Codeforces Round #821 (Div. 2) Editorial

#	User	Rating
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

#	User	Contrib.
1	cry	167
2	Um_nik	163
3	maomao90	162
3	atcoder_official	162
5	adamant	159
6	-is-this-fft-	158
7	awoo	157
8	TheScrasse	154
9	Dominater069	153
9	nor	153

Problem A is developed by mejiamejia.

1733A - Consecutive Sum

Hints

Solution

Solution Code

Challenge

1733B - Rule of League

Hints

Solution

Solution Code

Challenge

1733C - Parity Shuffle Sorting

Hints

Solution

Solution Code

Challenge

1733D1 - Zero-One (Easy Version)

Hints

Solution

Solution Code

1733D2 - Zero-One (Hard Version)

Hints

Solution

Solution Code

Challenge

1733E - Conveyor

Hints

Solution

Solution Code

Challenge

#include<bits/stdc++.h> using namespace std; const int N=5010; typedef long long ll; ll z0[N][N],z1[N][N],c[N]; int main() { ios::sync_with_stdio(0),cin.tie(0),cout.tie(0); int _; cin>>_; while(_--){ int n,x,y; cin>>n>>x>>y; string a,b; cin>>a>>b; for(int i=1;i<=n;++i) c[i]=a[i-1]^b[i-1]; for(int i=0;i<=n;++i) fill(z0[i],z0[i]+n+1,1LL<<60); for(int i=0;i<=n;++i) fill(z1[i],z1[i]+n+1,1LL<<60); if(c[1]) z1[1][1]=0; else z0[1][0]=0; int d=min(x,y); for(int i=2;i<=n;++i) if(c[i]){ for(int j=0;j<=i;++j){ if(j) z1[i][j]=min(z0[i-1][j-1],z1[i-1][j-1]),z0[i][j]=min(z0[i-1][j-1]+d,z1[i-1][j-1]+y); z0[i][j]=min(z0[i][j],z1[i-1][j+1]+d),z0[i][j]=min(z0[i][j],z0[i-1][j+1]+y); } } else{ for(int j=0;j<=i;++j){ z0[i][j]=min(z0[i-1][j],z1[i-1][j]); z1[i][j]=min(z0[i-1][j]+y,z1[i-1][j]+d); if(j>=2) z1[i][j]=min(z1[i][j],z1[i-1][j-2]+y),z1[i][j]=min(z1[i][j],z0[i-1][j-2]+d); } } if(z0[n][0]!=1LL<<60) cout<<z0[n][0]<<endl; else cout<<-1<<endl; } return 0; }

Comments (71)

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sary

2 years ago, # |

Can someone explain/share the O(N) solution of task D2?

→ Reply

Andalus

2 years ago, # ^ |

+34

If $$$x \geq y$$$, follow D1. Otherwise, for $$$x < y$$$, observe that for each mismatch, we can fix it by either performing a single $$$y$$$ operation with another non-adjacent mismatch, or we can perform a sequence of $$$x$$$-operations to either the next or previous mismatch. We can use a 1D DP as follows:

$$$dp[i]$$$ represents the minimum cost for dealing with the first $$$i$$$ mismatches alone. If $$$i$$$ is even, then all mismatches should be fixed. If $$$i$$$ is odd, then $$$i - 1$$$ mismatches should be fixed while one mismatch remains. The $$$mis[]$$$ array stores mismatch indices.

For even $$$i$$$, $$$dp[i] = \min (dp[i - 2] + (mis[i] - mis[i - 1])x, dp[i - 1] + y)$$$. We can fix the $$$i$$$-th mismatch through either an $$$x$$$-operation chain with the $$$(i - 1)$$$-th mismatch (first option), or through a $$$y$$$-operation with some earlier mismatch (second option).

For odd $$$i$$$, $$$dp[i] = \min (dp[i - 2] + (mis[i] - mis[i - 1])x, dp[i - 1])$$$. Here, for the $$$i$$$-th mismatch, the first option is the same as before ($$$x$$$-operation chain with $$$(i - 1)$$$-th mismatch), but the second option is to let the $$$i$$$-th mismatch be the unresolved one. Note that $$$dp[i - 1] \leq dp[i - 2] + y$$$ (see even formula), so there is no need to consider using a $$$y$$$-operation for the $$$i$$$-th mismatch for odd $$$i$$$ (but it wouldn't change the result if this was added as a candidate).

The last element in the $$$dp$$$ array (which must be an even index) is the answer.

MrRD

This solution looks more elegant!

nonrice

Very well explained, I understood it without trouble. This post should be in the editorial.

zero_bitches

Better than the editorial. Thanks!

daksh2021

How would you add the the y-operation for odd i, if we wanted to? Also please can you explain a little why there is no need to consider the y-operation in case of odd i?

If you wanted to consider the $$$y$$$-operation for the last element of odd $$$i$$$, it would have a cost of $$$dp[i - 2] + y$$$ (so the second-to-last element is the unresolved one). However, we know that $$$dp[i - 1] \leq dp[i - 2] + y$$$ (because applying the even formula on $$$i - 1$$$ yields $$$dp[i - 1] = \min (dp[i - 3] + (mis[i - 1] - mis[i - 2])x, dp[i - 2], dp[i - 2] + y) \leq dp[i - 2] + y$$$). Therefore, for odd $$$i$$$:

$$$dp[i] = \min (dp[i - 2] + (mis[i] - mis[i - 1])x, dp[i - 1], dp[i - 2] + y) = \min (dp[i - 2] + (mis[i] - mis[i - 1])x, dp[i - 1])$$$

Thanks for replying! But I still have one doubt... for y-operation for last element of odd i, why have you only tried — by keeping either last element unresolved or second last element unresolved? Why didn't you try keeping any other element unresolved?

All other options are implicitly covered by $$$dp[i - 2]$$$, which denotes the optimal way of handling the first $$$i - 2$$$ elements. I should correct my earlier statement actually: $$$dp[i - 2] + y$$$ does not only cover the case where the second-to-last element is unresolved, but it actually covers the cases where any of the first $$$i - 1$$$ elements are unresolved.

However, note that the different options is not about "which element is unresolved" but more about "how do we deal with the last element". We can break this down to four options:

Don't pair the last element: this costs $$$dp[i - 1]$$$. It would, of course, also be the unresolved element in this case.
Pair it with the second-to-last element using an $$$x$$$-operation chain: this costs $$$dp[i - 2] + (mis[i] - mis[i - 1])x$$$. Here, the unresolved element is whoever was unresolved in $$$dp[i - 2]$$$ (which is the optimal choice among the first $$$i - 2$$$ elements).
Pair it with the $$$(i - 1)$$$-th element using a $$$y$$$-operation (assuming they're not adjacent). The cost here is $$$dp[i - 2] + y$$$. Again, the unresolved element is whoever was unresolved in $$$dp[i - 2]$$$. The assumption of non-adjacency is not an issue, because if the last two elements were adjacent, then option 2 would win over option 3 (since $$$x < y$$$).
Pair it with one of the first $$$(i - 2)$$$ elements using a $$$y$$$-operation. The optimal way to deal with the first $$$(i - 2)$$$ elements is given with $$$dp[i - 2]$$$, which includes one unresolved element among them. Therefore, the optimal way to handle this case would be to pair the last element with whoever is unresolved from $$$dp[i - 2]$$$. This does mean that the $$$(i - 1)$$$-th element is not handled, so it would end up becoming the unresolved one. The cost here is also $$$dp[i - 2] + y$$$.

If the fourth option still confuses you, let me elaborate a little more: we are not specifically choosing that the $$$(i - 1)$$$-th element is the unresolved one, but rather, we are choosing to let the last element pair with some element from among the first $$$(i - 2)$$$ elements. The optimal choice here is to pair it with the unresolved element in $$$dp[i - 2]$$$. [Proof: if it's better to pair it with some different element $$$j$$$, this would imply that leaving $$$j$$$ unresolved would be better for the value of $$$dp[i - 2]$$$ than whichever element was actually left unresolved]. And this would naturally lead to the $$$(i - 1)$$$-th element being the unresolved one, since all other elements are handled.

In reality, there is no actual difference between options 3 and 4. They can both be combined into a single case of "use a $$$y$$$-operation on the last element". With an understanding of how a $$$y$$$-operation works, where the cost is the same irrespective of which element we pick (as long as it's non-adjacent, but option 2's superiority would dominate such scenarios), it should be intuitive to observe that it doesn't actually matter as to which element we are pairing it, a long as it's a $$$y$$$-operation.

Finally, given that the unresolved element would eventually have to be resolved in later even cases, either through an $$$x$$$-operation chain with the next element (which requires that the last element is the unresolved element) or a $$$y$$$-operation with some element (where there is no advantage for any particular element over another), it's easy to see that keeping the last element unresolved cannot be worse than to pair it up with a $$$y$$$-operation so that someone else can wait for a future $$$y$$$-operation, i.e., option 1 will never be worse than options 3 or 4.

Sorry for the wall of text; I think this scenario is actually really simple intuitively, but I can understand that one might have doubts about whether this intuition is rigorously proven, so I hope I was able to ease some of those doubts.

HowMoring

ebshu21

23 months ago, # ^ |

← Rev. 2 →

i have a little doubt in this solution, can you please explain? you defined your dp state as : dp(i) = minimum cost for dealing with the first i mismatches "ALONE". now let's take the even case only : suppose you do a y-operation on the i-th mismatch with some previous mismatch j, now after this transition, can the new state really be defined by dp(i-1) as per the rules because when we were at the ith state(precisely speaking when we started the state at dp(i), all the previous mismatches before ith one were not fixed yet, but when we perform a y-operation, with some jth mismatch, it has already been fixed, so its not correct to say this state is dp(i-1), furthermore say for example that we fixed the ith mismatch using a y operation with the (i-2)th mismatch, now when we use dp(i-1), there is an operation in the transition equation that is done with the just previous mismatch using an x operation: (mis[i]-mis[i-1])x but when we are dp(i-1) now, this would be (mis[i-1]-mis[i-2])x but why would we perform this x operation when the (i-2)nd mismatch has already been resolved)

← Rev. 3 →

I'm not entirely sure I understand your question. Let's say $$$i$$$ is even and we decide to resolve the $$$i$$$-th mismatch using a $$$y$$$-operation. There are $$$i - 1$$$ earlier mismatches that we could pair with the $$$i$$$-th mismatch. Let's assume that the optimal solution arises when we pair the $$$i$$$-th mismatch with the $$$j$$$-th mismatch ($$$j < i$$$).

Claim: The value of $$$dp[i - 1]$$$ is equal to the cost of dealing with the first $$$i - 1$$$ mismatches except for the $$$j$$$-th mismatch.

Proof: By definition, $$$dp[i - 1]$$$ is the optimal way to deal with the first $$$i - 1$$$ mismatches except with one left unresolved. Suppose this unresolved mismatch is the $$$k$$$-th mismatch. Consider three cases:

Case 1: The minimum cost when the $$$k$$$-th mismatch is unresolved is equivalent to the minimum cost when the $$$j$$$-mistamtch is unresolved. The claim holds in this case. This also includes the case of $$$k = j$$$ btw.
Case 2: The minimum cost when the $$$k$$$-th mismatch is unresolved is higher than the minimum cost when the $$$j$$$-mistamtch is unresolved. Then $$$dp[i - 1]$$$ is not the optimal way of dealing with the first $$$i - 1$$$ indices, which is a contradiction.
Case 3: The minimum cost when the $$$k$$$-th mismatch is unresolved is lower than the minimum cost when the $$$j$$$-mistamtch is unresolved. This means that when we need to deal with the $$$i$$$-th mismatch, it would be better to pair the $$$i$$$-th mismatch with the $$$k$$$-th mismatch as opposed to the $$$j$$$-mismatch. This contradicts the premise that the optimal way to deal with the $$$i$$$-th mismatch is to pair it with the $$$j$$$-th mismatch.

Therefore, only Case 1 is valid, proving the claim. This also means that $$$dp[i] = dp[i - 1] + y$$$, since $$$dp[i - 1]$$$ gives us the cost of dealing with the first $$$i - 1$$$ mismatches except for the $$$j$$$-th mismatch, which we pair with the $$$i$$$-th mismatch for a cost of $$$y$$$.

Does this resolve your confusion, or did I misunderstand your query?

this was my exact query, thanks for the explanation, its really very well explained.

cxm1024

← Rev. 4 →

+15

There is a solution with an $$$O(n)$$$ DP.

For $$$x\ge y$$$ the solution is obvious (it's Problem D1), so I'll assume that $$$x<y$$$ then.

First, pre-process $$$dif[i]$$$ which means the index of $$$i$$$-th difference between a and b.

Let $$$f[i]$$$ be the minimum cost to make first $$$i$$$ differences became same. It's impossible when $$$i$$$ isn't even, so if $$$i$$$ is odd, $$$f[i]$$$ becames the minimun cost to make that there's only one difference in the first $$$i$$$ differences.

Before we begin to solve, let $$$f[0]$$$ and $$$f[1]$$$ be $$$1$$$ first.

Then, we can calculate $$$f[i]$$$ for all $$$i$$$ from $$$2$$$ to $$$n$$$ in order:

If $$$i$$$ is even, $$$f[i]$$$ can be $$$f[i-2]+min(x\times(dif[i]-dif[i-1]),y)$$$. $$$f[i-2]$$$ means the cost when $$$dif[1\to i-2]$$$ are all solved, and $$$f[i]$$$ means the cost when $$$dif[1-i]$$$ are solved, so $$$f[i]$$$ need to add the cost to make $$$dif[i]$$$ and $$$dif[i-1]$$$ same.

$$$f[i]$$$ can also be $$$f[i-1]+y$$$. $$$f[i-1]$$$ means the cost when there's only one difference in $$$dif[1\to i-1]$$$, and $$$f[i]$$$ means there's no difference, so it has to add the cost to make $$$dif[i]$$$ and $$$dif[j]$$$ same. Although we don't know the real $$$j$$$, we know the cost is $$$y$$$, because the condition that $$$j=i-1$$$ has already solve by the last paragraph, so we can assume that $$$j<i-1$$$.

To sum up, the real $$$f[i]$$$ need to be the minimum cost between last two condition.

When $$$i$$$ is odd, the reasoning process is similar. $$$f[i]$$$ can be $$$f[i-2]+min(x\times(dif[i]-dif[i-1]),y)$$$, or $$$f[i-1]$$$. You can try to independently calculate.

Here is my code, you can refer to it.

Thanks for the explanation.

I have seen another tricky solution: We will exclude the x>=y as we can solve it with D1 solution and the case where only 2 nearby mismatches( ans=min(2*y,x) ),

(Option#1): So you will try to jump to any other mismatch with cost Y and goes to the next mismatch (dp[i] = dp[i+1] + y), I think this is valid because if you will take any mismatch with cost Y you will pair it with another mismatch.

The problem with option 1 is calculating Y for every mismatch, so you have to count the number of mismatches taken by the first option and divide them by 2 , to avoid that, you can just multiply the second option (x) by 2 and divide the final solution by 2.

(Option#2): The other transition is to chain with the next one(i+1) and transition to (i+2) so you solved i and i+1 mismatches with cost x*(mismatch[i+1] — mismatch[i+2]). option 2 is only valid if there exists (i+1<n). dp[i] = min(dp[i],dp[i+2] + x*(mismatch[i+1] — mismatch[i+2]))

solution credits to IsaacMoris

psychobot

let v be the sorted vector of indices where a[i]!=b[i] (let size of v be vn) then the answer of the problem is f(0)

where, cost(i,j)=min(y,|v[i]-v[j]|*x)

f(vn-1)=y,f(cn-2)=cost(vn-1,vn-2)

f(i) = min(((vn-i)%2)*y + f(i+1),cost(i,i+1)+f(i+2))

coderdhanraj

was this blog in 'draft' for 2 years?

blobugh

Yes.

_Shayan_

Hello Guys! For D2 I had this solution 172748609, I also don't no the reason that why its true.Can any one hack or say the reason behind it?

PsychicBoom_

First, we get all the positions which satisfy a!=b and save them in a new array C. Then we consider four positions which are adjacent in C. Let them be p1,p2,p3,p4 (p1<p2<p3<p4) . There is always min((p2-p1)*x,y)+min((p4-p3)*x,y)<=min((p3-p1)*x,y)+min((p4-p2)*x,y), which means that (p1,p3)+(p2,p4) can't be a part of answer. So you are right :)

K1ppy

Why do you say that a greedy solution does not work in D2. I think it's too bad that such tasks appear in the competition!

Edited the phrase more clearly.

lis05

-13

LoL, what did u expect from the fifth problem?

OleschY

I'm pretty sure the post was to be interpreted with a "/s", considering the "not so unusual" hint (as it was before 351F44 edited it).

MankS

Hello guys, would really appreciate a little help with the B question. I thought that if we can make a combination with the 2 values of x and y that sums up to length-1 then we will get the right answer(was not sure so just tried). So I ran binary search between 0 and length-1 to search for the number of players that had x wins to find the answer and then print the player x times, but I got WA. here is the code:https://codeforces.me/contest/1733/submission/172718129 kindly guide me, thank you. EDIT:I found the error,it was a silly mistake in the printing part.THe code is working fine now. Code:https://codeforces.me/contest/1733/submission/172869494

Citypop

Binary search only works on monotonic searching space. Simple words, if x works either y > x or y < x should work. I am not sure how this problem is related to binary search. Maybe explain why you think binary search will do here.

cosmere

Does anyone have a solution for D2 using DP with memoization?

thanhchauns2

You can check my solution: 172731715,

The main part:

void Process()
{
	ll n, x, y;
	cin >> n >> x >> y;
	string a, b;
	cin >> a >> b;
	vec<ll> C;
	FOR(i, 0, n - 1)
	{
		if (a[i] != b[i]) C.pb(i);
	}
	ll s = C.size();
	if (s % 2 == 1)
	{
		cout << -1 << '\n';
		return;
	}
	if (x >= y)
	{
		ll ans = y * (s / 2);
		if ((ll)C.size() == 2)
		{
			if (C[1] == C[0] + 1) ans = min(x, y * 2);
			else ans = min((C[1] - C[0]) * x, y);
			cout << ans << '\n';
			return;
		}
		if ((ll)C.size() == 0)
		{
			cout << 0 << '\n';
		}
		else
		{
			cout << (s / 2) * y << '\n';
		}
		return;
	}
	FOR(i, 0, s)
	{
		FOR(j, 0, s)
		{
			DP[i][j] = -1;
		}
	}
	function<ll(ll, ll)> dp = [&](ll l, ll r)
	{
		if (DP[l][r] != -1) return DP[l][r];
		if (l == r) return DP[l][r] = 0;
		if (r == l + 1)
		{
			if (C[r] == C[l] + 1) return DP[l][r] = min(x, y * 2);
			return DP[l][r] = min((C[r] - C[l]) * x, y);
		}
		DP[l][r] = 1e18;
		DP[l][r] = min(DP[l][r], dp(l, l + 1) + dp(l + 2, r));
		DP[l][r] = min(DP[l][r], dp(l + 1, r - 1) + y);
		DP[l][r] = min(DP[l][r], dp(l, r - 2) + dp(r - 1, r));
		return DP[l][r];
	};
	cout << dp(0, s - 1) << '\n';
}

rrdas

Let us say we have a vector of mismatched indices. We will be picking from this vector in pairs.

In the DP, at any mismatched index, we have following options:

We can pair this index with the next mismatched index by continuously changing everything in between. Cost would be distance between mismatched indices multiplied by x.

We can pair with any previously unpaired index. Cost would be y.

We don't pair with anything, just create another unpaired index. It will need to be paired with later. Cost is 0 as we have not resolved anything.

Here is my submission https://codeforces.me/contest/1733/submission/172860412

In D2(editorial), how is it possible to have $$$i+1$$$ ones in the first $$$i$$$ elements of $$$c$$$?

i is correct. Now edited. Thanks for noticing me.

I think there is one more problem with it.

You recalculate dp for all $$$0 \le j \le i$$$, but in formulas where $$$c_i=1$$$ there is a condition $$$j=i+1$$$.

Edited, thanks. There was some mistake while translating 0-based MCS into 1-based editorial.

29_2002

What would be the expected rating of the first 2 questions?

We estimated A as *800, B as *1000 or *1100. Let's wait.

flakes24

and C?

I think *1500.

nk3

Are you sure? I normally can't solve 1500s but C and even D1 were pretty solvable for me

Same, I was guessing around 1300.

Well, standing says it can be easier than *1500. It is what I estimated before the contest, maybe real difficulty is more easier? :/

Yeah could be. Thank you for the contest tho! Great problems.

pwned

+38

Great Problem E! Though its implementation isn't complicated, its idea is quite thought-provoking and requires insight. This is how CF problems must be. Truly nice problem!

lucasxia01

Is the challenge for E solvable?

I don't know.

freehandle

I think it's plausible. Consider the state of the conveyer after using exactly k slimes and look at each diagonal, regarding right arrow as 0 and down arrow as 1. For example, when k = 0, all conveyers have right arrows, so all diagonal's values are 0. When k = 1, all conveyers except the first row have right arrows, so all diagonal's values are 1.

As k increases,

1st diagonal's values are: 0, 1, 0, 1, ....

2nd diagonal's values are: 00, 01, 11, 10, 00, ...

3rd diagonal's values are: 000, 001, 011, 001, 101, 100, 110, 100, 000, ...

4th diagonal's values are: 0000, 0001, 0011, 0111, 0011, 0001, 0011, 0111, 1111, 1110, 1100, 1000, 1100, 1110, 1100, 1000, 0000, ...

It's not something I really proved, but it seems there's an obvious pattern here. I guess this pattern will allow us to find a path for k-th slime in O(N log k) for a N*N conveyer system.

Ghaith

I'm curious to know how to solve problem c challenge, any hints ?

Aurnob_syc

Please, share solution for Problem C challenge. Thanks.

dulatcodes

You can always make the whole array equal to the last element of the array in n-1 steps.

First, apply the operation on the first and the last element. After this the first and the last element will become equal.

Now just iterate through the array starting from the second element:

Case 1. If the current element has the same parity as the last element apply operation to the current and the last element. This will make the current element equal to the last.

Case 2. Otherwise, apply the operation to the current and the previous element. This will make the current element equal to the previous. But we know that the previous element is already equal to the last because we already iterated through it, so this will make the current element equal to the last.

khmndr

9 months ago, # ^ |

Have you solved the "Challenge" given , finding minimum operations ?

Did you get it ?

Yahia_Emara

+13

Difference between E's solution I submitted in last two minutes and the AC solution

helloworld1102

Hello, blobugh can you please tell in the question D1 & D2, how you even have a intuition of starting with the XORs of both the binary strings a and b. At my last thought after so long I could only think of working on b trying making equalto a. But actually of no use because it's the same thing.

So how you even got the intuition of working on XORs of both the strings??

It is one of the result of an observation. To do such observation, check if it can be transformed to be structurally identical.

CF114514ancestor

There was an incorrect solution get accepted on D2 during the contest, and unfortunatly didn't get hacked. The idea was use range(or interval?) DP with some strange strategies to enumerate the decision. Code link

Hope to strengthen the data soon.

Ryu204

A typo

Edited. Thanks!

yeetholmes619

For the problem E, is it possible to find out the periodicity of a particular roller in $$$O(1)$$$ time or precompute them after which we can simulate the process for the slime that arrived at $$$t-x-y$$$ in $$$O(x+y)$$$ ?

I don't have a concrete idea yet on how would one calculate the periodicity tho :( Any suggestions would be helpful! I feel one could exploit this to solve the challenge as well?

So I tried to print the pattern of the state of each roller form t = 1 till t = 50 and 0 represents right and 1 represents down and in the beginning between {1,0} and {0,1} you see a good pattern. but in the diagonal (x+y = 2) There doesn't seem to be a pattern that is being followed, but there seems to be some similarity in {0,2} and {2,0} perhaps but {1,1} seems to show no similarity ? And when we try talking about the diagonal (x+y = 3) I can't spot any good pattern. So I'm skeptical of being able to find a pattern in the periodicity although I'm almost certain that the states of all rollers are periodic in nature.

Pattern in the state of each roller

Enter the coordinate
0 0
01010101010101010101010101010101010101010101010101
Enter the coordinate
0 1
00110011001100110011001100110011001100110011001100
Enter the coordinate
1 0
00011001100110011001100110011001100110011001100110
Enter the coordinate
1 1
00001000100010001000100010001000100010001000100010
Enter the coordinate
2 0
00000011110000111100001111000011110000111100001111
Enter the coordinate
0 2
00011110000111100001111000011110000111100001111000
Enter the coordinate
3 0
00000000000111111110000000011111111000000001111111
Enter the coordinate
1 2
00000111011110001000011101111000100001110111100010
Enter the coordinate
2 1
00000010001111011100001000111101110000100011110111
Enter the coordinate
0 4
00000111111111111111100000000000000001111111111111
Enter the coordinate
2 2
00000001100111001100000110011100110000011001110011
Enter the coordinate

gs15120

At E, I could think everything in time but the most important part, the simulation with x slime balls processes. How can I approach that process from the hints?

UNRULY_HORSE

I wonder for D2, if we can just use DP to solve all the conditons? So if x>y, can we use DP instead of greedy algorithm? I try it, but got wrong answer.

ivan100sic

Shout out for problem E. You broke CF's top rated member, and it took me a full day to solve it, well done

1838641320

Can somebody hack my solution? https://codeforces.me/contest/1733/submission/179012720 I think it might be wrong.

23 months ago, # |

can anyone elaborate the proposed solution of D2 in deep details, about the transitions given(how did we get those transitions). Actually i am unable to make out much from the proposed solution

not_ahmed_hamed

21 month(s) ago, # |

For D1 consider this test case:

a and b are different in [1,6,7,10] indices ** according to editorial's solution, [1,7] with cost 8 and [6,10] with cost 8 so ans = 8+8 = 16. but the min cost is to take [6,7] with cost 2 and [1,10] with cost 8 so ans = 2+8 = 10.

Do I miss something?

21 month(s) ago, # ^ |

Sorry I just realized that x>=y :(

vedansh.srivastava6

20 months ago, # |

why was n less 3000 in D1 given the editorial provides a O(N) solution?

naman_dwivedi

19 months ago, # |

/* in this question we have to make two strings equal by performing some operations and we have some cost associated with the two operations x when r-l = 1 y when r-l != 1 we can basically change two numbers simultaneously either zero to one or one to zero significance of it is that we choose two numbers so numbers are basically form a pair */

/* when x >= y (easy version) we can find the indices where we have different bits in the two strings and if the count is odd then we cannot make pairs such that all elements are paired and no one is left unpaired so the answer is -1

when we have the indices of bad elements ( not equal bits positions ) so if there are only two postions so we have to check if the two elements are consecutive or not if consecutive then find minimum of min(x,2*y) because we directly change two by using x or we can change one by one by pairing with some far element

and if there are more elements the answer will be count/2 * y because in this case we can always pair the elements without being making them consecutive elements in the pair */

/* when there is no boundation on x ( hard version ) we have discussed the case when x >= y now we will discuss for x < y here greedy will not work so we have to use dp here instead but question is how

now we have bad elements (indices of not equal bits )

lets take an example x = 2 and y = 11 and s1 = 000001 and s2 = 100000

indices of bad elements -> 1 and 6

so what we can do now either pick 1 and 6 and take a cost of y which is 11 or we change basically flow the change like , making changes in s1 s1 becomes 000010 then 000100 then 001000 then 010000 then 100000 now both s1 and s2 are equal so total cost is distance between two bits (distance that flow of change has covered) * x that is x*(6-1) 2*5 = 10 so optimal will be to use flow technique for minimum cost so answer will be 10

so now let's think we have two indices so we can either use dis * x or y which is minimum so will use dp here

we will use dp on bad elements (indices of not equal bits) so we will have an array of bad elements

now we will generate the minimum answer for all subarrays of the array of bad elements

let's take an example: array is -> 1 4 5 6 7 10 and x = 1 and y = 3 so answer for 2 size subarrays will be 3 1 1 1 3 now we will generate answer for all 4 size subarrays we will generate all even size subarrays as we will add a pair of elements in our calculated subarray

so answers will be 1 4 5 6 -> 4 , 4 5 6 7 -> 2 , 5 6 7 10 -> 4 no the answer for full array means 6 elements 1 4 5 6 7 10 6 -> 6

now what can be the transitions : lets say you are forming the 6 length subarray 1 4 5 6 7 10 (previous example ) you can generate it from 4 length and 2 length subarrays ( previously calculated )

so there are three possibilities 1: 1 4 + 5 6 7 10 2: 1 4 5 6 + 7 10 3: 1 + 4 5 6 7 + 10 in the first and second case we can add the answer of two subarrays let it be ans1 = 1 4 + 5 6 7 10 -> 3 + 4 = 7 ans2 = 1 4 5 6 + 7 10 -> 4 + 3 = 7 and for the last case the answer will be answer of subarray + min(dis*x,y) so ans3 = 1 + 4 5 6 7 + 10 -> 2 + min((10-1)*1,4) => 2 + 4 = 6 that is the answer and store the answer as minimum of all the three answers i.e. min({ans1,ans2,ans3}) in dp

in this way the answer for the whole array is calculated now you will say that r-l = 1 condition is not mentioned it is not needed indeed in this case as is distance is 1 and x is smaller then x*1 will be smaller than y so it is chosen automatically and answer will be in dp[1][array.size()-1]

the base condition will be to calculate the answer for all 2 size subarrays */

/* hope you get the solution if have any doubt you can message me i am free to help

<--------------------------------- THANK YOU ------------------------------------> */