351F44's blog

By 351F44, 2 years ago,

Problem A is developed by mejiamejia.

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1733E - Conveyor

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• +124

 » 8 days ago, # |   +1 Can someone explain/share the O(N) solution of task D2?
•  » » 8 days 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.
•  » » » 8 days ago, # ^ |   -6 This is same as tourist solution.
•  » » » 7 days ago, # ^ |   +3 This solution looks more elegant!
•  » » » 7 days ago, # ^ |   +3 Very well explained, I understood it without trouble. This post should be in the editorial.
•  » » » 5 days ago, # ^ |   +3 Better than the editorial. Thanks!
•  » » 8 days 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 •  » » 8 days 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 •  » » 8 days 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))  » 8 days ago, # | +15 was this blog in 'draft' for 2 years? •  » » 8 days ago, # ^ | +10 Yes.  » 8 days 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? •  » » 6 days 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  » 8 days 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! •  » » 8 days ago, # ^ | 0 Edited the phrase more clearly. •  » » 8 days ago, # ^ | -13 LoL, what did u expect from the fifth problem? •  » » » 8 days 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).  » 8 days 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 •  » » 8 days 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. •  » » » 8 days ago, # ^ | 0 thank you for replying.Actually as i said in the post,I worked on the logic that if i can find a value k such that (k*x)+((length-k)*y) can form length-1(as we will have total wins as length-1) then we will have an answer. So the value of k can range from 0 to length-1. therefore to find that k i used binary search. I know I many have made some mistakes,therefore any tips are appreciated. Thank you •  » » » 8 days ago, # ^ | 0 Thanks a lot for your help. I had made a minor error in my code in the printing the other logic was correct. The solution do works correctly now. Code:https://codeforces.com/contest/1733/submission/172869494 •  » » 8 days ago, # ^ | 0 Take a look at Ticket 16198 from CF Stress for a counter example.This should help you figure out why binary search wouldn't work. •  » » » 8 days ago, # ^ | 0 Thanks a lot for your help. the code had a minor error in the printing the other logic was correct. Here is the solution Code:https://codeforces.com/contest/1733/submission/172869494 •  » » » » 8 days ago, # ^ | 0 You got it!  » 8 days ago, # | +1 Does anyone have a solution for D2 using DP with memoization? •  » » 8 days 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'; }  •  » » 8 days ago, # ^ | ← Rev. 2 → 0 you can see my solution of D2 using recursion+memoizatoin: https://codeforces.com/contest/1733/submission/172789187 •  » » 8 days 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 •  » » 8 days ago, # ^ | 0  » 8 days ago, # | 0 In D2(editorial), how is it possible to have$i+1$ones in the first$i$elements of$c$? •  » » 8 days ago, # ^ | 0 i is correct. Now edited. Thanks for noticing me. •  » » » 8 days 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$j=i+1$. •  » » » » 8 days ago, # ^ | 0 Edited, thanks. There was some mistake while translating 0-based MCS into 1-based editorial.  » 8 days ago, # | 0 What would be the expected rating of the first 2 questions? •  » » 8 days ago, # ^ | 0 We estimated A as *800, B as *1000 or *1100. Let's wait. •  » » » 8 days ago, # ^ | 0 and C? •  » » » » 8 days ago, # ^ | 0 I think *1500. •  » » » » » 8 days ago, # ^ | 0 Are you sure? I normally can't solve 1500s but C and even D1 were pretty solvable for me •  » » » » » » 7 days ago, # ^ | 0 Same, I was guessing around 1300. •  » » » » » » » 7 days 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? :/ •  » » » » » » » » 7 days ago, # ^ | 0 Yeah could be. Thank you for the contest tho! Great problems.  » 8 days 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!  » 8 days ago, # | 0 If anyone feel that the editorial is too complex and you have solved longest palindromic subsequence or longest palindromic substring or https://codeforces.com/contest/1728/problem/DThen see my submission for problem D: https://codeforces.com/contest/1733/submission/172855405Very easy to understand and O(N*N) DP •  » » 8 days ago, # ^ | 0 O(N) DP solution using Memoization: https://codeforces.com/contest/1733/submission/172865814Idea is to calculate how much we can save if we take or don't take x. Then subtract it from max cost possible i.e; y*(n/2)-dp[n]  » 8 days ago, # | 0 My code for problem D2 is giving Time Limit Exceeded at test case 3, Please help me in optimising my Approach, suggest me with this intution only, please help. ll f(ll i, string s, string s1, ll n, ll x, ll y,map&dp) { if(s==s1) return 0; if(i>=n) return 1e10+7; if(dp[s]!=0) return dp[s]; if(s[i]==s1[i]) return dp[s]=f(i+1, s, s1, n, x, y,dp); ll p=(ll)1e10+7, q=(ll)1e10+7; for(int j=0;j>n>>x>>y; string s,s1;cin>>s>>s1; mapdp; ll ans = f(0, s, s1, n, x, y,dp); if(ans>=(ll)1e9+7) cout<<-1; else cout<  » 8 days ago, # | 0 hats off...!!  » 8 days ago, # | +3 Is the challenge for E solvable? •  » » 8 days ago, # ^ | 0 I don't know. •  » » 7 days 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.  » 8 days ago, # | +1 I'm curious to know how to solve problem c challenge, any hints ?  » 7 days ago, # | 0 A video editorial explaining the dp cases of D2 would be very helpful. Or maybe if someone can explain the cases here?  » 7 days ago, # | 0 I think stress plays a lot of role during contests. I had a whole 1 hour and 22 mins. left to do problem C during contest but was not able to do it, today tried to upsolve it and solved it in 15 mins without looking at the editorial.  » 7 days ago, # | +2 Please, share solution for Problem C challenge. Thanks. •  » » 7 days 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. •  » » » 7 days ago, # ^ | 0 Oh! I see. So the editorial's solution of n-1 operations is the optimal solution. Thanks. •  » » » 7 days ago, # ^ | 0 This is the same solution as the orginal solution , however he asked about the solution of the challenge which asks to find minimum number of operations Test Sample : 2, 1, 3 according to your solution will cost 2 operations while the minimum cost is 1 operation we can apply the operation on second and third elements  » 7 days ago, # | +13 Difference between E's solution I submitted in last two minutes and the AC solution  » 7 days ago, # | 0 Hello, 351F44 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?? •  » » 7 days 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.  » 7 days 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.  » 7 days ago, # | 0 172746684 else cout << min((r — l) * x, y) << endl; isn't it redundant to check for the minimum?  » 7 days ago, # | 0 any Recursive approach for D2?  » 6 days ago, # | 0 A typoNow, find out which cell contains clime ball among the diagonal. •  » » 5 days ago, # ^ | 0 Edited. Thanks!  » 5 days 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?
•  » » 5 days 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 
 » 4 days 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?
 » 3 days 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<
 » 43 hours ago, # |   0 if the problem C ask us that they want to minimize the number of operation , how we can solve it ?
 » 31 hour(s) ago, # |   0 Can anyone explain that in question B Rule of League, in sample test case 4 ans is 2 but it can also be 1 right. As this was stated there Each player has either won x games or y games in the championship. i can't understand it plz help. 5 5 2 0 8 1 2 3 0 0 2 0 1 6 3 0 
•  » » 27 hours ago, # ^ | ← Rev. 7 →   0 yes , in test case 4 it can be 1 or 2 , both of them are correct . one of them won just one times and the second one won 0 times
•  » » » 27 hours ago, # ^ |   0 173387621 I have tried it but they say its wrong this is my submission.
•  » » » » 27 hours ago, # ^ | ← Rev. 2 →   0 if(mi != 0) cout<<-1<
•  » » » » » 24 hours ago, # ^ |   0 Thanks man that's a really rookie mistake.