All ideas belong to MikeMirzayanov.

1249A - Yet Another Dividing into Teams

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int n;
cin >> n;
vector<int> a(n);
for (int j = 0; j < n; ++j) {
cin >> a[j];
}
bool ok = true;
for (int p1 = 0; p1 < n; ++p1) {
for (int p2 = 0; p2 < p1; ++p2) {
ok &= abs(a[p1] - a[p2]) != 1;
}
}
cout << 2 - ok << endl;
}
return 0;
}
```

1249B1 - Books Exchange (easy version)

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int n;
cin >> n;
vector<int> p(n);
for (int j = 0; j < n; ++j) {
cin >> p[j];
--p[j];
}
for (int j = 0; j < n; ++j) {
int cnt = 0;
int k = j;
do {
++cnt;
k = p[k];
} while (k != j);
cout << cnt << " ";
}
cout << endl;
}
return 0;
}
```

1249B2 - Books Exchange (hard version)

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int n;
cin >> n;
vector<int> p(n);
for (int j = 0; j < n; ++j) {
cin >> p[j];
--p[j];
}
vector<int> used(n);
vector<int> ans(n);
for (int j = 0; j < n; ++j) {
if (!used[j]) {
vector<int> cur;
while (!used[j]) {
cur.push_back(j);
used[j] = true;
j = p[j];
}
for (auto el : cur) ans[el] = cur.size();
}
}
for (int j = 0; j < n; ++j) cout << ans[j] << " ";
cout << endl;
}
return 0;
}
```

1249C1 - Good Numbers (easy version)

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int n;
cin >> n;
while (true) {
bool ok = true;
int cur = n;
while (cur > 0) {
if (ok && cur % 3 == 2) ok = false;
cur /= 3;
}
if (ok) break;
++n;
}
cout << n << endl;
}
return 0;
}
```

1249C2 - Good Numbers (hard version)

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const long long INF64 = 1e18;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
long long n;
cin >> n;
vector<int> vals;
int pos2 = -1;
while (n > 0) {
vals.push_back(n % 3);
if (vals.back() == 2) pos2 = int(vals.size()) - 1;
n /= 3;
}
vals.push_back(0);
if (pos2 != -1) {
int pos0 = find(vals.begin() + pos2, vals.end(), 0) - vals.begin();
vals[pos0] = 1;
for (int i = pos0 - 1; i >= 0; --i) {
vals[i] = 0;
}
}
long long ans = 0;
long long pw = 1;
for (int i = 0; i < int(vals.size()); ++i) {
ans += pw * vals[i];
pw *= 3;
}
if (vals.back() == 1) ans = pw / 3;
cout << ans << endl;
}
return 0;
}
```

1249D1 - Too Many Segments (easy version)

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 250;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, k;
cin >> n >> k;
vector<pair<int, int>> segs(n);
vector<int> cnt(N);
for (int i = 0; i < n; ++i) {
cin >> segs[i].first >> segs[i].second;
++cnt[segs[i].first];
--cnt[segs[i].second + 1];
}
for (int i = 0; i + 1 < N; ++i) {
cnt[i + 1] += cnt[i];
}
vector<int> ans(n);
for (int i = 0; i < N; ++i) {
while (cnt[i] > k) {
int pos = -1;
for (int p = 0; p < n; ++p) {
if (!ans[p] && (segs[p].first <= i && i <= segs[p].second) && (pos == -1 || segs[p].second > segs[pos].second)) {
pos = p;
}
}
assert(pos != -1);
for (int j = segs[pos].first; j <= segs[pos].second; ++j) {
--cnt[j];
}
ans[pos] = 1;
}
}
cout << accumulate(ans.begin(), ans.end(), 0) << endl;
for (int i = 0; i < n; ++i) {
if (ans[i]) cout << i + 1 << " ";
}
cout << endl;
return 0;
}
```

1249D2 - Too Many Segments (hard version)

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 200200;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, k;
cin >> n >> k;
vector<pair<int, int>> segs(n);
vector<int> cnt(N);
vector<vector<int>> ev(N);
for (int i = 0; i < n; ++i) {
cin >> segs[i].first >> segs[i].second;
++cnt[segs[i].first];
--cnt[segs[i].second + 1];
ev[segs[i].first].push_back(i + 1);
ev[segs[i].second + 1].push_back(-i - 1);
}
for (int i = 0; i + 1 < N; ++i) {
cnt[i + 1] += cnt[i];
}
vector<int> ans(n), sub(N);
set<pair<int, int>> curSegs;
int curSub = 0;
for (int i = 0; i < N; ++i) {
curSub += sub[i];
for (auto it : ev[i]) {
if (it > 0) {
curSegs.insert(make_pair(segs[it - 1].second, it - 1));
} else {
auto iter = curSegs.find(make_pair(segs[-it - 1].second, -it - 1));
if (iter != curSegs.end()) curSegs.erase(iter);
}
}
while (cnt[i] - curSub > k) {
assert(!curSegs.empty());
int pos = curSegs.rbegin()->second;
curSegs.erase(prev(curSegs.end()));
++curSub;
--sub[segs[pos].second + 1];
ans[pos] = 1;
}
}
cout << accumulate(ans.begin(), ans.end(), 0) << endl;
for (int i = 0; i < n; ++i) {
if (ans[i]) cout << i + 1 << " ";
}
cout << endl;
return 0;
}
```

1249E - By Elevator or Stairs?

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, c;
cin >> n >> c;
vector<int> a(n - 1), b(n - 1);
for (int i = 0; i < n - 1; ++i) {
cin >> a[i];
}
for (int i = 0; i < n - 1; ++i) {
cin >> b[i];
}
vector<vector<int>> dp(n, vector<int>(2, INF));
dp[0][0] = 0, dp[0][1] = c;
for (int i = 0; i < n - 1; ++i) {
dp[i + 1][0] = min(dp[i + 1][0], dp[i][1] + a[i]);
dp[i + 1][0] = min(dp[i + 1][0], dp[i][0] + a[i]);
dp[i + 1][1] = min(dp[i + 1][1], dp[i][1] + b[i]);
dp[i + 1][1] = min(dp[i + 1][1], dp[i][0] + b[i] + c);
}
for (int i = 0; i < n; ++i) {
cout << min(dp[i][0], dp[i][1]) << " ";
}
cout << endl;
return 0;
}
```

Thanks to neal for the additional editorial of this problem and providing the linear solution!

**Tutorial**

Tutorial is loading...

**Solution (Vovuh, n^3)**

```
#include <bits/stdc++.h>
using namespace std;
const int N = 210;
int n, k;
vector<int> a;
vector<vector<int>> g, dp;
void dfs(int v, int p) {
dp[v][0] = a[v];
for (auto to : g[v]) {
if (to != p) dfs(to, v);
}
for (int dep = 0; dep < N; ++dep) {
if (dep == 0) {
for (auto to : g[v]) {
if (to == p) continue;
dp[v][dep] += dp[to][max(0, k - dep - 1)];
}
} else {
for (auto to : g[v]) {
if (to == p) continue;
int cur = dp[to][dep - 1];
for (auto other : g[v]) {
if (other == p || other == to) continue;
cur += dp[other][max(dep - 1, k - dep - 1)];
}
dp[v][dep] = max(dp[v][dep], cur);
}
}
}
for (int dep = N - 1; dep > 0; --dep) {
dp[v][dep - 1] = max(dp[v][dep - 1], dp[v][dep]);
}
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
cin >> n >> k;
++k;
a = vector<int>(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
g = vector<vector<int>>(n);
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);
}
dp = vector<vector<int>>(n, vector<int>(N));
dfs(0, -1);
cout << dp[0][0] << endl;
return 0;
}
```

**Solution (PikMike, n^2)**

```
#include <bits/stdc++.h>
using namespace std;
#define forn(i, n) for (int i = 0; i < int(n); ++i)
const int N = 200 + 13;
const int INF = 1e9;
int n, k;
int a[N];
vector<int> g[N];
int dp[N][N];
int d[N];
int tmp[N];
void dfs(int v, int p = -1){
d[v] = 1;
dp[v][0] = a[v];
for (auto u : g[v]) if (u != p){
dfs(u, v);
int nw = max(d[v], d[u] + 1);
forn(i, nw)
tmp[i] = -INF;
forn(i, d[v]) for (int j = max(0, k - i); j < d[u]; ++j)
tmp[min(i, j + 1)] = max(tmp[min(i, j + 1)], dp[v][i] + dp[u][j]);
forn(i, d[u])
tmp[i + 1] = max(tmp[i + 1], dp[u][i]);
forn(i, d[v])
dp[v][i] = max(dp[v][i], tmp[i]);
for (int i = d[v]; i < nw; ++i)
dp[v][i] = tmp[i];
d[v] = nw;
for (int i = d[v] - 1; i > 0; --i)
dp[v][i - 1] = max(dp[v][i - 1], dp[v][i]);
}
}
int main(){
scanf("%d%d", &n, &k);
forn(i, n)
scanf("%d", &a[i]);
forn(i, n - 1){
int v, u;
scanf("%d%d", &v, &u);
--v, --u;
g[v].push_back(u);
g[u].push_back(v);
}
dfs(0);
printf("%d\n", dp[0][0]);
}
```

Thanks for the quick editorial.

[Deleted]

i didn't get the output of C1. Can someone make me understand

For C2, why does a position of 2 in the ternary representation need to be replaced?

What does it mean a position of 2 in the ternary representation? It means that the corresponding power of 3 is used 2 times. But in the problem, it is said that there should not be any power repeated twice or more. That's why the position of 2 in the ternary representation replaced.

Makes sense, I didn't see it that way (different approach to the solution). Thanks for clarifying.

why cant we replace the 2 in the ternary representation to 1 ? why do we have to replace it with 0 ?

Let's assume we have a ternary representation of 200 (it's decimal equivalent is 2*(3^2)+0*(3^1)+0*(3^0)=18) now if we replace 2 with 1 and replace all 0s to the right of 2 with 1 then we get 111 which is ternary representation of 13. So you see we get a number which is less than 18 but we want a number greater than 18 so we have to set one more bit to 1 to the left of 2 which makes out ternary representation to 1111 (it is decimal equivalent of 40) but now if we make all 1 to 0 except the leftmost 1 then we get 1000 (it's decimal equivalent is 27 ) which is the least number we can get which is a good number and greater than 18. Hope this example will help you to understand the solution better.

i am very confused what the turoial want to say,

I Did This :- it works :-

https://mirror.codeforces.com/contest/1249/submission/251409990

Precomputation Although :-

For C2, I used a simpler greedy approach. First, I find the smallest number that is the sum of all powers of 3 till the sum becomes >= n. Let us call the sum Z. So this will lead to a number of form 111111 where 1 => The corresponding power of 3 is added to the sum. Now, suppose the highest power of 3 that was included in the sum is m. I go from mth to 0th power and try to subtract corresponding power from current Z. If it leads to a number that is still >= n, we reduce Z by the current power. Else, we continue. Intuition is that if we subtract a larger power, its sum is anyways going to be larger than the sum of all lower powers combined. So it will be a better choice for sure.

My Submission: https://mirror.codeforces.com/contest/1249/submission/63202537 (Got AC, I hope no hack :P )

i submitted same logic but got wa , i changed the order too.

That's a pretty elegant solution. It does seem quite intuitive, but, do you have any sort of proof of why this greedy approach works?

My solution the same. I think we MUST include x=10000 from y=11111 form if n<=y because it's guaranteed that n strictly between 1111 and 11111 (because of previous step, where n>1111) and 10000 smallest option for this 63155044

i saw you code,It was the easiest approach. Thank you so much.

I have an $$$\mathcal{O}(n \log n)$$$ solution to problem F: 63206899.

The key idea is to merge smaller DP states into bigger. Would have been a nice harder Div. 1 problem. :)

Update: it's actually $$$\mathcal{O}(n)$$$.

It's also funny because I submitted $$$\mathcal{O}(n^4)$$$ in contest: 63153425

Can you please explain your approach in detail? thanks.

Is it possible for you to provide a quick explanation of your approach ? This looks interesting. I could only come up with O(n2) but O(n) is really impressive.

Since it seems like there's plenty of interest, I wrote a post to explain my full optimization process from $$$\mathcal{O}(n^4)$$$ to $$$\mathcal{O}(n)$$$: https://mirror.codeforces.com/blog/entry/70822

Thanks a lot neal :)

It was amazing contest!

My thoughts exactly after seeing B. Lmao XD

I did the same as you. Despite of the different of implementation, I think the author's ideal is the same, the way he implemented like using dfs :v

Difference is that a simple vector/ArrayList would suffice for finding the cycle, so the image still applies. I don't expect a Div3B problem to require the participants to be familiar with DSU.

I use Tarjan to slove problem B2.....

Wow, you definitely win, I don't know Tarjan's lol

It is a algorithm that you can find all the strongly connected components in a directed graph in O(n). Then you can count the size of each strongly connected component.

That's definitely an overkill. I did the same as Spheniscine. Maybe I should look into Tarjan's. That definitely will come in handy.

Orz wqyzstql,you are so strong !

Thank you very much!

Tarjan is much more overkill =)))

If DSU is a sword, Tarjan's is a chainsaw

But I think DSU is a very common algorithm in some Algorithmic competition...And Tarjan is rarely used.

I didn't understand the tutorial of C1 here, more explanation please?

Convert the number to base 3 (ternary). A number is good iff there is no $$$2$$$ digit in it.

For the easy version, incrementing the number until it's good would be sufficient. For the hard version, a more-analytic solution is required.

Why is it that 2 shouldn't be present in the number?

Converting a number $$$n$$$ to base 3 is a way of representing $$$n$$$ as the sum of powers of 3. e.g. if $$$n = 1201_3$$$, $$$n = 1\cdot3^3 + 2\cdot3^2 + 0\cdot3^1 + 1\cdot3^0 = 3^3 + 3^2 + 3^2 + 3^0$$$. The problem requires the powers of 3 be distinct, which means a good number should have no $$$2$$$ in its base-3 representation, as that would mean more than one instance of that power.

why add a power of 3 at pos0 will make remaining element 0 as said in the tutorial of problem c2. I'm not getting it would you please explain it also.

I'd explain it like this:

First, convert $$$n$$$ to base 3, and prepend a leading $$$0$$$. Store this in array $$$s$$$.

Now we want to get rid of all instances of $$$2$$$ in $$$s$$$. How do we do that? Let's just focus on

oneinstance, we'll call its index $$$i$$$.Now we want to increment $$$n$$$ until $$$s[i]$$$ changes. What happens then? You can imagine $$$s$$$ as being like a car odometer, except in base 3. Thus, the $$$2$$$ will roll up to become a $$$0$$$. When this happens, all digits to the right of $$$s[i]$$$ will be $$$0$$$, and $$$s[i-1]$$$ would increment.

Since getting rid of one $$$2$$$ this way zeros out all digits to its right, we can now fix $$$i$$$ to be the index of the leftmost $$$2$$$.

We're still not done yet though, because if $$$s[i-1] = 1$$$, it's now $$$2$$$, which we don't want. So we zero it out too and continue leftward until we find a $$$0$$$, which is guaranteed because we prepended a leading zero. We can then increment that $$$0$$$ to $$$1$$$, then convert the array back into the final answer $$$m$$$.

thank you so much, i was stuck on this problem for hours.Then i read the turorial still i cant understand it.But you explained so well.Your hardwork is appreciated

I tried to solve B question both parts using dfs but getting the wrong output due to bug somewhere in the code. could anyone please help me debugging.it would be great help for me.

## include <bits/stdc++.h>

using namespace std; vectorv[100005]; bool vis[100005]; int size=0; void dfs(int n) { vis[n] = true;

size++; for(int x : v[n]) { if(!vis[x]){ dfs(x); }

} } int main() { //int q; //cin>>q; { int n; cin>>n; int arr[n+5],hello[n+5]; for(int i=1;i<=n;++i){ cin>>arr[i]; if(i==arr[i]){ hello[i]=1; vis[i]=true; continue; } v[i].push_back(arr[i]); v[arr[i]].push_back(i); } for(int i=1;i<=n;i++) {

if(!vis[i]){ size=0; dfs(i); //cout<<size<<endl; for(int j=1;j<=n;j++){ if(vis[j]){ hello[j]=size; //cout<<hello[j]<<" "; }

}

I have for a while commented on the query part for more simplicity.

-_-

I thought of a greedy for problem D1 which consisted of looking for the segment that had the most points that currently belonged to more than k segments and was updating. Can someone explain to me why this greedy doesn't work.

Look at the testcase

Now you would choose the [10; 20] segment, as it contains 6 points. But then you need to remove 2 more, where you could've deleted segment 4 and 5.

In generał, you can try to disprove greedy by creating example in which the first choice ruins the rest of solution because now you have to pick lots of small elements to solution.

Can you explain why the approach in the editorial is correct

It's also a greedy approach, but instead we step through the segments in the order of their $$$l$$$, and preferentially remove the segments with the greatest $$$r$$$ when too many cover the current $$$l$$$ value (because removing those segments will lower the count more permanently for future values of $$$l$$$)

See this submission for an example: 63225521

By the way, this problem would be a perfect use-case for a double-ended priority queue (e.g. MinMaxPriorityQueue from the Guava library), but I don't know of any languages that have that in its STL. TreeSet/ordered set works though, with a slightly worse constant factor.

Can you help me understand why my implementation does not work? My submission is 63270199

I did the following: 1. sort all segments first by their r then by l then by index. 2. starting from the minimum to the maximum integer point, do the following. (a).as long as the current segment has not been removed and covers the current integer i, increment the counter cnt for i. (b).if cnt <= k, do nothing, move on to the next integer; else mark all the remaining segments that covers i as removed and increment the deletion counter that needs to be output. If the segment has been removed, do not increment this deletion counter.

I thought this should work as the segments are sorted by their r first. So I am always processing segments that have smaller r. It turns out that I must be wrong somewhere...

I'd appreciate your help.

You shouldn't sort segments by $$$r$$$ first; you should sort them by $$$l$$$ first, then "sweep" from left to right, maintaining current covering segments in a TreeSet (or a MinMaxPriorityQueue if you're adventurous enough to try implementing one lol). It's the TreeSet/DEPQ that should maintain order by $$$r$$$, because you want to remove the minimum $$$r$$$ entries from the structure that are less than the current $$$l$$$ value (because they aren't covering our "cursor" anymore), as well as preferentially drop the maximum $$$r$$$ entries (and adding them to the answer set) if there are too many segments.

For a TreeSet, a secondary comparator for $$$id$$$ is necessary because TreeSet will assume that entries are equal if the comparator returns 0, and so won't add "duplicates". Since $$$id$$$ is contextually guaranteed to be unique, it's a sufficient tiebreaker.

This "sort by $$$x$$$ first, then put it into a priority queue/set that sorts by $$$y$$$" is actually a very common pattern in CP problems involving segments or time-intervals.

Thank you for your quick reply! I implemented a solution based on your suggestion and it was accepted.

As for the adventurous MinMaxPriorityQueue implementation, now I wouldn't ask you how to solve problem D easier version had I know how to implement that data structure, would I? :)

I am pretty new to competitive programming so it has been quite a struggle here for me in codeforces contests. Mind if I add you as a friend?

Certainly, though the "friend" list on this platform is more like a watch/follow; it doesn't even tell me who added me as a "friend".

I'm glad it helped though! I've been programming for a long time, but it's only recently that I've started to pick up advanced optimization / data structure techniques. I started learning about them while solving the puzzles on http://adventofcode.com , then moved on to here after I've solved them all

Yeah, the UI here is not the best to navigate. There is a send message feature, maybe it is like sending emails?

I've been programming for a few years too but never paid attention to developing my algorithmic skill. Decided to form a habit of solving problems/joining contests to improve my problem solving skill. I first started on LeetCode a few months ago then decided to make it more painful by signing up here. The problems here are definitely a lot challenging.

Yes; "send message" is like private/direct messaging systems on forums

Hey , can you help me in D2 of problem of this round?? My approach was to sort by r and then traverse the array and keep a segment tree for counting how many current elements are there in any range. If i can add a segment , I add 1 in range (li to ri) . Then check for next segment if max value between (li+1,ri+1) is less than k , I add it too and keep on doing same . This approach works when k = 1 as it is a standard max activity problem but for larger k it isnt working and I am unable to understand why this doesnot work. Link to my submission

Hi, mate i find jmrh's idea is similar to mine, but not completely the same. i thought that looking for the longest segment that is consisted by more than k points,or the segment consist the points(using the testcase upon, that more than K points would be {10 min,22 max}.)is the most possible one to be pop out,and so on . (first try,it would be seg[10;20],then left with point 21 and point 22 . sec try it would be the seg[21,30] cuz its longest by now and contains 21 and 22,and it's done.)

so is this kind of way could work? how s it compare to the approach in the editorial?

Hi ;) Sorry for late response ;) I don't quite understand, after removal of [10; 20] and [21; 30] point with x = 8 is still present in [1; 9] and [8; 12].

Crap! U got me . Thx bro.i just realized

Np ;) Good luck :)

Still don't know why problem F is O(n^3) but not O(n^4)

What does dp[u][dep] represent in problem F?

I think it can be derived like this: Let $$$child_i$$$ denote the number of children of the $$$i$$$-th node.

The total time taken at the $$$i$$$-th node is $$$O(N * (child_i)^2)$$$.

So, the time complexity of the algorithm becomes:

$$$\sum\limits_i (N * (child_i)^2) = N * \sum\limits_i (child_i)^2 \le N * (\sum\limits_i child_i)^2 = N * N^2 = N^3$$$

So, the time complexity is $$$O(N^3)$$$.

I think you've forgotten to multiply by the depth , am I wrong :/ ?

so you should write N^4

The 'N' he's taken is denoting the depth, its not for number of vertices because he's already taken sigma for summation over all vertices.

Why this py3 solution for B1 got TLE hacked... I think it is almost same with the tutorial...

This is for B1? how does it get hacked :thinking:

yep,63138123,as you can see... Also, I noticed there are many py3 solutions like my submission hacked as well...

I see the hack case now，a simply bad condition which cost 200*200*200 times....sadly for python3,it got TLE. R.I.P

I saw some people solved B2 using DFS. Can anyone explain how to solve B2 using DFS?

For each position ; answer is the size of the connected component in which that element is.

Only those elements are connected which form a cycle.

And size of the connected component can be solved using DFS !

Why doesn't greedy work for problem E? I am thinking, if we use the elevator we continue to use it if we can, If we don't use it, we try to take elevator as much as possible as in next step we won't have to add 'c' if we get the elevator

Maybe $$$b_i$$$ is small but $$$b_{i+1}$$$ is very very large. In that case it is better to pay $$$c$$$ cost and get off the elevator to take the smaller $$$a_{i+1}$$$ cost stairs rather than continue with the elevator. It is better to keep best case scenarios for getting to the floor by both elevator and stairs and keep comparing as you go along.

https://mirror.codeforces.com/contest/1249/submission/63192936

What you said is what I wanted to say, meaning we will take the best out of the two but prefer to use elevator if both costs are same

That will be decided by dp, it will factor in the cost of $$$c$$$ against the benefit of $$$a$$$ over $$$b$$$.

Can Anyone Explain why won't we have to check for all previous (1 to i-1) cases for particular i? In this case, it will take O(n**2).

This would be because while travelling from 1 to (i-1), you would have already taken the minimum to reach there. So to reach from (i-1) to (i) , you just need to add the stairs/elevator of the cost from (i-1) to (i) and hence you can ignore all others for particular i

This is not exactly correct. The DP works because of the structure of the distances, as prefix sums of increasing arrays. In general, you will not be right.

This is because of the structure of the distances to the 1st floor (prefix sums of strictly increasing arrays).

I have submitted two solutions for B1 in Python3 and Pypy3 with the same code. Python3 got TLE but Pypy3 AC. Why there is so much difference between these two? If one logic is accepted in one language then it must be accepted in all available languages given there otherwise giving all languages during submission doesn't make sense. vovuh can you please look into it?

AC solution(Pypy) : 63211941 WA solution(Python) : 63211890

It’s somehow similar to using O3 flag when compiling cpp? I guess it’s just more optimised (most of the time)

Edited. I read the code again and find that it's not the input()'s fault. Sorry for my last post.

What if

going downwas also allowed inE, what difference would it create? How'd we solve it then? Can someone please throw some light on it?Going down

iscurrently allowed right now, it just won't create an advantage. Because you'll have to go up to the floor, cross it, and then come down, which will always be larger than going up to the floor and stopping right there. So, going down will not give any advantage because of the structure of the problem.Thanks! Got it now

In this problem you are

allowedto go down, but this is not optimal at all.Thanks!

Anyone else who did B2 with DSU :') ?

For Problem D I created no more than k non-overlapping segments greedily based on shortest right position. This approach seems to fail for some reason, any ideas why it's failing? code

Regarding the C2 question test case 9.

Can someone tell me how can the input of 450283905890997363 expect itself as an answer? the ternary representation of this input is : 10000000000000000000000000000000001200 which has a 2 in it so the correct answer should be 10000000000000000000000000000000010000 convert into decimals, right?

you are wrong, 450283905890997363 representation of 3^37

upd: you can check diff between 450283905890997363 and 450283905890997444 (your answer) is 81 (fifth bit that equals to 3^4)

You are right, thanks for pointing out 3^37. my conversion function was wrong I guess

In the contest, after long time debug on problem D1 (stupid update error), I even don't read statement E, now I solved it easily without Editorial. I'm very regret about it.

Here is an $$$O(n^2)$$$ solution for problem F: 63223147 , I used a greedy approach to solve it and have not find any hacks. I think it is a correct approach.

Someone also mentioned this in his/her's blog.

https://www.cnblogs.com/wrjlinkkkkkk/p/11726342.html

Oh yeah, thank you !

Can someone explain to me the code for B2?

It is obvious that we have bunch of cycles that are not connected.

I keep an array(S) for size of each cycle and an array(N) for each node that shows what cycle they belong to.

for each node i see if it belongs to some cycle . if not i go traverse that until i reach the same node and number all nodes the same(in N) .

like this :

so after i checked all nodes for the node

iits output is S[N[i]]can someone help me with this?

my failed submission

my accepted submission

the only change in the both is that i commented out

ios_base::sync_with_stdio(false);

cin.tie(NULL);

how does this affect the output?????

I think your vector is out of range , you can test when $$$n$$$ is $$$1e18$$$ ,the variable $$$cnt$$$ is 40 ,but the size of the vector is 39 . Obviously, it is out of range.

But I don't understand why removing the fast i/o lines fixes the issue.

I also don't understand. Maybe it's an undefined behavior.

The cnt can't be 40..the input size is max 10^18...so cnt can be max 38

You can debug ,and output the

`cnt`

.My complier is visual stutdio ,when n is 1e18,there will be an error that the vector is out of rangeThanks for pointing it out..My last for loop was buggy. Fixing the loop made the solution get accepted without removing ios_base::sync_with_stdio(false)

But i still can't see how removing ios_base::sync_with_stdio also removed the bug XD

I also don't know , we need professionals to answer this strange problem.

endl flushes the output

Kindly explain dp[v][dep] in probelm F (Maximum weight subset)?

Too good contest, thanks

Another (easier?) implementation of D2

Editorialist made D2 implementation way too complicated and for no reason. Here is my submission

Can some one explain Editorial of D2 , i did not got clear understanding .

If you've understood idea behind D1, then its the same thing done efficiently or say smartly ; observe this code, may be you'll get this from it.

Yeah , i was trying to do without using the set (STL) and that's why i was facing difficulty. I upsolved it finally.

Can you please explain in detail your approach now @rananjay

My solution for 1249B2 — Books Exchange (hard version) was giving wrong answer using disjoint set with path compression. Any idea what was wrong ?

I am getting WA in D1. Link to solution: https://mirror.codeforces.com/contest/1249/submission/63245610

Please help!

As usual, I don't understand the editorial.

In E. What does it mean by "Time overhead" ? Time to close the door? Time to open the door?

Every time you use the elevator, you need to pay $$$c$$$ in time cost once, in addition to the $$$b$$$ time costs for the floors traveled. (You can imagine it as time spent waiting for the doors to open. Don't think too hard about the real-world logic of it though; this is CP-land) You can ascend several consecutive floors per use, but if you get off to use the stairs, then go back to the elevator, you have to pay $$$c$$$ again.

I got another solution for C2 which does not involve any case handling (or dealing with annoying carries!).

We can easily write a function $$$f(i)$$$ that returns the i-th good number. It can be implemented by representing $$$i$$$ in base 2, then treat the resulting digits as a base 3 number. The function runs in $$$O(\log i)$$$.

Then we use binary search to find the smallest $$$i$$$ that satisfy $$$f(i) \geq n$$$. If $$$s$$$ is the smallest $$$i$$$ that satisfy the condition, the answer would be $$$f(s)$$$. The total time complexity would be $$$O(\log^2 n)$$$.

My submission: 63244719

Nice approach!

Btw, I think you mean, "express $$$i$$$ in base 2" instead of $$$n$$$. And "$$$f(i)$$$ can be found in $$$O(\log i)$$$".

Thanks for your correction ;)

Beautiful :)

deleted

Is it ok, what a few participants 1600+ rating got it updated after contest?

Can problem F be done for general graphs in polynomial time?

Nope. Even the case with $$$k = 1$$$ and all $$$a_i = 1$$$ is NP-complete (it is maximum clique problem).

what is the meaning of curSegs.erase(prev(curSegs.end()));

Erasing the last element of curSegs

Problem E

Why the jury's answer 14th number is 23 instead of 24? The next number by stairs will have +3 and by elevator either +1 or +3 (1 + 2),

The previous value was 21 so it can be either 24 or 22..

Here is pastebin what I'm talking about: https://pastebin.com/FmZyit6Y.

Hey did you figure out the problem? I have the same doubt.

I think you didn't use 2 states in your dp, i did the same. Why it will fail is, suppose you can go from the i'th floor to x'th floor by stairs, but you find that, it's optimal to reach the (x+1)th floor by elevator, from the i'th floor. (Note that this is because the cost 'c' is added just once). so obviously you need 2 states of dp for this.

vovuh For problem F. The recurrences you mention work if computing the maximum value of a set when distance between any $$$2$$$ vertices is $$$\ge k$$$. But, the problem asks to compute the answer when the distance between any $$$2$$$ vertices is $$$> k$$$. When I read your code, I saw that you are actually incrementing $$$k$$$ beforehand. So, I think you should mention in the tutorial that you change the problem first to the $$$\ge k$$$ version or update the recurrences as:

If $$$dep = 0, dp_{v, 0} = a_v + \sum\limits_{to \in children(v)} dp_{to, k}$$$.

Otherwise, $$$dp_{v, dep} = max(dp_{v, dep}, dp_{to, dep-1} + \sum\limits_{other \in children(v) \setminus \{to\}} dp_{other, max(k-j, j-1)})$$$

Also, there's a typo:

"we can notice that this is. I think you meantnowwhat we wanted"not.Yes, I forgot to mention that I increase $$$k$$$ in the beginning of the program to compute the subset with distances $$$\ge k$$$, thank you. And thanks for mentioning the typo, it will be fixed now.

what is life

In C2 tag is given meet in the middle but it is giving TLE . Dd anyone solve this using meet in the middle and binary search , please reply ...

How to optimize this meet in the middle + binary search to C2? 63895945

Yes I am also asking for that

by increasing the size of the vector on which binary search is performed, thus reducing the size of the other vector. accepted code : 67982667

Can someone explain the recurrence of problem F in detail ?

Could you please explain the recursive relation in your tutorial of problem F, and what does dep-1 and k-dep-1 mean ? neal

Sorry for hitting this old post.

I wanted to share my thoughts for an alternative solution to E using recursion dp.

For the E elevator and stairs I wrote a recursive approach so initially it struck me that I will end up running my dp solution from top to every other floor which will end up giving a bad complexity of

O(N^2).Well to tackle that we could simply run from the last floor to the top and then we need to take max of the dp values from each floor to the first floor. (Max because along a route with the same minimum time if we go from the top to the floor we take the overhead charges before and as a result we took minimum back then and now starting from that floor on our way to first floor we take overhead charges from that floor already so for the floors where took minimum of the dp values now we have to take maximum instead)

An explanation of also how we could use to move the other way around besides moving from lower floor to higher floor as stated in the problem.

My submission. https://mirror.codeforces.com/contest/1249/submission/74881991

P.S- If my solution is incorrect please do let me know. (If it can get hacked)

Hello I got WA on the test case 1 for the final case, is this because I am only using long?

The last four digits are the only difference. Here is the link https://mirror.codeforces.com/contest/1249/submission/90984427.

Thanks

can any one tell what's wrong in my solution to c1 , my idea : to generate all subset (sum it ) and put it in set and check whether it is present or not 114425157

115143692 Modified code

https://mirror.codeforces.com/contest/1249/submission/165470986

Can anyone help me what wrong I have done?

for D2, I use 2 segment trees, the first one to find the smallest position which is covered by strictly more than k segments and the second one is find the segment which covers this position and its right bound is the biggest. the TC is O(NlogN) code : https://ideone.com/KsuR2Q