Thanks to Rox and _overrated_ for help with problem ideas and preparation!

1294A - Коллекционирование монет

Idea: MikeMirzayanov

**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 a[3], n;
cin >> a[0] >> a[1] >> a[2] >> n;
sort(a, a + 3);
n -= 2 * a[2] - a[1] - a[0];
if (n < 0 || n % 3 != 0) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
}
}
return 0;
}
```

Idea: MikeMirzayanov

**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 t;
cin >> t;
for (int tt = 0; tt < t; tt++) {
int n;
cin >> n;
vector<pair<int, int>> a(n);
for (int i = 0; i < n; ++i) {
cin >> a[i].first >> a[i].second;
}
sort(a.begin(), a.end());
pair<int, int> cur = make_pair(0, 0);
string ans;
bool ok = true;
for (int i = 0; i < n; ++i) {
int r = a[i].first - cur.first;
int u = a[i].second - cur.second;
if (r < 0 || u < 0) {
cout << "NO" << endl;
ok = false;
break;
}
ans += string(r, 'R');
ans += string(u, 'U');
cur = a[i];
}
if (ok)
cout << "YES" << endl << ans << endl;
}
return 0;
}
```

1294C - Произведение трех чисел

Idea: MikeMirzayanov

**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;
set<int> used;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0 && !used.count(i)) {
used.insert(i);
n /= i;
break;
}
}
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0 && !used.count(i)) {
used.insert(i);
n /= i;
break;
}
}
if (int(used.size()) < 2 || used.count(n) || n == 1) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
used.insert(n);
for (auto it : used) cout << it << " ";
cout << endl;
}
}
return 0;
}
```

Idea: vovuh

**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, x;
cin >> q >> x;
vector<int> mods(x);
set<pair<int, int>> vals;
for (int i = 0; i < x; ++i) {
vals.insert(make_pair(mods[i], i));
}
for (int i = 0; i < q; ++i) {
int cur;
cin >> cur;
cur %= x;
vals.erase(make_pair(mods[cur], cur));
++mods[cur];
vals.insert(make_pair(mods[cur], cur));
cout << vals.begin()->first * x + vals.begin()->second << endl;
}
return 0;
}
```

Idea: vovuh

**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 n, m;
cin >> n >> m;
vector<vector<int>> a(n, vector<int>(m));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
cin >> a[i][j];
--a[i][j];
}
}
long long ans = 0;
for (int j = 0; j < m; ++j) {
vector<int> cnt(n);
for (int i = 0; i < n; ++i) {
if (a[i][j] % m == j) {
int pos = a[i][j] / m;
if (pos < n) {
++cnt[(i - pos + n) % n];
}
}
}
int cur = n - cnt[0];
for (int i = 1; i < n; ++i) {
cur = min(cur, n - cnt[i] + i);
}
ans += cur;
}
cout << ans << endl;
return 0;
}
```

Idea: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
#define x first
#define y second
vector<int> p;
vector<vector<int>> g;
pair<int, int> dfs(int v, int par = -1, int dist = 0) {
p[v] = par;
pair<int, int> res = make_pair(dist, v);
for (auto to : g[v]) {
if (to == par) continue;
res = max(res, dfs(to, v, dist + 1));
}
return res;
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
p = vector<int>(n);
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);
}
pair<int, int> da = dfs(0);
pair<int, int> db = dfs(da.y);
vector<int> diam;
int v = db.y;
while (v != da.y) {
diam.push_back(v);
v = p[v];
}
diam.push_back(da.y);
if (int(diam.size()) == n) {
cout << n - 1 << " " << endl << diam[0] + 1 << " " << diam[1] + 1 << " " << diam.back() + 1 << endl;
} else {
queue<int> q;
vector<int> d(n, -1);
for (auto v : diam) {
d[v] = 0;
q.push(v);
}
while (!q.empty()) {
int v = q.front();
q.pop();
for (auto to : g[v]) {
if (d[to] == -1) {
d[to] = d[v] + 1;
q.push(to);
}
}
}
pair<int, int> mx = make_pair(d[0], 0);
for (int v = 1; v < n; ++v) {
mx = max(mx, make_pair(d[v], v));
}
cout << int(diam.size()) - 1 + mx.x << endl << diam[0] + 1 << " " << mx.y + 1 << " " << diam.back() + 1 << endl;
}
return 0;
}
```

An even simpler solution to A; first check that a+b+c+n is divisible by 3, then check that (a+b+c+n)/3 is no less than a, b, and c.

What is the reason behind this,

`if m < a or m < b or m < c`

?You cant give negative coins to make the equality condition.

m represents the total coins that person would get at the end of the entire transaction.Since we are only adding coins and not removing coins from any person the final sum present with each of them should be greater or equal to the initial amount of coins

because if A + a = B + b = C + c ,then 3 | （A + B + C + n）,m = (a + b + c + n)/3,So m is what happens when they finish n（sorry，my English is so bad）

i can't understand can u explain to me more

yea i'm sure Now Alice has a coins ,Bar has b coins , Cer has c coins we will divide n to three interger A,B and C so that a + A = b + B = c + C , and A + B + C = n define k = a + A So (a + A + b + B + c + C) = 3*k = (a + b + c + n) 3*k % 3 == 0 <=> (a+b+c+n)%3 == 0 and guarantee k <= a && k <= b && k <= c

(I already said my English is so bad

Here is a typo in explanation of problem E, answer for each column is "min(n — cnt[i] + i)"

Thanks, fixed.

Rohit_varma**Can someone tell me the validity of solution approach of problem C given in editorial ,I want to know the intuition behind how this approach is leading us to the correct solution?**

Instead of three distinct number you can consider two number

aandd(b.c)whereais the least factor ofnandd(b.c)is another or greatest factor ofn. Then try breakingdintobandcsuch thata,bandcare not equal. For example24. Smallest factor isa=2and another factor isd=12. Further breakdinto it's own factorb=3andc=4such that they are not equal.thanks:)

Although my approach is a bit slower than editorial's, it is very intuitive. Find all distinct factors of N other than 1 and N. This can be done in O(sqrt(N)). Sort these factors. Run two nested loops over the factors. Outer loop picks the value for A. Now we have to find two distinct factors of N / A. Since the factors are sorted, we can use a two pointer approach and find it in one iteration of the inner loop. The idea is something like this: Let L = 0, R = size of factors array — 1. Repeat while L < R: If factors[L] * factors[R] < N / A, increment L. If factors[L] * factors[R] > N / A, decrement R. If factors[L] * factors[R] == N / A, we have found our answer. Make sure to not use the same factor that was picked by the outer loop.

Thanks:)

Anyone give me the intuition that how to solve E.

I also want to know how to solve E.

I think this problem is the most difficult one of all the six problems.

First you have to see that for each column minimum operation have to be calculated now for each column you have to see how much circular rotation is needed for each element in the column now u have the count of each type of cyclic rotation some elements need 1 rotations some 2 etc.. now using this data we can say that n — (count of ith rotation) + i is needed to set the column calculate minimum of this now add minimums of all column

Sry i am bad in explanation

Lets just take a look at only one column. We can rotate the column or modify one elememnt per move.

We should admit that these moves are REVERSIBLE, so it's the same problem as: "We need to rotate the disordered column and THEN modify its element to get the standard column". So we create an array Moves[N] (means if we firstly rotate the column i times, and the total moves we need to get the standard column is Moves[i] , Moves[i] is initally assigned n + i since we suppose that every elements need to be modified), but some elements (for example: element "x") don't need modify IF (x is in the standard column). Thus we need to do Moves[the right the rotating times]--; And the answer for this column is min{Moves[i] (0 <= i < n) } The final result is the sum of each column's answer.

Hope i could help you. qwq.

For E you can divide it into subproblems 1. Solve Each column as an independent entity 2. How to solve? 3. Notice columns are in form of m*i+(j+1), here 0<=i<=n-1, jth column number , remove (j+1) all column are in same form. 4. Each column- Now make a dictionary to count number of elements in that rotation(how far that element is from it's actual position) (you can see in this solution 69363795 ) and first check element is valid or not. now run through the dictionary (d[x]=y) ans = min(x+(n-d[x]), ans) (ans = n initially)

Maybe I am too late to answer, the obvious observation is all columns are independent. Now, to solve each column, you need to find a rotation, which sets most of the numbers at their position and rest will be fixed by modifying. How do we get our best move ? We may have best number for rotations from 0 to n-1, where n is the number of elements in that column. We run a loop from 0 to n-1, and calculate the element required at ith position, in column. Now we need to know, where is this element in original column, and hence the required number of rotation. We store these required number of rotations by each element. Now, if there is a rotation value 2, which appears 3 times, that means, by rotating 2 times, we will get three elements at their position. So if a rotation x appears y times, that means x moves will fix y elements at their position, and we will need n-y moves to fix rest of elements. We need the minimum value for this. That query for original positions of values can be done by storing original values and their positions in a multimap.

well explained. Thanks, But how would you account for duplicates values in the same column, which position would you consider.

We will consider all the duplicates, that's why I mentioned a multimap.

Nice explanation. Thanks.

Nice explanation!!

Wow, test cases on F is very tight.

Anyway, I'm very impressed by Mike's sol. It's neater than DP

In problem D I dont understand this

"Firstly, let's understand what the operation does. It changes the element but holds the remainder modulo x. So we can consider all elements modulo x."

In case 1: x = 3 then: if ($$$cnt_0 == k$$$) ($$$cnt_0$$$ declared above) you can take the elements in this $$$set = {0,3,6,9,12,...}$$$ by increasing x or decreasing x. if ($$$cnt_1 == k$$$) you can take the elements in this $$$set = {1,4,7,10,13,...}$$$ by increasing x or decreasing x. if ($$$cnt_2 == k$$$) you can take the elements in this $$$set = {2,5,8,11,14,...}$$$ by increasing x or decreasing x. Suppose $$$y = min(cnt_0 = 5,cnt_1 = 3,cnt_2 = 4)$$$ Obveriously $$$cnt_1 = 3$$$ is the least. then $$$set_0 = {0,3,6,9,12}$$$ $$$set_1 = {1,4,7}$$$ $$$set_2 = {2,5,8,11}$$$ $$$mex(set_0 \cup set_1 \cup set_2) = ((x = 3) \times (cnt_1 = 3) + 1) = 10$$$ So the answer is 10. So at the begining you can consider all elements modulo x.Because this operation has no effect on the final answer Sorry for my bad English

Thank you from 2023 :)))

Can anyone please explain D in a better manner?

In one move you can choose any element $$$a_i$$$ and change it to $$$a_i:=a_i+x$$$ or $$$a_i:=a_i−x$$$ any number of times only with the condition that $$$a_i$$$ never becomes negative. If you have an element of the form $$$kx + c$$$ in which $$$0 \leq c < x$$$, it could change to $$$c, x + c, 2x + c ... (k-1)x + c, kx + c, (k+1)x + c$$$ and so on. For example, if you have $$$0$$$ as an element, then, it can stay at $$$0$$$ or change to $$$x, 2x, 3x$$$ and so on. As a consequence, in this particular case, all the numbers $$$a_i \equiv 0 (mod x)$$$ can take the place of any of them. In general, if $$$a_i \equiv b (mod x)$$$, you can change that $$$a_i$$$ to any number that has the same remainder $$$b$$$ after dividing it by $$$x$$$. So your $$$mex$$$ start with the number $$$0$$$ because the first number you need. Then, on each query you are going to add on an array one more element with the remainder that the number of that query has. Let's suppose you are still in $$$0$$$ and you read a number whose remainder is $$$0$$$, you decrement the frequency in your array with index $$$0$$$ by one and while your $$$mex$$$ can increase, you repeat that before going to the next query because you can have many numbers with remainder $$$1$$$, $$$2$$$, even until $$$x-1$$$ that can increase the $$$mex$$$. The time complexity is $$$O(n)$$$ because there will be at most $$$n$$$ times in which you could substract one to the frequency of a remainder in the array.

Can anyone explain Problem D. I can't get it

check this link

For Problem

F:How to prove that 2 endpoints of a diameter will result an optimal answer ?

You can check that the optimal answer always goes through the midpoint of diameter

Update: What I am writing is serious! Not for joking

proof by contradiction .Suppose take any three paths not involving the diameter .Then take the diameter and connect one of the vertices of diameter with vertex on three paths and take one of the branch .Since the branch left has length shorter than the diameter hence contradiction .

can you explain it in a diff. way? I am not able to get it.

you can check this out. It could help you.

Can someone explain it some other way ? i still cant understand why its optimal to choose the 2 endpoints of a diameter of the tree

Suppose, you have selected two leaf nodes(It is always better to take leaf nodes than internal nodes, obviously) and now you will find the farthest node from the nodes in the path between selected leaf nodes. Now think, which node you will get? It will be obviously one of the endpoints of one of the diameters. So now, you can think of getting a more optimal answer. It would be better if you have selected the other endpoint of the diameter than the initially selected leaf nodes. If you still didn't get this then you can ask me, which part is not understandable for you !! (:

I have a somewhat different approach to F. We store the maximum , second maximum and 3d maximum for all subtrees. This can be easily done using dfs. Then we loop for all nodes. Answer is either m1 + m2 + m3 or m1 + m2 + max(some m2 in the subtree of the node). All this info can be maintained in a single dfs. https://codeforces.com/contest/1294/submission/69415213 If anyone is further interested in the approach, I would be happy to elucidate it further.

Hi can you please explain what maximum here means? maximum distance node in each subtree?

Yes, the maximum distance node from the current node considering only the current node's subtree. Same goes for second and third maximums.

Who knows where to read about multi-source bfs?

Link : This isn't too detailed, but you get the gist.

It would also be interesting to see that we can also do multi-source shortest path using Djikstra's, for which I have a gfg link : Link

What is the DP approach for F?

just like dp finding diameter of tree

https://codeforces.com/contest/1294/submission/69449118 You can view my code.

dp[i]->the i-th vertex is the root,and choose the vertex as a/b/c,another two is under the i-th vertex.

dp[i]

=max(

dp[j]+1

there is a path between the i-th vertex and the j-th vertex,the biggest two deep[j] + 2

deep[j] is the number of paths between j and i)

You can change the i-th vertex to its kids,and update the value in O(1)!

Total complete in O(n).

What is max_deep array?

The maximum distance between the i-th node and one of the leaf under it.

Probably overkill for C, but a randomized solution over divisors works and is really easy to code. The sieve in my solution is unnecessary (but can help if n was bounded above by say, 1e7).

why is there a stress on finding the third vertex c in F when we already know the edges in a diameter will already contain the maximum number of edges as asked??

because you have to find maximum no. of unique edges between a-b, b-c, and c-a. so, if your diameter have n-1 edges then you can pick any vertex, but if less than n-1 then you have to find 3rd vertex in order to make unique edges maximum.

Ok got it..thanks!

In problem F Why always take diameter?

Seen lot of solutions up here for "A" . Well I think I did the better implementation (no sort needed just the max).

And "E" was a good greedy question I must say :)

I have used the same approach as the editorial. Why am i getting WA on E. link to my solution. Thank you for your efforts

In problem E Can anyone tell me why this is true, I'm not able to understand?

" if ai,j%m≠j (% is modulo operation) then there is no such cyclic shift."

Thank you!

Means that particualr aij value does not fit in the jth that column, aij value must be change in that case

It is because all numbers in the j-th column must have j as a reminder when divided by m. It's clear to see why this is true looking at the image in the problem. So, numbers that don't have this property don't belong to the j-th column, therefore there's no cyclic shift that will put them in place because they don't belong to that column.

69433414 for part A ,my code was running fine on IDE but didn't work on codeforces . Can someone suggest me some correction?

For dist == n, your code is printing YES two times

[DELETED]

what is the "obvious dynamic programming solution" for problem F?

Oh, I got it. You just have to root the tree and for each node check if this node is the node with degree = 3 in the answer. (in case that the tree has no node with degree >= 3 we can just print two nodes with degree 1 and another arbitrary node).

For each node u with degree >= 3 we will maximize between the two following cases :

1- sum of the maximum length of a path to a leaf passing through 3 different children of u starting from u (if has at least 3 children)

2- sum of the maximum length of a path to a leaf passing through 2 different children of u starting from u and the maximum length of a path going to a leaf from parent of u starting from u (if u has parent)

Of course we need to maintain the leaves that give us optimal answer.

https://codeforces.com/blog/entry/73274?#comment-575441

i did the exact same thing but I'm getting WA on TC 22

explanation of my submission->

1) firstly i found any node with degree 1 and made it the root of the tree and also we will check if there is any node with degree>=3 or not.

2) then comes the dfs here i find the height of every node h[node], maximum depth i.e distance from root denoted it by maxxh[node] and a vector maxh[node] which will have the all the depths from the children of this and also its own depth from the root in decreasing order.

3) now i calculate the optimum node and did another dfs to find the 3 leaf nodes.

How to approach E if you could shift the columns downwards also ?

This is a good question.

What we are doing at the moment is calculating the cost for all $$$n$$$ rotations.

I think we would do the same thing except that we would calculate the cost for all $$$2n$$$ rotations.

Promble F:

Help! Can anyone prove why we should get the diameter first? Why we will get the best answer in this way?

I can't prove it.

We can try an exchange argument.

Suppose we have chosen 2 nodes $$$u, v$$$. If $$$u, v$$$ lie on the diameter, then we can extend $$$u, v$$$ to the diameter ends and get a larger answer.

If $$$u, v$$$ do not lie on the diameter, then we can get a bigger answer by choosing the diameter. (Since the definition of the diameter is the two nodes who are furthest apart.)

I very nice contest. The problems were very Mathematical and interesting.

Here are my solutions

multi-source BFSfrom every vertex on the diameter and find out the furthest vertex from the diameter.Thanks for the useful editorial and new codeforces problems

Solution for 1294D - MEX maximizing in O(n) 69491985

Could you explain your approach or intuition behind your solution too?

Let $$$MEX$$$ the current MEX. If $$$y_{i} = MEX$$$ then the $$$MEX$$$ increases by one, if it is smaller or larger, then can we use it to improve our $$$MEX$$$ soon, how do we do this?

Let $$$D$$$ be an unused $$$y_{i}$$$, if $$$MEX \equiv D \pmod{x}$$$ then we can use $$$D$$$ to improve $$$MEX$$$. The numbers we use are not going to be changed in a future query because if we do this our $$$MEX$$$ would go down since the smaller number not used would be less than the $$$MEX$$$

So in each query we try to improve the $$$MEX$$$ as much as possible. For this we have an $$$ST$$$ array of size $$$x - 1$$$, so that $$$ST_{i}$$$ is the number of times we have the unused element $$$i$$$.

Sorry for my English (Google Translator). XDXDXD

My approach in c :

First let's analyze the problem. We have to find out if n can be expressed as a multiple of 3 distinct values. Suppose we have a number with prime factorization of n = 2^2 * 3^3 * 5^1. Now if you distribute the prime factors into three numbers (for example a = 2 * 3^2, b = 3, c = 2*5, Here no new power of any factor has been created but we have just distributed the existing powers), without knowing anything else we can say a*b*c = n. So this problem can be translated to, "can we distribute the prime factors of n into a,b,c such that a!=b!=c."

Solution plan: Suppose given number is n = 2^2 * 3^3 * 5. We will select a = 2 (shortest prime factor), b = 3 (second shortest prime factor) and c = n / (a*b). (all the rest prime factors). (Thus we don't actually need the powers of all prime factors rather all the distinct prime factors).

Algorithm:

1. Find out NUM = distinct prime factors of n. Store them in a vector V.

2. if(NUM == 0) n is prime, ans is no.

3. if(NUM == 1) a = V[0], b = V[0]*V[0], c = n / (a*b). if(a*b*c == n && a!=b && b!=c && c!=a && c > 1) ans is yes. print a,b,c.

4. if(NUM == 2) a = V[0], b = V[1], c = n / (a*b). again check the same condition as 3.

5. if(NUM >= 3) a = V[0], b = V[1], c = n / (a*b). print a,b,c.

When NUM >= 3 there always exists an answer because there are already at least 3 distinct prime factor.

In problem F, we can use dfs instead of multi-source bfs.

Let $$$u$$$ and $$$v$$$ be the endpoints of a diameter. Then, it's always optimal to take vertex $$$w$$$ which $$$dist(u, w)+dist(v, w)$$$ is maximal as the third vertex ($$$dist(x, y)$$$ is distance between $$$x$$$ and $$$y$$$).

Let $$$r$$$ be the intersection of $$$w$$$ and the diameter. Then, the answer can be expressed as follows: $$$dist(u, v)+dist(r, w)=dist(u, v)+\frac{dist(u, w)+dist(v, w)-dist(u, v)}{2}=\frac{dist(u, v)+dist(u, w)+dist(v, w)}{2}$$$

Since $$$dist(u, v)$$$ is the diameter, we should maximize $$$dist(u, w)+dist(v, w)$$$.

Anyone can offer a DP solution for problem F? Thanks!

https://codeforces.com/blog/entry/73274?#comment-575441

Let's take a moment to appreciate the elegance of the rounds prepared by vovuh.

in Question 1 why condition 2c-b-a exist? I mean why 2 multiplied by c and the 3 elements need to be sorted?

## I'll tell you a simple solution, which might give you an intuition

## $$$a + A = k$$$ (some value)

## $$$b + B = k$$$

## $$$c + C = k$$$

## add all these three equations

## $$$a+b+c+(A+B+C)= 3*k$$$

## $$$a+b+c+n= 3*k$$$

## $$$\frac{(a+b+c+n)}{3}= k$$$

## So check if $$$(a+b+c+n)$$$ is divisible by $$$3$$$

## now $$$k= \frac{(a+b+c+n)}{3}$$$ and since

## $$$a + A = k$$$ this implies $$$k-a=A$$$

## $$$b + B = k$$$ this implies $$$k-b=B$$$

## $$$c + C = k$$$ this implies $$$k-c=C$$$

## Since the above 3 should be +ve, check if $$$k > max(a,b,c)$$$

## So effectively check 2 things

## 1. if $$$(a+b+c+n)$$$ is divisible by $$$3$$$

## 3. if $$$\frac{(a+b+c+n)}{3} > max(a,b,c)$$$

69553041 (Problem -E) Can someone please tell me whats wrong in this solution... I am getting wrong ans on test case 62

Submission : https://codeforces.com/contest/1294/submission/69645922

My Approach: I am just trying the brute force approach by taking maximum distance separated nodes from any one of the leaf nodes.

Then, I am finding all the vertices between both the maximum separated vertices. After that, I am finding the node which is farthest from any of the middle vertexes that are not in the path of our pre-chosen 2 vertices.

Can anyone help me figure out why I am going wrong?

Thank you :)

How to do F with DP?

what is the dp approach for problem F?

thanks in advance.

In the Question E,it should be clearly mentioned that those operations are separate.It created a confusion because it was written "one move" as :

In one move, you can:__choose any element of the matrix and change its value to any integer between 1 and n⋅m, inclusive;take any column and shift it one cell up cyclically (see the example of such cyclic shift below).It seemed as if we Change any element of matrix then its necessary to do cyclic shift.even a simpler soln for F is just run 4 dfs

2 dfs to find diameter

3rd dfs to mark all the vertices in the path b/e end points of diameter

4th dfs to calculate the minimum distance of each node from diameter.