Все идеи принадлежат MikeMirzayanov.
1385A - Три попарных максимума
Разбор
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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--) {
vector<int> a(3);
for (auto &it : a) cin >> it;
sort(a.begin(), a.end());
if (a[1] != a[2]) {
cout << "NO" << endl;
} else {
cout << "YES" << endl << a[0] << " " << a[0] << " " << a[2] << endl;
}
}
return 0;
}
1385B - Восстановление перестановки после слияния
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--) {
int n;
cin >> n;
vector<int> a(2 * n);
for (auto &it : a) cin >> it;
vector<int> used(n);
vector<int> p;
for (int i = 0; i < 2 * n; ++i) {
if (!used[a[i] - 1]) {
used[a[i] - 1] = true;
p.push_back(a[i]);
}
}
for (auto it : p) cout << it << " ";
cout << 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 t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(n);
for (auto &it : a) cin >> it;
int pos = n - 1;
while (pos > 0 && a[pos - 1] >= a[pos]) --pos;
while (pos > 0 && a[pos - 1] <= a[pos]) --pos;
cout << pos << endl;
}
return 0;
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int calc(const string &s, char c) {
if (s.size() == 1) {
return s[0] != c;
}
int mid = s.size() / 2;
int cntl = calc(string(s.begin(), s.begin() + mid), c + 1);
cntl += s.size() / 2 - count(s.begin() + mid, s.end(), c);
int cntr = calc(string(s.begin() + mid, s.end()), c + 1);
cntr += s.size() / 2 - count(s.begin(), s.begin() + mid, c);
return min(cntl, cntr);
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n;
string s;
cin >> n >> s;
cout << calc(s, 'a') << endl;
}
return 0;
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
vector<int> ord;
vector<int> used;
vector<vector<int>> g;
void dfs(int v) {
used[v] = 1;
for (auto to : g[v]) {
if (!used[to]) dfs(to);
}
ord.push_back(v);
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n, m;
cin >> n >> m;
g = vector<vector<int>>(n);
vector<pair<int, int>> edges;
for (int i = 0; i < m; ++i) {
int t, x, y;
cin >> t >> x >> y;
--x, --y;
if (t == 1) {
g[x].push_back(y);
}
edges.push_back({x, y});
}
ord.clear();
used = vector<int>(n);
for (int i = 0; i < n; ++i) {
if (!used[i]) dfs(i);
}
vector<int> pos(n);
reverse(ord.begin(), ord.end());
for (int i = 0; i < n; ++i) {
pos[ord[i]] = i;
}
bool bad = false;
for (int i = 0; i < n; ++i) {
for (auto j : g[i]) {
if (pos[i] > pos[j]) bad = true;
}
}
if (bad) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
for (auto [x, y] : edges) {
if (pos[x] < pos[y]) {
cout << x + 1 << " " << y + 1 << endl;
} else {
cout << y + 1 << " " << x + 1 << endl;
}
}
}
}
return 0;
}
Разбор
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Решение
#include <bits/stdc++.h>
using namespace std;
int n, k, ans;
vector<set<int>> g;
vector<set<int>> leaves;
struct comp {
bool operator() (int a, int b) const {
if (leaves[a].size() == leaves[b].size()) return a < b;
return leaves[a].size() > leaves[b].size();
}
};
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
cin >> n >> k;
g = leaves = vector<set<int>>(n);
for (int i = 0; i < n - 1; ++i) {
int x, y;
cin >> x >> y;
--x, --y;
g[x].insert(y);
g[y].insert(x);
}
for (int i = 0; i < n; ++i) {
if (g[i].size() == 1) {
leaves[*g[i].begin()].insert(i);
}
}
set<int, comp> st;
for (int i = 0; i < n; ++i) {
st.insert(i);
}
int ans = 0;
while (true) {
int v = *st.begin();
if (int(leaves[v].size()) < k) break;
for (int i = 0; i < k; ++i) {
int leaf = *leaves[v].begin();
g[leaf].erase(v);
g[v].erase(leaf);
st.erase(v);
st.erase(leaf);
leaves[v].erase(leaf);
if (leaves[leaf].count(v)) leaves[leaf].erase(v);
if (g[v].size() == 1) {
int to = *g[v].begin();
st.erase(to);
leaves[to].insert(v);
st.insert(to);
}
st.insert(v);
st.insert(leaf);
}
ans += 1;
}
cout << ans << endl;
}
return 0;
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int cnt0, cnt1;
vector<int> col, comp;
vector<vector<pair<int, int>>> g;
void dfs(int v, int c, int cmp) {
col[v] = c;
if (col[v] == 0) ++cnt0;
else ++cnt1;
comp[v] = cmp;
for (auto [to, change] : g[v]) {
if (col[to] == -1) {
dfs(to, c ^ change, cmp);
}
}
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<vector<int>> a(2, vector<int>(n));
vector<vector<int>> pos(n);
for (int i = 0; i < 2; ++i) {
for (int j = 0; j < n; ++j) {
cin >> a[i][j];
--a[i][j];
pos[a[i][j]].push_back(j);
}
}
bool bad = false;
g = vector<vector<pair<int, int>>>(n);
for (int i = 0; i < n; ++i) {
if (pos[i].size() != 2) {
bad = true;
break;
}
int c1 = pos[i][0], c2 = pos[i][1];
if (c1 == c2) continue;
int r1 = a[0][c1] != i, r2 = a[0][c2] != i;
g[c1].push_back({c2, r1 == r2});
g[c2].push_back({c1, r1 == r2});
}
col = comp = vector<int>(n, -1);
int cnt = 0;
vector<pair<int, int>> colcnt;
int ans = 0;
for (int i = 0; i < n; ++i) {
if (col[i] == -1) {
cnt0 = cnt1 = 0;
dfs(i, 0, cnt);
++cnt;
colcnt.push_back({cnt0, cnt1});
ans += min(cnt0, cnt1);
}
}
for (int i = 0; i < n; ++i) {
for (auto [j, diff] : g[i]) {
if ((col[i] ^ col[j]) != diff) {
bad = true;
}
}
}
if (bad) {
cout << -1 << endl;
} else {
cout << ans << endl;
for (int i = 0; i < n; ++i) {
int changeZero = colcnt[comp[i]].first < colcnt[comp[i]].second;
if (col[i] ^ changeZero) {
cout << i + 1 << " ";
}
}
cout << endl;
}
}
return 0;
}
vovuh There's a typo in idea of E "Otherwise, the answer is always "YES".... Its not always YES.
It should always be YES because if there exists an undirected edge where no matter the direction you choose, it creates a cycle, then that means there was a cycle in the graph to begin with which is a contradiction.
I think u r right. I had a doubt that in ur statement that there was a cycle in the graph to begin with..... Can u please calrify this statement.
It's simply this. Ignoring the undirected edges, if there is already a cycle in the graph, the answer is NO. Otherwise we can always specify directions for the undirected ones and get a YES.
Great contest with amazing set of questions, just a bit inclined towards DIV 2 rather than DIV 3.
Only for last three questions.
There's already Div.4, no need for Div.3 to be too easy,i think.
Is it first time that ratings have been given to problems before the main testing ??
I think the rating of some Div. 3 problems are assigned manually, it's mentioned in this blog.
Great contest indeed and great editorial . . . learnt new things in this contest. . . Make more of these. Thanks
Was this test rated, if it was how much time will it take to update the ratings ?
It is rated and you'll get the result after system testing(no longer than a day).(Forgive my poor English)
Why would someone have to forgive your English? It's just a language to express your thoughts. This is not the first language of most of the people here. So, the possibility of making mistakes is too high. No one should ever bother about these kind of things.
at least on this site, if you are to able to communicate, that should be pretty much it.
https://mirror.codeforces.com/contest/1385/submission/87162089 can someone explain what is the error in this code for problem E ? I can't understand the diagnostics message
Hi, I see issues with your dfs function, but not sure if this is the cause of runtime error.
visitedarray to check if you have to perform the dfs on child nodes. You're only using the value of parent node. This suffices in case of trees, but for general graph you may get a cycle which your function is not figuring out.set<int> recurby value, so you may end up using O(n^2) memory in the worst case. As this recur set will be copied for each function call. Pass the recur set by reference:set<int>& recur. If a child is completed but the parent is still there, you don't want the recur set to get cleared. Dorecur.erase(node)on the last line of dfs instead. Detect Cycle in a Graphhttps://mirror.codeforces.com/contest/1385/submission/87191537
Can you Please explain why i am getting tle on test case 3 in question D? I have used recursion.
Pass
string sby reference in your recursive function. Likestring& s.Thanks. Got it. Can you please explain why passing by reference does not give tle? and we should always use pass by reference or it was specific for some question?
When a function is called, the arguments get copied, unless they are passed by reference. So here your string s, which can be of length 10^5, was being copied in each recursive function call. When we pass by reference the very big string does not get copied in each function call, only a reference to string s is being passed in each function call.
When to use pass by reference, and when to use pass by value depends on what you want your code to do. Pass by reference vs value
What about this then? Edit: He just needed not to pass the whole string, it's feasible if we pass s/2 every round, even with pass by value.
I was dealing with the same problem, I noticed that many people have submitted kind of the same code as I wrote and got ac..but I was getting tle for no good reason.. but now after passing by reference it got accepted... here is my code which got ac.. and here is the one getting tle :/
maybe its because you were passing substrings.. and as per nilayjain said.. "very big string does not get coppied in each function call" your strings were getting smaller in each call, while mine was as big as the original one in the every call.. thats why i was having tle :(
It doesn't make sense to me why one would pass a value that does not change in a recursive function. I mean, just declaring it global makes it a lot sensible.
ya.. maybe we should always declare large data structures globally, just to avoid these situations :/ and ya.. it still doesn't make any sense :/
Global variable aren't a good practice, one should refrain from using them. They are the places where most of the bugs tend to reside. Pass by reference is the key!
"Global variable aren't a good practice, one should refrain from using them. They are the places where most of the bugs tend to reside." ???????
Yes, you are right, it is a good idea to declare large structures globally, especially when you're using those structures in different functions. But this is not a rule of thumb, it all depends on situation. And if it is still not making sense, think about the complexity of the recursive function for 2 cases we are discussing above:
Passing the substring by value: complexity is = n + n / 2 + n / 4 + ... + 1 = 2 * n. (since you divide the string in 2 halves and then pass it)
Passing the full string by value for all recursive calls: complexity is = n + n + n ... (n times) = n^2. This is because even though your recursive function is taking lesser time n, n/2, n/4 etc, but copying the string still takes n time for each recursive call. This will be TLE for sure.
Thanks !! Got it :D
Thanks!
Anyone else used binary search for C?87114284
i did it
I still didn't get the Binary Search approach, can you explain? Thanks
let's say $$$P(x)$$$ tells us whether by erasing the prefix of length $$$x$$$ the array becomes good (this is the subarray $$$[x+1\dots n]$$$). The key is that if $$$P(x) = true$$$, then $$$P(x+1) = true$$$. So, apply binary search to find the lowest $$$x$$$ such that $$$P(x) = true$$$. Now for a fixed $$$x$$$, to know if the array becomes good we do following:
Let's say $$$L[i]$$$ tells us whether we can put all elements from left side of $$$i$$$ before $$$i$$$ in the array $$$C[]$$$ or not.
Let's say $R[i]$ tells us whether we can put all elements from right side of $$$i$$$ before $$$i$$$ in the array $$$C[]$$$ or not.
Now, if there exists some $i \in [x+1\dots n]: L[i] = R[i] = true$, then $$$P(x) = true$$$; otherwise $$$P(x) = false$$$.
Me too
I also did it
How to solve D in bottom up manner?? And/or Using iterative only??
Look at this submission. Here $$$dp[i][j]$$$ is the minimum cost of make good the range starting at position $$$i$$$ having length $$$2^j$$$. The transitions are the same as recursive approach.
$$$dp[i][j] = \min(dp[i][j-1] + \text{cost}(i + 2^{j-1}, i + 2^j-1), dp[i+2^{j-1}][j-1] + \text{cost}(i, i + 2^{j-1}-1))$$$
Thanks for the reply. But I can't understand fully. Can you please explain how you are finding which character should come in the range I to 2^j-1 and what dp[0][I] means? And what does the j-i+1 in the cost function mean?
Let's say $$$k=\log_2 n$$$. Then, if the range is of length $$$2^p$$$, the character here must be $$$\text{'a'}+k-p-1$$$, cause each time we half the length of range, the character augments in 1. Now, the cost at some range is the amount of characters on the range that are different of the right character.
For 1385E - Directing Edges, an alternative solution can be as described below with
DFS,DSU.Strategy:
Creating edge from SMALLER value node to BIGGER value nodeObservation:
If the directed edges make any cycle, then there is no answer. Else, there is always an answer.Solution Construction:
If u is a normal node and v is a node that participated in a direct edge, then always make edge from u to v
If u and v both participated in directed edges and both belong to separate dsu component, then make an edge u->v if root(u) < root(v), else v->u
If u and v both participated in directed edges and both belong to same dsu component, then make an edge u->v or v->u in such a way that no cycle occurs.
Solution: 87185374
(This strategy essentially behaves like the topsort solution from editorial. The editorial's cpp code's equivalent strategy is such that it tries to make edge from bigger value node to smaller value node)
I think problem D can add a modify operation — let $$$s_i\gets c$$$ ($$$c$$$ is a character).
Then we can use a segment tree to solve it.
Actually, the
calcfunction in problem D's code is just like apushupfunction in a segment tree.i am still having difficulty for understanding complexity of prob D depth of recursion is log n
there are n possibilities if i am not wrong
e.g. aaaabbcd aaaabbdc aaaacdbb aaaadcbb the rest four with aaaa at the end... like we try all n possiblities but doesn the editorial do an additional step of counting at each step?
Approach the runtime of this problem in this way: For given string of length N and target of C-good string(C represents a generic character). We have 2 choices:
Either we make the 1st half C-good and the 2nd half (C + 1) good, or we make the 1st half C + 1 good and the 2nd half C good. Let's denote T(N) as the runtime to solve length N. Then to solve the smaller subproblem it takes T(N / 2) time. It also takes O(N / 2) to count the cost of making half of the string all character C. So we have the following formula:
T(N) = 2 * (T(N / 2) + O(N / 2)) = 2 * T(N / 2) + O(N).
Applying the master theorem, the run time is O(N * logN). This is exactly the same with how we analyze merge sort's runtime.
pefect!! can you please help me in one more thing?
https://mirror.codeforces.com/contest/1385/submission/87155344
i had precalculated the count...
isnt my solution O(n)?
I think you are right. T(N) = 2 * T(N / 2) + O(1), which is O(N).
The preprocessing takes O(26 * N), so it is O(N) in total. Nice work by the way to pre-compute costs!
T(N) = 2 * T(N / 2) + O(1) has O(N). not O(logN)..
Yeah, you are right, T(N) = a * T(N / b) + O(N^d), if a > b^d, T(N) = O(N^logb(a)) a = 2, b= 2, d = 0, so T(N) = O(N). I edited my comment.
You explained in a good manner.Thanks!
Can't we count the cost of making a substring into a letter in constant time by pre-calculating the cost to convert the string into in each letter. This will reduce time complexity to O(N) and 26*N needed for pre-calculation. Correct me if I am wrong. I know that O(N *logN) is actually smaller than O(26*N)
I think you are right. It is asymptotic[user:Shameek]ally better than O(N * logN). You can check out 87155344 by Shameek. He solved this problem by pre-computing costs.
I did solve D using recursion. Actually, at each step, you will calculate which half of the string you need to make a 'c'-good string; but this doesn't mean that the half which will require the minimum number of changes will give the least total number of moves ultimately.
Therefore, you need to call on both the halves and get the value of making these halves a 'c+1'-good string and finally you will compare the value in the whole of both these halves with each other and return the minimum value. So, all in all, you will be traversing the string twice (i.e 2n) and yeah the depth of recursion will be long n only.
link to my solution in C++: https://mirror.codeforces.com/contest/1385/submission/87154630
link to my solution in python: https://mirror.codeforces.com/contest/1385/submission/87147603
vovuh For problem F, the editorial states: We can notice that all leaves are indistinguishable for us. I think this is correct conceptually but do not know how to prove it. Can you share how you come up with this fact? When I first read this problem, I was thinking about the order of choosing different leaves to remove is going to have an impact on the final answer. Obviously this is not true.
I cant understand why problem E's solution needs to --x; --y;? Hoping for reply:)
well input is based on indexing from 1 to N, where as vuvoh's code is reading input from 0 to N minus 1. like input will consider nodes 1, 2, 3, 4 but vuvoh's code will consider them as 0, 1, 2, 3 this is the reason he did x + 1 and y + 1 in output statements.
so it is not necessary,right? thanks
https://mirror.codeforces.com/contest/1385/submission/87187974
anyone ? this python code is getting runtime error on 27th test (last one). i am new to python and have no idea why is this happening.
I'm also having runtime error on 30th test case in python..don't know why https://mirror.codeforces.com/contest/1385/submission/87401446
May be they don't want us to use Python... XD
Do we need to memoize in problem D i passed it using simple recursion.
We do not need to memoize because there is no overlapping subproblems. Simple recursion gives you O(N * logN) runtime already.
Why the complexity is O(NLogn)? It should be exponential. Can you please prove the complexity. Thanks in advance. PS: I read the prove using master's theorem. But is there any other way around to calculate it without using theorem.
You can go descending by the recurrence, and see a pattern, then you can proof the pattern by using induction.
Can anyone help me for Problem D. I am getting TLE for Test Case 3. I have followed the approach similar to that given in editorial 87191656
Difference between pass by reference and pass by value. Your same code I made the string as a pass by reference and got an AC. sub link. Pass by value creates copies each time in the recursion stack while pass by reference doesn't.
Shouldn't the editorial be the same as the tag given, for example, Problem C has the tag Binary Search whereas the explanation is pure implementation. I mean explain the approach that requires algorithms rather than intuitive ones.
Edit — Same goes for D no explanation for the DP approach
Check the submissions, or the comments section already has multiple implementation of all problems
Editorial Answer is very good for C. But i implemented it using binary search.
If we think this question as a binary search question then according to me it is a very good binary search question.
Thats why i thought to share my idea.
If we consider f(x) as a boolean function that tells whether the array is good after xth index,then obviously f(x) will become a monotonic function by intution. As by intution we can see that if we take x1>x then for every x1 the array is good. So just find minimal value for x so that the array is good after index x and finally your answer will be n-(total elements in good array).
Hope this solution helps someone.
What problems should I practice to solve D easily ? Any specific concepts ?
https://mirror.codeforces.com/contest/1285/problem/D
This is one of my favorite recursion problem.
you should practice problem related to divide and conquer, recursion and backtracking .
In problem E, is this a standard way of checking if there's a cycle in the directed graph? I did a quick google search and couldn't find such implementation.
I usually go about coloring the vertices and check if I revisit a vertex in current traversal.
Yes using colors is pretty standard. You can check for the cycle while in the same dfs you do to construct the topological sort.
I think D can be further optimised to O(N) by making a prefix array that stores count of a,b,c....z in any prefix. In this way the time of count function will drastically reduce to O(1). And final complexity will become exponential(logn) i.e 2^(logn) which is almost equivalent to O(n).
This way count(l,r,c) can be written as pre[r][c]-pre[l-1][c]
Did anyone solve G by 2-SAT ??
https://mirror.codeforces.com/contest/1385/submission/87187380
This I guess.
No, that is dsu + bipartite colouring, same as editorial.
Ohk.
I read 2-SAT :).
https://mirror.codeforces.com/contest/1385/submission/87222125
I have done it using 2-SAT. Thanks.
Can you explain more? I was thinking of using 2-SAT as well but 2-SAT's assignment won't guarantee minimum number of columns swapped (afaik), how do you get around this?
That is actually not 2-SAT. You are making components for a vertex and it's complement and taking the component with smaller size. Please correct me if I am wrong.
This is what I was doing-
Consider each column as a node.
If two nodes have same number in any row then one of them should be flipped, this gives us the relation of (x^y) which can be written as ((x or y) and (not x or not y)).
If two nodes have same number in opposite rows then either both should be flipped or none should be flipped. So that gives not(x^y).
not(x^y)
not((not x and y) or (x and not y))
((not(not x and y)) and (not(x and not y))
(x or not y) and (not x or y)
Now both the conditions are reduced to CNF so maybe we can apply 2-SAT. I could not implement this idea and also problem requires that there should be minimum number of 1's in the answer. I am not sure how to do that either.
I am not very good in 2-SAT so it is possible I might have missunderstood you. Please correct me if I am wrong.
It is 2-SAT brother according to my understanding . Because For every column it has 2-states . either swap or no swap. So I took the other state as the complement. and then i am uniting those states. Making relations between states.
https://en.wikipedia.org/wiki/2-satisfiability
See this. I assume i did this only . Boolean satisfiability.
Complement in my sol means that it is a swap. So i have taken that states size as 1 and 0 otherwise.
Otherwise if you are not satisfied , then I can't say. Because it is a new topic for me , I just read it today. But accroding to me this is 2 — SAT.
I was thinking of same approach for optimizing number of 1's I was thinking to construct a graph with 12..n numbers as nodes and they are connected if a they share a column.I wanted to take each component of this graph and do 2SAT on all columns involved with the numbers of that component. They must have only 2 possible solution states so minimum can be taken.( Last part somehow made sense to me and i didn't prove it so not sure )
Yeah TopGun__ has done that. I was not sure about only 2 solution states but his solution does that and is correct. The assignment of values could have been done greedily. Thanks __paryatak__ and TopGun__.
IN ques 1385C — Make It Good
eg .
1
4
1 3 2 4
soln:
step 1 : 3 2 4 c=[1]
step 2 : 2 4 c=[1 , 3]
step 3 : 2 c =[1 , 3 , 4]
the ans is 1 but A/C to this ques ans is 2 how?? please any one hep
Read the problem statement carefully.
Why am I getting a TLE in D when I have used the same logic, and the code has same time complexity as mentioned in the tutorial?
It would be really kind of any one of you if you could help me out
My Submission : https://mirror.codeforces.com/contest/1385/submission/87198924
Pass string by reference.
Why does not passing string by reference cause a TLE? Or in other words, how does passing string by reference optimize the time complexity?
Does it decrease the constant in the Time Complexity expression? If yes, it would be kind of you to explain
If you don't pass the string by reference then the recursion call creates a whole copy of that string and takes O(n) time and that is the reason for TLE.
And this is not only for string actually, but it holds for any type of argument.
How to solve problem F if we can remove any k leaves at a time(i.e not only the leaves from a particular nodes but we can pick leaves from any node)?
why E's solution code will get MLE at test1 using Clang++17 while AC on g++17
Need help. I am getting TLE on testcase 3 for the problem D. I have tried similar approach but i am not sure why i am getting TLE. MySolution. Any help would be appreciated.
Try passing the string by reference
https://mirror.codeforces.com/contest/1385/submission/87155412
Can anyone Please explain why i am getting tle on test case 3 in question D? I have used recursion.
https://mirror.codeforces.com/contest/1385/submission/87198479
in the above submission, I just write both the parts separately which were included in minimum in the very first submission and this time I did not get the tle but I am unable to understand what was the problem
Don't pass string by value, pass by reference.
N.B. Why are you passing an unnecessary 2D vector to function? And it is strange that you pass that vector by reference but pass string by value.
yeah i get it but then why my second submission got accepted there also I did not pass string by reference?
and that 2D vector is prefix array to calculate the required number of changes between l to r
I still got TLE using 2D prefix array. Guess yours was just a little faster to beat the time limit.
You accepted solution is also almost close to time limit. I think when you are using function directly then compiler has to perform some overhead computation to save calculated value and finally clear them which results in tle.
And 200ms is not much time. You can't perform so many overhead computation in this time.
Problem D — Not able to understand what is causing TLE https://mirror.codeforces.com/contest/1385/submission/87200081
string fk=ss will copy full string from ss to fk. It will happen on each iteration which will lead tle.
Anyone else solved E using analysis of DFS coloring states? (0 for not seen, 1 for discovered, 2 for visited)
My Solution:
Maintain a global vector of pairs for all the edges in the final solution
For every node, sort its edges in the following manner (directed edges comes before undirected)
Do a DFS starting from any node.
Color current node 1
While considering its neighbors:
A)If directed: (Standard methods used for detecting a cycle)
1) Check if the next node is colored 1; That implies a back edge is present, thus cycle formed
2) If the next node is colored 0, then recursively call DFS on it, push {current node, next node} as pair in edges
3) If the next node is colored 2, push {current node, next node} as pair in edges
B)If undirected:
1) Check if the next node is colored 1; implies a back edge would exist if {current node, next node} is considered an edge, thus convert it into a forward edge and push back {next node, current node} as an edge
2) Check if the next node is colored 0 (I might get a little sloppy with my explanation for this, so please bear with me :D)
Now I choose to always select {next node, current node} as my edge in this case due to the following:
Lemma: If I visit all the directed edges before undirected edges for a node, then conversion of all undirected edges to incoming directed edges for the current node {next node, current node}, will never result in the formation of a cycle
Proof:
If all undirected edges are converted into incoming edges. A cycle consisting of any pair these edges and the current node wouldn't be formed (as both edges are in the same direction). The same applies to all incoming directed edges.
Now for all outgoing directed edges, as per our sorting before we would have already visited these directed edges before encountering any undirected edge for our current node. Let's assume that along the path of the above outgoing directed edge, we reach a node 'u' and encounter an undirected edge linked to node 'v', which we convert into an incoming directed edge ('v' -> 'u'). Now if a cycle were to be formed using 'v', 'u', 'current node' and 'next node', then it wouldn't be possible as both of them are incoming edges ('v' -> 'u') & ('next node' -> 'current node').
Thus for all next nodes colored 0, we will choose the {next node, current node} as an edge.
3) If the next node is colored 2, we do nothing as we established above, it would already would have been treated as an edge {current node, next node} while we observed DFS for the next node (As its already been visited (colored 2)).
After visiting its neighbors, color the current node as 2 (visited and explored).
Finish DFS. Finally print edges.
My Solution: (https://mirror.codeforces.com/contest/1385/submission/87158004)
(Here I created a map of the managed undirected edges so as to only treat the edge for only one node, that implies in case of undirected edges, we wouldn't need to check if the next node is colored 2 as if it would, we would have already skipped past it, thus without checking for anything we can treat the undirected edge as an incoming directed edge)
You can do F using Tree DP in 2 DFSes in $$$\mathcal O (n)$$$ time.
Arbitrarily root the tree.
Let $$$p(u)$$$ be $$$u$$$'s parent and $$$C(u)$$$ be the set of $$$u$$$'s children.
Let $$$\operatorname{in}(u)$$$ be the max number of moves we can do if we only consider the subtree of $$$u$$$.
Let $$$\operatorname{out}(u)$$$ be the max number of moves we can do if we remove the subtree of $$$u$$$.
Let $$$\operatorname{can}_i(u)$$$ be $$$1$$$ if we can remove all the nodes from the subtree of $$$u$$$, $$$0$$$ otherwise.
Let $$$\operatorname{can}_o(u)$$$ be $$$1$$$ if after deleting the subtree of $$$u$$$, we can remove all nodes except $$$u$$$'s parent, $$$0$$$ otherwise.
Let $$$\operatorname{cnt}_i(u)$$$ be the number of children $$$v$$$ of $$$u$$$ which we can turn into a leaf if we restrict ourselves to subtree of $$$v$$$.
Let $$$\operatorname{cnt}_o(u)$$$ be the number of nodes connected to $$$p(u)$$$ which we can turn into leaf after deleting the subtree of $$$u$$$.
Now, $$$\operatorname{in}(u) = \sum_{v \in C(u)} \operatorname{in}(v) + \left\lfloor \frac{\operatorname{cnt}_i(u)}k\right\rfloor$$$; $$$\operatorname{can}_i(u) = 1$$$ iff $$$\operatorname{cnt}_i(u) = \left\lvert C(u)\right\rvert$$$ and $$$k \mid \operatorname{cnt}_i(u)$$$; $$$\operatorname{cnt}_i(u) = \sum_{v \in C(u)} \operatorname{can}_i(u)$$$.
We can compute this in the first DFS.
$$$\operatorname{out}(u) = \operatorname{out}(p(u)) + \operatorname{in}(p(u)) - \operatorname{in}(u) - \left\lfloor \frac{\operatorname{cnt}_i(p(u))}k\right\rfloor + \left\lfloor \frac{\operatorname{cnt}_o(u)}{k}\right\rfloor$$$; $$$\operatorname{cnt}_o(u) = \operatorname{cnt}_i(p(u)) - \operatorname{can}_i(u) + \operatorname{can}_o(p(u))$$$; $$$\operatorname{can}_o(u)$$$ is $$$1$$$ iff $$$\operatorname{cnt}_o(u) + 1 = \deg(p(u))$$$ and $$$k\mid \operatorname{cnt}_o(u)$$$ which we can compute in the second DFS.
Now the answer will be $$$\max_{u}\left(\operatorname{in}(u) + \operatorname{out}(u) - \left\lfloor\frac{\operatorname{cnt}_i(u)}{k}\right\rfloor + \left\lfloor\frac{\operatorname{cnt}_i(u) + \operatorname{can}_o(u)}{k}\right\rfloor\right)$$$.
Code
Can you someone tell me why am I getting TLE for the 4th question (a-good string)? My solution is very similar to the editorial. Here's my submission Your text to link here...
Pass string by reference
My video editorial for C and E.
for Q.3- Make it good, can someone tell why this is an incorrect logic? Looking for a counterexample.
for(int i=1;i<=n-2;i++) { if(a[i]<a[i-1] && a[i]<a[i+1]) { c=i; }
Read the editorial carefully. You don't even explain your logic and expect others to find errors.
5 2 1 1 3 2 what would be the answer for this testcase.
2
ans is 1. if we remove 2 at index 0,we can make it good.
I thought 5 was part of the list.
https://mirror.codeforces.com/contest/1385/submission/87201950 can anyone explain why my code giving tle for problem D. it is similar to editorial solution.
This mistake is very common, I made it too. You are passing arguments by value which is very costly when the function gets called recursively. Just change the string argument to
string& ain count function and it should just work fine.try string & a,not string a.
Can anyone tell what is this code wrong for solution of Problem B? I iterate from left to right and take all integers in permutation1 until the first integer of permutation1 appears again. Then I do the same for second interger. But it throws wrong ans.
change
vector<int> A(2*n),B,C;tovector<int> A(2*n),C; vector<int> B(n);change
B.emplace_back(A[i]);toB[j]=A[i];because here:
if( A[i] !=B[j] )B vectoratjindex may overflow.Try again.Thanks, it worked. It was indeed overflowing.
congratulation!
I am
WAon test case 2 forE. I have used the same topo sort approach. Where am I getting wrong? 87207905Detailed editorials with appropriate proofs and implementations on my YT channel
Editorial for problem C: https://youtu.be/vjRHaZb16bU
Editorial for problem D: https://youtu.be/9gVpnosPKec
Editorial for problem E: https://youtu.be/yn6ZPaqwlso
Enjoy watching!!
The logic is clearly explained! Good work!
If anyone need, Detail Explanation(not a video tutorial) for D Here
If anyone is interested in rerooting and DP on trees method to solve Problem F: here is my submission :) 87210583
Please explain it as well.
Sure ^_^
TL;DR: if you have calculated the maximum number of moves for the tree rooted at a vertex, find a formula to calculate it for its children.
Let's notice, first of all, that if we choose a vertex $$$v$$$ and consider it as a root, then we can calculate the answer for that vertex: that is, we can find the maximum number of moves if at each move we remove $$$k$$$ leaves such that each leaf was a child of it's adjacent vertex for the graph rooted at vertex $$$v$$$. A point to note: in particular, this means that we never remove vertex $$$v$$$, as it never becomes a child of another vertex, as it is the root itself.
Now another lemma: for a fixed rooted tree, the order of operations is not important. As long as there are leaves in the rooted tree (note again, that we don't count the root as a leaf), we can keep removing them, and any sequence of moves leads to the same number of moves.
So for a given vertex, we can calculate this by DFS. Let's calculate it for vertex $$$1$$$.
But we also maintain a map: $$$map \lt pair \lt int, int \gt , bool \gt ok$$$, which I define as $$$map[{v, par}]=true$$$ if in the tree, if $$$v$$$ is child of $$$par$$$, can we remove all children of $$$v$$$, hence making it possible for $$$v$$$ to be removed from $$$par$$$.
We also store $$$count[v]$$$: number of direct children of $$$v$$$, each denoted by $$$u_i$$$, such that $$$ok[{u_i, v}]=1$$$
Now let's say we have calculated the maximum number of moves $$$moves[v]$$$ in rooted tree $$$v$$$. How to do calculate $$$moves[u]$$$ for each child $$$u$$$ of $$$v$$$?
Here's where re-rooting comes in. We have the recurrence relation
$$$moves[v]=moves[par]-count[par]/k+(count[par]-ok[{v, par}])/k-c/k+(c+ok[{par, v}])/k$$$ where $$$c$$$ is the number of vertices adjoint to $$$v$$$, except $$$par$$$, such that $$$ok[{c, v}]=1$$$.
The formula may look a bit involved but it is really intuitive so I recommend you to come up and prove it yourself. Lemme know if you have any doubts :)
The answer is, of-course, the maximum over all $$$moves[v]$$$
Can anyone help me out with this solution for C — Make it good? Link to solution. It is giving correct solution on test 2 on my local machine, but wrong answer on the site. I changed Language from GNU C++14 to GNU C++17(64), it passed a few more test cases but again it gave Wrong answer. Can anyone tell me why this is happening? Thank you.
de
please don't copy your whole code here, it might spoil the solution.
sry im new to codeforces
if you do want to do it, there is a part under the codeforces icon when you write your comment that supports spoilers.
Can someone explain why this wont pass third test case for problem B? 87097426
ll a[52];not enough,tryll a[200];somebody please explain me the tc 3 of Problem D :: 8 bbaaddcc
how the ans is 5.
Thanks in adv
aaaabbcd , total changes = 5
aaaa bbcd , left half has all chars = 'a'
aaaa bb cd , left half has all chars = 'b'
aaaa bb c d , both halves are of length 1
So this is a a-good-string with 5 changes
В задаче Е опечатка в разборе. В Третьем слове вместо «если» написали «ели»
Can i get a author's testcase, which doesn't fit completely in CodeForces tasks set of testcases? (for example, the 2nd testcase from task F)
I can give you the generator's code. Note that you need to have testlib library locally to compile it.
Thank you I just have some trouble with this line in solution:
if (leaves[a].size() == leaves[b].size()) return a < b;
I don't understand, why code without this line gives WA and with this line gives AC
Can you explain?
If two vertices have the same size of $$$leaves_v$$$ and you don't compare them by some other metric, it's undefined behavior. Standard set can't handle it (because can't compare which one is less than other) and you don't know what it will return, so it can lead to RE or WA easily. This is a pretty standard bug.
Is rating updated ??
I participated the contest but then I had to go somewhere else, so i leaved the contest without single submission.
So is this some kind of feature of codeforces to not involve
registered but not participated contestantin rating changes?If you participate without a single submission, you are considered to not participate in that contest. Simply, you won't gain nor lose any rating.
What if I participated in the contest and submitted a solution which got WA on the first testcase itself. So will that affect my rating?
try it.
jk, don't.
Thanks for information.
you're welcome :))
Can someone explain how the complexity of the 4 th solution is (nlog(n)) ?
Look up the master theorem.
For 1385F - Removing Leaves, there exists a O(n) solution based on rerooting of trees concept.
Check this out : 87161106
i am kinda having hard time understanding B . can anyone plz explain the problem and tutorial to me . thanx in adv.!!!!!
In B you are given a permutation of length n. Now the question is saying that they are merging the permutation of n with itself. So now you just have to find the original order of the permutation. Merging is done in such a way that the relative order is maintained in the merged permutation.
Like for e.g- original permutation is 5 3 2 4 1 and the merged permutation is 5 3 5 2 4 3 2 1 4 1
the answer to the question is simply 5 3 2 4 1 and you can see that the relative order is not distorted 3 if always coming after 5, 2 after 3 and soon.
Is there any iterative way to solve problem D?
using unordered_set in Problem B. gives WA. why?
There might be some issues with your code. My solution is working perfectly. Link to my solution
Hi Guys, I couldn't solve question E during the contest.So I thought to give it a try after reading the editorial. I have implemented recursive method of topological sorting before , so i tried to do it by iterative method, but somehow i couldn't get the logic right .
This is the case on which my code fails 3 3 1 1 2 1 1 3 1 3 2
Here is the my code : https://mirror.codeforces.com/contest/1385/submission/87232399
Can somebody help.
IN Codeforces Round #656 (Div. 3) Question C If I use the logic: grp=2; if(n>2) for(int i=n-3;i>=0;i--) { if(a[i]>a[i-1]&&a[i-1]<a[i-2]) break; grp+=1; } cout<<grp<<endl;
How is the logic wrong?? Can anyone give me a corner case?
I believe you wanted to write ((a[i]>a[i+1])&&(a[i+1]<a[i+2]))... And here is a test case 11 5 8 2 7 5 5 6 7 6 Your code gave the output 3 while the answer is 5...
Hello,Please tell me why my code is not getting submitted,ques C,make it good. link:https://mirror.codeforces.com/contest/1385/submission/87280302
I guess you understood that we need to find the last index in the array such that the values to the right of it are greater and those to the left are smaller and are our answer will be that index, but the problem with your code is that the immediate value to the right of desired index might be equal to the value at that index and your code will in that case skip that position and give the wrong answer,
I made some changes in your code, check this submission 87360403
I just used a variable flag that will indicate that there are values to the right of current position that are greater than the value at current index, and if it is true and the value at current position is less than the value to the left of it then we have found our answer.
Thanks...Are you connected in any social networking site like facebook,instagram. or r in in telegram so that we can discuss problems there.Please share those linkes if possible.Thank You
I have sent you a personal message....
For G, after the contest, I found a different, possibly simpler, greedy algorithm.
I haven't proved that this always gives the best result, or worked out its theoretical performance, but my Python implementation of this passed all the tests well within the time limit.
See https://mirror.codeforces.com/contest/1385/submission/87241176
87210508
I seem to be getting an error for my solution of Problem F on some test of testcase 19 and can't seem to figure out where the error is. Can someone here help me out ? Thanks in advance
https://mirror.codeforces.com/contest/1385/submission/87250962
Please anyone explain me why my code of question 4 is resulting in TLE on testcase 3.
vovuh Both solutions are the same while one is giving the wrong answer for problem F. Wrong Answer Accepted
Those solutions are not exactly the same:
In your WA solution, you call leaves[v].erase(leaf) and then s.erase(v)
In your AC solution, you call s.erase(v) before you call leaves[v].erase(leaf), and since v points to s.begin(), the operation leaves[v].erase(leaf) will use a different value of v in your AC solution.
I don't get what difference does it make changing the order of execution of leaves[v].erase(leaf)
v is a pointer to the first value of s (s.begin()). If v was defined as v=s.begin(), rather than v=*s.begin(), then there would be no difference between the two solutions. But since v points to s.begin(), when you remove v from s, then call leaves[v].erase(leaf), the value of v here is different from the value of v in the previous line
Hello! For problem C, I saw 'binary search' as a problem tag. If anyone here is able to produce a solution as to how one uses binary search to solve the problem, I will greatly appreciate it.
Problem E was easily searchable, I think. A quick search on
how to convert undirected graph to directed acyclic graphgot me this result.https://stackoverflow.com/q/45752331/7121746
Can somebody elaborate on this part in F please:
Shouldn't we insert v when we are done removing all the k leaves? Also what is the part if(leaves[leaf].count(v)) for?
https://mirror.codeforces.com/contest/1385/submission/87268081 can anyone explain what this code get Memory limit exceeded on test 7 !!
Problem F: http://mirror.codeforces.com/contest/1385/submission/87375086 wrong answer test case 20. I can't find my error.what's wrong in this. I used connect[] and leaf[] array for counting number of node connected with j node and number of leaf node connected with j node...help me..thank you
Here is my solution for problem D https://mirror.codeforces.com/contest/1385/submission/87450593 and its showing time limit error. Can anyone please tell me the mistake (I have used recursion as explained in tutorial).
Why is the time complexity of Problem D not exponential but O(N * log N)?
Write the recurrence relation out. It will look a lot like merge-sort. T(n) = 2*T(n/2) + n
This submission might help see the recursion better.
Can someone help me in problem F. Why does the solution in the tutorial add the leaf again into the set st? I fell it to ensure that st doesn't become empty. Any other reasons for it?
Because the set with the custom comparator doesn't update the order of elements itself, you need to update it "manually". But during one move very few elements change so we can just remove them and then insert back with updated values.
I have tried to make editorial for questions A-E . please have a look. Language :- Hindi
https://www.youtube.com/playlist?list=PLrT256hfFv5UWlctr-HughML5i0U-zFYy
For Problem G, can someone explain how did you think of modeling the problem as a graph?
I think the modeling process is like 2-SAT
shouldn't the complexity of A-good string be O(n)? As we are doing constant amount of calculations for each node and there are a total of 2*n nodes in segment tree.
In the problem F, if two nodes have same no of leaves then the order of them can be anything right, why do we need the if statement in the comparator function of the problem F. Please anyone clarify it.
``` struct comp {
};
code for Problem C, please explain what is wrong, same logic as of editorial
can anyone explain what i am doing wrong for Problem C, its getting failed at 888th test-case,
i used the same logic as in tutorial, but the code is bit different but i didn't get it, please help