummy nodes either. Is it the most efficient solution?

I solved D in this way after the round, but my implementation was slower than I expected.

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tiojuan

5 years ago, # |

Can anyone help me with problem D. I sumbited the same code twice, and in the constest y got TLE, but after resumbitting i got AC. https://mirror.codeforces.com/contest/1245/submission/64045463 https://mirror.codeforces.com/contest/1245/submission/64035229

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BumbleBee

5 years ago, # ^ |

+13

Your AC code was just 4 ms away from TLE. It just got lucky. The execution time of the same code can differ by a few ms in different submissions. That is the reason behind TLE. Your code was not just lucky enough to get AC in contest time!

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ZED_

5 years ago, # |

D is very much similar to this problem : https://vjudge.net/problem/LightOJ-1059

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5 years ago, # |

Although the contest was very well written and prepared, but still after seeing the next contest, I am like — "Finally, a div. 3 round...., Yay!"

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MijPeter

5 years ago, # |

← Rev. 2 →

Hello my solution to problem D is either bugged or incorrect. With bugs I can cope but I worry it might be simply incorrect. It fails to pass test 13 (wrong answer).

My idea:

Firstly, take each city and make network out of it, making total n networks. Assume that every network doesn't have electricity supplied to it.

while (there exists network that doesnt have electricity supplied) {
  take any network P that doesnt have electricity supplied

  if (its cheaper to build plant inside P, than to connect P to any other network
      (OR) its impossible to connect P to any other network (there are noo other networks)) {
    build plant (cheapest possible amongst members in P) inside P supplying electricity to it
  }
  else {
    build connection between P andd other, closest possible network (closest = cheapest road)
    merging two networks into one
  }
}

I'm not worried about time complexity as above can be implemented in a fairly fast way. Just whether this should work or not.

Thanks for help in advance.

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GLAYS

5 years ago, # ^ |

Hello :)

I didn't read your method, all I could do to help was stresstest, and I found a case where your code(I used the last submission) fails compared to my accepted solution.

test case

your answer

correct answer

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MijPeter

5 years ago, # ^ |

← Rev. 2 →

Thanks!

Turns out my solution was bugged :)

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Agnimandur

5 years ago, # |

How are you supposed to solve D using Kruskal's Algorithm?

When I sorted the 2,000,000 edges, I got TLE on Test 7. Is this only because I used Java, or is Kruskal too slow in general for this problem?

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