### blobugh's blog

By blobugh, 4 years ago,

Problem A is developed by mejiamejia.

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#### 1733E - Conveyor

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 » 23 months ago, # |   +1 Can someone explain/share the O(N) solution of task D2?
•  » » 23 months 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.
•  » » » 23 months ago, # ^ |   +3 This solution looks more elegant!
•  » » » 23 months ago, # ^ |   +3 Very well explained, I understood it without trouble. This post should be in the editorial.
•  » » » 23 months ago, # ^ |   +3 Better than the editorial. Thanks!
•  » » » 22 months ago, # ^ |   0 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?
•  » » » » 22 months ago, # ^ |   0 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 $i$:$dp[i] = \min (dp[i - 2] + (mis[i] - mis[i - 1])x, dp[i - 1], dp[i - 2] + y) = \min (dp[i - 2] + (mis[i] - mis[i - 1])x, dp[i - 1])$
•  » » » » » 22 months ago, # ^ |   0 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?
•  » » » » » » 22 months ago, # ^ |   +1 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.
•  » » » 22 months ago, # ^ |   0 NB
•  » » » 20 months ago, # ^ | ← Rev. 2 →   0 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)
•  » » » » 20 months ago, # ^ | ← Rev. 3 →   0 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?
•  » » » » » 20 months ago, # ^ |   +3 this was my exact query, thanks for the explanation, its really very well explained.
•  » » 23 months ago, # ^ | ← 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 •  » » 23 months ago, # ^ | ← Rev. 2 → +1 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 •  » » 23 months ago, # ^ | ← Rev. 2 → 0 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))  » 23 months ago, # | +15 was this blog in 'draft' for 2 years? •  » » 23 months ago, # ^ | +9 Yes.  » 23 months ago, # | +3 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? •  » » 23 months ago, # ^ | ← Rev. 2 → 0 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  » 23 months ago, # | ← Rev. 4 → 0 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! •  » » 23 months ago, # ^ | 0 Edited the phrase more clearly. •  » » 23 months ago, # ^ | -13 LoL, what did u expect from the fifth problem? •  » » » 23 months ago, # ^ | ← Rev. 2 → 0 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).  » 23 months ago, # | ← Rev. 3 → 0 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.com/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.com/contest/1733/submission/172869494 •  » » 23 months ago, # ^ | 0 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.  » 23 months ago, # | +1 Does anyone have a solution for D2 using DP with memoization? •  » » 23 months ago, # ^ | 0 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 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 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'; }  •  » » 23 months ago, # ^ | 0 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.com/contest/1733/submission/172860412  » 23 months ago, # | 0 In D2(editorial), how is it possible to have$i+1$ones in the first$i$elements of$c$? •  » » 23 months ago, # ^ | 0 i is correct. Now edited. Thanks for noticing me. •  » » » 23 months ago, # ^ | 0 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$j=i+1$. •  » » » » 23 months ago, # ^ | 0 Edited, thanks. There was some mistake while translating 0-based MCS into 1-based editorial.  » 23 months ago, # | 0 What would be the expected rating of the first 2 questions? •  » » 23 months ago, # ^ | 0 We estimated A as *800, B as *1000 or *1100. Let's wait. •  » » » 23 months ago, # ^ | 0 and C? •  » » » » 23 months ago, # ^ | 0 I think *1500. •  » » » » » 23 months ago, # ^ | 0 Are you sure? I normally can't solve 1500s but C and even D1 were pretty solvable for me •  » » » » » » 23 months ago, # ^ | 0 Same, I was guessing around 1300. •  » » » » » » » 23 months ago, # ^ | 0 Well, standing says it can be easier than *1500. It is what I estimated before the contest, maybe real difficulty is more easier? :/ •  » » » » » » » » 23 months ago, # ^ | 0 Yeah could be. Thank you for the contest tho! Great problems.  » 23 months ago, # | +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!  » 23 months ago, # | +3 Is the challenge for E solvable? •  » » 23 months ago, # ^ | 0 I don't know. •  » » 23 months ago, # ^ | ← Rev. 2 → 0 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.  » 23 months ago, # | +1 I'm curious to know how to solve problem c challenge, any hints ?  » 23 months ago, # | +1 Please, share solution for Problem C challenge. Thanks. •  » » 23 months ago, # ^ | ← Rev. 2 → 0 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. •  » » » 5 months ago, # ^ | 0 Have you solved the "Challenge" given , finding minimum operations ? •  » » 5 months ago, # ^ | 0 Did you get it ?  » 23 months ago, # | +13 Difference between E's solution I submitted in last two minutes and the AC solution  » 23 months ago, # | 0 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?? •  » » 23 months ago, # ^ | 0 It is one of the result of an observation. To do such observation, check if it can be transformed to be structurally identical.  » 23 months ago, # | 0 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 linkHope to strengthen the data soon.  » 23 months ago, # | 0 A typoNow, find out which cell contains clime ball among the diagonal. •  » » 23 months ago, # ^ | 0 Edited. Thanks!  » 23 months ago, # | 0 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?
•  » » 23 months ago, # ^ |   0 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 rollerEnter 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 
 » 23 months ago, # |   0 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?
 » 22 months ago, # |   0 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. #include 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<
 » 22 months ago, # |   0 Shout out for problem E. You broke CF's top rated member, and it took me a full day to solve it, well done
 » 21 month(s) ago, # |   0 Can somebody hack my solution? https://codeforces.com/contest/1733/submission/179012720 I think it might be wrong.
 » 20 months ago, # |   0 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
 » 18 months ago, # |   0 For D1 consider this test case: 1 10 2 8 1111111111 0111100110 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?
•  » » 18 months ago, # ^ |   0 Sorry I just realized that x>=y :(
 » 16 months ago, # | ← Rev. 2 →   0 why was n less 3000 in D1 given the editorial provides a O(N) solution?