Внезапно, все идеи задач этого контеста принадлежат мне. И я очень сильно прошу прощения за отсутствие одного маленького теста в задаче B. Это было самой крупной ошибкой, которую я допустил во время подготовки этого раунда.
1324A - Yet Another Tetris Problem
Разбор
Tutorial is loading...
Решение
for i in range(int(input())):
n = int(input())
cnto = sum(list(map(lambda x: int(x) % 2, input().split())))
print('YES' if cnto == 0 or cnto == n else 'NO')
1324B - Yet Another Palindrome Problem
Разбор
Tutorial is loading...
Решение
for i in range(int(input())):
n = int(input())
s = list(map(int, input().split()))
ok = False
for i in range(n):
for j in range(i + 2, n):
if s[i] == s[j]: ok = True
print('YES' if ok else 'NO')
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
string s;
cin >> s;
vector<int> pos;
pos.push_back(0);
for (int i = 0; i < int(s.size()); ++i) {
if (s[i] == 'R') pos.push_back(i + 1);
}
pos.push_back(s.size() + 1);
int ans = 0;
for (int i = 0; i < int(pos.size()) - 1; ++i) {
ans = max(ans, pos[i + 1] - pos[i]);
}
cout << ans << endl;
}
return 0;
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
vector<int> a(n), b(n);
for (auto &it : a) cin >> it;
for (auto &it : b) cin >> it;
vector<int> c(n);
for (int i = 0; i < n; ++i) {
c[i] = a[i] - b[i];
}
sort(c.begin(), c.end());
long long ans = 0;
for (int i = 0; i < n; ++i) {
if (c[i] <= 0) continue;
int pos = lower_bound(c.begin(), c.end(), -c[i] + 1) - c.begin();
ans += i - pos;
}
cout << ans << endl;
return 0;
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
bool in(int x, int l, int r) {
return l <= x && x <= r;
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, h, l, r;
cin >> n >> h >> l >> r;
vector<int> a(n);
for (auto &it : a) cin >> it;
vector<vector<int>> dp(n + 1, vector<int>(n + 1, INT_MIN));
dp[0][0] = 0;
int sum = 0;
for (int i = 0; i < n; ++i) {
sum += a[i];
for (int j = 0; j <= n; ++j) {
dp[i + 1][j] = max(dp[i + 1][j], dp[i][j] + in((sum - j) % h, l, r));
if (j < n) dp[i + 1][j + 1] = max(dp[i + 1][j + 1], dp[i][j] + in((sum - j - 1) % h, l, r));
}
}
cout << *max_element(dp[n].begin(), dp[n].end()) << endl;
return 0;
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
vector<int> a;
vector<int> dp;
vector<int> ans;
vector<vector<int>> g;
void dfs(int v, int p = -1) {
dp[v] = a[v];
for (auto to : g[v]) {
if (to == p) continue;
dfs(to, v);
dp[v] += max(dp[to], 0);
}
}
void dfs2(int v, int p = -1) {
ans[v] = dp[v];
for (auto to : g[v]) {
if (to == p) continue;
dp[v] -= max(0, dp[to]);
dp[to] += max(0, dp[v]);
dfs2(to, v);
dp[to] -= max(0, dp[v]);
dp[v] += max(0, dp[to]);
}
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
a = dp = ans = vector<int>(n);
g = vector<vector<int>>(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
if (a[i] == 0) a[i] = -1;
}
for (int i = 0; i < n - 1; ++i) {
int x, y;
cin >> x >> y;
--x, --y;
g[x].push_back(y);
g[y].push_back(x);
}
dfs(0);
dfs2(0);
for (auto it : ans) cout << it << " ";
cout << endl;
return 0;
}
Can someone suggest some basic dp problems like E (same or little higher difficulty) please.
You can always look at the problemset and turn on the
dptag and sort by rating or number of solves in ascending order. I'd recommend looking around a rating of 1600-2000 for some basic (or a bit harder) dp tasks.DP tasks from 1600-2000 rating
I'm aware of that feature but in practice, I found that most 1600-2000 problems tagged with dp don't have their intended solutions with dp. Rather they are some greedy problems which have in their editorials "This problem can be solved using dp also", which is not helpful when I'm unable to solve the problem. This is why I'm asking :)
I see. In that case, here are some USACO Gold problems I would recommend that aren't just greedy.
Thanks very much
Good DP problems for all rating range.AtCoder Educational DP
I think this might be helpful. DP PROBLEMS AND TUTORIALS
E: A different dynamic programming approach, with runtime
O(n*h).Let
dp(i,j)be the maximum number of good sleeping times if Vova slept theitontimes with the starting timej. Then, the valuedp(1,0)will be the answer. Initially, all dp(i,j) = 0.Transitions (to be done from i = n to i = 1, with j being all possible values of time (0-h)):
Let
t1 = (j + a_i) % handt2 = (j + a_i - 1) % hThen,
dp(i,j) = max(good(t1) + dp(i+1, t1), good(t2) + dp(i+1, t2))are all the cases covered in such? can you elaborate? thanks in advance
I think my solution will be clear to you. Just have a look, basic recursive function with memoization. https://ideone.com/i53rk9
can you explain your approach a little bit ??
Since, we know that we can either select 'a[i]' or 'a[i]-1'. So, I called the recursive function two times, one with 'a[i]' as sum and the other with 'a[i]-1' as sum and hence returns the maximum of that. Do the same for all i from 1 to n-1. It is just the same as calling 'fib(n-1)' and 'fib(n-2)' in fibonacci series. 2-d DP works because 'n' is in the range [1,2000] and every time we are doing 'sum%h' the range of value will be [0,h) and h ranges from [3,2000]. I recommend you to make a recursive tree of the given sample input (take the help from my code). This will help you to see the repeated calculation which I am storing in array(dp).
https://codeforces.me/contest/1324/submission/73769097
Can you please see to my code . why my code fails for testcase 40
Your code fails for input when L contains 0 i.e, when L,R is [0,..]. This happens beacause at the time of calling you are passing sum as 0. For example try for input (1 44 0 14 23), the answer should be 0 as neither 23 nor 22 lies between L,R but your code gives 1. To remove this problem, you just need to call the recursive function two times : (1. index=1,sum=arr[0]) (2. index=1,sum=arr[0]-1). And finally returns maximum of 1 and 2. Hope it helps !!!
Thank you so much man. I was completely frustrated as my code was not working but your comment was here to help me. :-)
What does the dp[i][j] state signify ?? Thanks
dp[i][j] signifies the total number of good sleeps from index i to n.
Plzz can you explain about the state transitions.
I used the exact same approach using index and time as state transitions with recursive memoized dp, in O(n*h) complexity. It’s cool that someone else had this exact same idea 6 years ago :)
Hi I used similar approach getting WA on test 40 can u please help me https://codeforces.me/contest/1324/submission/73656520
Read the previous comment. Your code also behaves similar.
Essentially, what my state is how many good sleeps Vovu can have by starting from this hour and disregarding the previous i-1. Now, we can only disregard the first 1,2,3,...n and starting hour can only be in 0,1,2,...,h-1 so we have n*h states which cover all possible cases.
Hey, i have solved question E. but i have done it in different way !! can you please explain, what you have done? i can't understand what you are trying to say !! please reply, i am right now focusing on DP.
Well I don't get what exactly u didn't understand, so I am copying the main logic of my code which got AC in hopes you can get the algorithm I used:
hmm, got it...you have done push dp and i have done it in reverse way.
Simpler way to solve C is to calculate the longest contiguous sequence of L's and then output it plus one.
Or you can find out farthest 2R's and print the difference in their positions. Edit : The editorial says the same . Sorry I hadn't read editorial before commenting here . XD
Still though, there is no need to remember the whole array of distances between R's. Instead, it is enough to remember the max distance.
Alternate solution to task E:
Considering all times of the day modulo h, let
dp[i][j]be the maximum number of good sleeping times assuming that Vova has slept the firstitimes and that the current time of day isj.Also, let's maintain boolean array
pos[i][j]such thatpos[i][j]is true IFF it is possible to end up at timejafter sleepingitimes.Now, transitions can be calculated by setting
dp[i+1][(j+ai)%h]tomax(dp[i+1][(j+ai)%h, dp[i][j] + ((j+ai)%h >= l && (j+ai)%h <= r ? 1 ; 0)and applying a similar transition forai-1.Now, the answer is just the max
dp[i][j]such thati=n.I noticed a comment denoting a similar solution although my submission was in-contest unlike the other comment.
i thought exactly the same but mine seems doesn't work
can you please help me
my submission
Thanks in advance
EDIT:
Now it is working but why do we need to maintain a boolean array of possibilites?
Shouldn't it be handled automatically by setting to zero the dp matrix initially
please explain.
During the contest, my intuition was thinking that not all
(i,j)pairs would be possible during DP, so I included the boolean matrix as an extra precaution.I knew that the extra space wouldn't really matter and played it safe. IDK if it makes a difference but I wasn't worried in-contest because I wanted to make sure I just got the task right.
Can anyone suggest where can we study and practice rerooting question like F. Thanks in advance : )
You can solve the following problems using the rerooting idea:
Thank you ^_^
luciocf Can you tell how to solve the first one — Save Nagato!
Do the same as said, in the above editorial !
Instead of keeping the black and white count keep a multiset for each vertex storing the depth using each branch, now reroot for each vertex and then erasing the largest depth kept in the multiset,putting the largest depth into each children multiset and then finally erasing it from each children multiset & putting it back in afterwards in the vertex' multiset.
Have a look here at my submission : https://dmoj.ca/src/1961273
I am unable to see your solution. It is showing access denied
http://collabedit.com/qfhfs
you can see my code : https://dmoj.ca/src/1961373. I think i write a similar structure to the solution given by vovuh.
I am also unable to see your solution
https://pastebin.com/SVGg3f0H Sorry.
Yes I will dm you
ok will wait for your message
Akshay184 luciocf A subtree of a tree T is a tree S consisting of a node in T and all of its descendants in T. Can you please tell me whether this problem follows this standard definition? Because I dont see it following. Or may be I am wrong.
No, in this problem, a subtree means a connected subgraph of the tree. It does not include all its descendants.
What was the case that wasn’t included for B? I got hacked and I just wanted to know what the case was.
I think your mistake is that you are keeping track of the last seen element rather than the first seen.
I just know a bit of cpp so I'm unable to run your code, but I can tell you the test case on which most of the codes where hacked It was: 1 3 1 1 1, try running this and the output should be "YES"
Incase you want the specific test case on which your code was hacked, you can check it at https://codeforces.me/contest/1324/submission/73037905, you will be getting an option of view test (right under Hacked written in red text)
Thank you, I'm new to CodeForces so I didn't know how to view the test that hacked my code.
To view the hack:
Your Profile -> Submissions -> Hacked Submission ->
#Button -> View HackYour Hacked Submission
@vovuh I tried to solve F using two DFS but kept failing test case 3 and I couldn't figure out why it fails on this test case. Can you help me point out my mistake? My submission is 73111564 and I've added comments to make it easy to read.
the problem is on how you build the tree, i changed your buildTree method, check https://codeforces.me/contest/1324/submission/73114269
Oh, yeah, I definitely did not build tree correctly :). Thank you very much!
A can also be done by checking if difference of adjacent elements mod 2 equals 0 or not
Why 6 downvotes?? :(
Codeforces community is kinda weird and specific to it's liking. Don't pay attention to upvotes or downvotes on your comments. Do check the FAQ on this to know more.
Can somebody explain why the above solution for D works? Once we sort all
Cithere is no track of their original indexes anymore.. so how is it correct?The indexes don't matter. The condition i<j is given just so that you don't count (1,2) and (2,1) as different pairs. Ex. If the array is a, b, c, Then the pairs are ab, ac and bc. But if it is b, a, c, Then also the pairs are ba, bc and ac.
Thanks got it now. The pairs will be unique anyways as per the above solution.
When will the ratings get updated... Generally how much time it takes after the contest ends .?
Question D can be solved in O(n) after sorting also using two pointers, which seems a bit efficient.
Link to my solution
The sort itself takes O(nlogn)
I clearly said after the sort, it takes O(n). It is just an efficient way to do the same thing. Suppose the sorted array is given, my logic would take O(n), but the given would still take O(nlogn).
And yet, the solution still has a time complexity of O(NlogN)...
Lol , If the array is sorted you say,its the same as saying that you would have done it in O(1) time if they gave you a table with sum of pairs which is O(n^2) or an array which is sorted(which is what you are asking). But O(1) doesnot matter because of the O(n^2) process , similarly , The very thing that your O(n) approach depends on an O(nlogn) process (doesnt matter if it is sorting) is what makes your solution O(nlogn) .
Can somebody explain palindrome problem? How does it check the palindrome by removing characters in between?
the question says we need a palindromic sub sequence of length greater than or equal to 3, So the easiest palindrome that can be formed by sub seq is of length 3 which has 1st and 3rd elements same, the 2nd element won't matter . Simply search for 2 same elements in the array but they should not be consecutive (so that you can insert any 1 element between the 2 similar elements).
If the array has 2 same elements that don't occur together then the answer is "YES".
The solution given by vovuh for E is giving WA on tc 2. Any suggestions on what could be wrong in it?
I copied the code from local directory, not from the polygon, the local code contained one bug which was fixed on polygon. You can realize what is this bug yourself (there is some kind of hint in the tutorial). And now you can copy and submit the correct solution but I don't know why do you need it.
this contest be like:
Can anyone explain O(N) solution for problem B ??!
The only thing important is the indexes of repeated elements in the array. so what we could do was to store the min index and the max index where each element is in the array and then just check is the diff of min index and max index is greater than 1. If it is then we could include any element between these indices to make a palindrome of length 3.
Link to my submission — https://codeforces.me/contest/1324/submission/73029984
P.S this is my first post so please ignore the informalities.
Thanks!
$$$O(N*LOG)$$$
Yes , im sorry . its n*log(n).
You can maintain an auxiliary array whose index will be the numbers present in the original array. That auxiliary array may be initialized by -1. So when you traverse the array if the corresponding index in aux array is -1 that means it is the first occurence of that number in the main array and you can replace -1 with that first index of that number. Now if you encounter that number again you can simply take the difference between current index and first index(stored in the aux array). if that difference is >1 then just print yes and return. Or if you cant find any such index print NO
$$$O(N)$$$ 73087070
Perhaps I am very poor at parity-trick questions, but problem A is very non-intuitive to me, and I was very shocked to see the number of submissions, problems B,C,D were much more straightforward for me.
I still don't think I understand problem A fully.
OK, but this can prove that answer is "NO" when parity of any two numbers is different. But how do I prove that it has to be "YES" when all numbers have same parity?
It's somehow very non-intuitive to me :(
The way Tetris(the game) works is the bottom most row gets cleared if there is a block present in every column. Since the block we can add spans 2 rows and 1 column, the height increases by 2 when we add it to a column. If the parity of each column is same then you can add those blocks to make the height of each column equal to, say, x. Therefore, you can clear the bottom-most row x times to get a fully empty board. Lets say your numbers are 1 3 5 9 3. Then you can add some number of blocks to each of the columns to get 9. After doing that, you can clear the whole board. Hope this helps :)
When parity of all numbers is same, in one step you can increment the minimum element by 2 and then decrease all elements by 1. So the overall effect is that minimum element got increased by 1 and rest all elements were decreased by 1. Now it's easy to see why after a finite number of steps all elements in the array will become equal.
I don't know it will help you this way or not but here's how I looked at it. Consider the cases when when all the adjacent numbers differ by an even counts(which means all the numbers have the same parity). Like 3 5 7 5 3 1 or 2 4 2 16 4. The only motto of the question is to make all numbers equal(Obviously, after that you can turn them all to zero).You start by adding +2 to the minimum number possible and then decrease the numbers till at least one of the numbers reaches to zero. Let's say p q r s t (numbers differs by an even count) and let's say t is the smallest number , you add +2 to t and let's say (for simplicity) t+2 is again the smallest number in the array so what happens if you actually perform operation 2nd described in the question — you subtract (t+2) to all numbers because t+2 is still the smallest number (so it's the first number that will be turned to zero) so the numbers become p-(t+2) q-(t+2) r-(t+2) s-(t+2) 0. Now you add +2 to zero and repeat the process and now it's easy to see that you can turn down them to same number if and if only they have same parity.
Thank you all for your attempts. I think I have a somewhat better understanding now.
The solution provided for E is not running. I think the issue is that we are running inner loop from 0 to n which includes the case of dp(i,j) where j>i which is invalid. Please correct if I am wrong.
I have an interesting approach to question D,Time complexity: O(n).
You can define c[i] = a[i] — b[i],then sort them,finally while(l<r){ if(c[r]+c[l] > 0){ ans+=r-l; r--; } else { l++; } I think many people will use it to pass;
Your approach is
O(N Log N)$$$-$$$ sort functionoh,sorry I forgot it,haha!
Where can I practice tree dp problems like https://codeforces.me/contest/1324/problem/F. Thnaks!
there you go 1092 pF
I never saw that kind of lengthy system testing. Did it happen like this before?
After 100% it went to 91% .WTH
My F question uses two DFS, the complexity should be no problem, why is it TLE?73153261
I tried the same approach with question D as given in tutorial, still i am getting a wrong ans. Can someone please spot where i might have gone wrong? It goes wrong in test case 12. Thanks in advance.
/* arr[i]= a[i]-b[i]; ->as in the tutorial
only that arr is sorted in DESCENDING order
and i have used a 2-pointer technique */
static int findPairs (Long arr[] , int n )
{ int l = 0, r = n-1; int result = 0; while (l < r) { if (arr[l] + arr[r] > 0L) { result += (r - l); l++; } else r--; } return result; }change the data type of result as long. Overflow is happening.
Wtf!!! Such a silly mistake. I wasted an hour almost to debug my solution. Thanks a lot
2 pointer is good approach ............
I think problem E can be solved by greedy Did anyone solved it by greedy ? I have tried , but i failed
No, you can't solve E by greedy.
Can someone help me find the error in my solution of E https://codeforces.me/contest/1324/submission/73208155 It gives wrong answer verdict on test 40
Replace the line
if(sleep>=l && sleep<=r)withif(sleep>=l && sleep<=r && i>0)This is because both
landrhave lower limit 0 in the question. So you need to omit that case.See my solution for clarity.
P.S — Assuming you don't have made any silly mistakes elsewhere.
Thanks it worked.
P.S I love pizza as well
Someone please explain me the rerooting technique used in problem F. I am unable to understand it's editorial. Please help.
If you don't understand the editorial, just see others code, i think it is more straight forward, when you understand the code, then you can go back see if you can understand the editorial.
I hope this helps
someone plz drope solution for A by using eazy c++
here is my solution in c++
Why is this code of 1324D — Pair of Topics getting TLE?
Idea of code in short :- BIT / Fenwick tree is implemented using unordered_map (C++) to query smaller values, of type (b[ j ] — a[ j ]), than (a[ i ] — b[ i ]) for all j < i.
Any suggestions to optimize this code using the same idea?
where does my code goes wrong for PROBLEM E link of solution:-
solution here...
.thank in advance:)
1 1 2 2 3 3 4 4 5 5 14 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 6 15 15
the checker is saying this is a palindrome. can anyone explain how? this is test 5,1st problem
as you can see number 6 occurs 2 times => 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 6 so you can choose any number of those 6 {7 7 8 8 9 9 10 10 11 11 12 12 13 13 14} 6
Is there any other DP solving for E problem?(thanks in advance)
Can anyone explains why my solution for the Problem-F is giving TLE. According to me its complexiety should be O(2*(n+v)) in worst case?
vector mv[200005];
int a[200005];
int dfs(int x,int p)
{
int newcost=0; for(int i=0;i<mv[x].size();i++) if(mv[x][i].first!=p) if(mv[x][i].second!=-2) { if(mv[x][i].second>0) newcost+=mv[x][i].second; } else { mv[x][i].second=dfs(mv[x][i].first,x); if(mv[x][i].second>0) newcost+=mv[x][i].second; } if(a[x]) newcost++; else newcost--; return newcost;}
void solve()
{
int n; cin>>n; int flag=0; for(int i=1;i<=n;i++) { cin>>a[i]; if(a[i]==1) flag=1; } for(int i=0;i<n-1;i++) { int u,v; cin>>u>>v; mv[u].push_back(mp(v,-2)); mv[v].pb(mp(u,-2)); } if(flag==0) { for(int i=0;i<n;i++) cout<<"-1 "; return; } int i=1; while(i<=n) { if(mv[i].size()==1) break; i++; } mv[i][0].second=dfs(mv[i][0].first,i); for(int i=1;i<=n;i++) { int newcost=0; for(int j=0;j<mv[i].size();j++) { if(mv[i][j].second==-2) { mv[i][j].second=dfs(mv[i][j].first,i); } if(mv[i][j].second>0) newcost+=mv[i][j].second; } if(a[i]) newcost++; else newcost--; cout<<newcost<<" "; }}
This gives TLE : * Used Arrays A,B,C and 2 loops to input A and B and form C * Sorted C * This finally used lowerbound thing https://pastebin.com/icGaeuH8
This doesn't give TLE : * Used vectors A,B,C and 3 loops to input these * Sorted *Lower bount thing https://pastebin.com/9fajyGEi
Can anyone help me out ?
Tutorial Question E :
loop 1: i=0;i<n;i++ loop 2: j=0;**j<=n**;j++
-> i is number of sleeps -> j is number of sleeps he goes one hour earlier
So j <= i for all j Thus, j<=i rather than j<=n.
can anybody explain to me that in problem d -c[i]+1 logic? I didn't get it
I hope you got c[i] + c[j] > 0. Now for this inequality , either c[i] must be positive , or c[j] or both. So we consider c[i] to be positive for all cases. Now c[j] has to be greater then -c[i] atleast to be 0 or positive. However c[i]+c[j] > 0 and not equal to zero and thus c[j] >= -c[i]+1.
For problem D, I still can't understand why we can skip it when c[i]<=0? Can someone help me?
We sort c: c[i-1] <= c[i]
We need c[i] + c[j] > 0 (j < i). So, if c[i] <= 0 with all j = 0..i-1 we can't get c[i] + c[j] > 0 :)
Question (B) says that we can solve it by $$$O(n^2)$$$ or by $$$O(n)$$$. I cannot understand the writer's explanation of how we will do it in $$$O(n)$$$. Can anyone please provide code for the same. Thank you.
Try this 74688485
In problem D, doesn't sorting the c[] array mean losing information about i < j?
iam not able to understand F problem please help
Check it out
Also, what does this even mean?
The subtree of the tree is the connected subgraph of the given tree.I feel the english is bad and it should be:
A subtree of the tree is a connected subgraph of the given tree.can somebody explain why in problem F the dp[v] = a[v] + max(0,dp[to]) as in subtree all nodes are included so it should be dp[v] = a[v] + dp[to] why we are taking max over 0 . can somebody explain 1st testcase how for node 1 ans is 2.
In E can anyone please point the mistake in dp equation :
if(j>i) dp[i][j]=0;if(j==0) dp[i][j]=dp[i-1][j]+is((sum[i]-j)%h,l,r);else dp[i][j]=max(dp[i-1][j],dp[i-1][j-1])+is((sum[i]-j)%h,l,r);solved.. it may be an interesting task to find out bug in this solution.
there should be an else if instead of if in the second line of logic code mentioned.
Sorry to bother you vovuh, but it seems like problem B is broken. Validator is crashing and solutions are not being evaluated. "Denial of Judgment" is being shown on evaluator
Can anyone tell me why this submission is going wrong on TC 12 for problem D? Submission Link. Its very odd, I see that the Editorial solution isn't too far off from my method (It uses binsearch while I use a ptr method).
Can anyone explain me why he took max(0,dp[child]) everywhere?
Assuming you're talking about problem F, we don't want to add $$$dp[child] \lt 0$$$ to $$$dp[parent]$$$ since we want to maximize $$$dp[parent]$$$.
mphillotry A subtree of a tree T is a tree S consisting of a node in T and all of its descendants in T. But doesnt the condition
max(0, dp[child])violate the above standard definition? Please reply.The tree is unrooted therefore
descendantsdon't make sense in this context. The editorial only fixes the tree on some root to help solving the problem.Perhaps it wasn't the best choice to call them
subtreesas confusions like yours may result, but the problem statement does give its own definition of a "subtree":the subtree of the tree is the connected subgraph of the given tree.Why does this solution for Problem F get timed out ..The Time complexity is O(n).. I guess its just traversing the tree 3 times... (https://codeforces.me/contest/1324/submission/86646387)
If someone comes across this, can you please tell why is this solution failing at case 50.
Для задачи "E. Режим сна" более лаконичным является решение с конца (вычитаем часы начиная с последнего элемента входного массива). Получается тоже ДП, но не требуется совершать проверки "переходов из недостижимых состояний".
По итогу, выводим в качестве ответа значение 0-элемента (То есть, сколько будет хороших раз, если начинать с 0-часа).
In problem B, it says an array A is a palindrome if A[i] = A[n — i — 1] for all values i from 1 — n. However it should be A[i] = A[n — i + 1].
can anyone pls tell where my code fails for {1324B — Yet Another Palindrome Problem}176153984. vovuh
Can someone clear me up in the E problem why 1 24 0 0 3
this gives ans answer as 0 even though 1st sleep can give an optimal sleep