This round was mostly made possible by problemsetters from the CerealCodes initiative!

CerealCodes is an organization based in the United States dedicated to running high quality competitive programming contests. You can check out our previous contest here.

To make $$$a$$$ not sorted, we just need to pick one index $$$i$$$ so $$$a_i \gt a_{i + 1}$$$. How do we do this?

To make $$$a$$$ not sorted, we just have to make $$$a_i \gt a_{i + 1}$$$ for one $$$i$$$.

In one operation, we can reduce the gap between two adjacent elements $$$i, i + 1$$$ by $$$2$$$ by adding $$$1$$$ to $$$1 \dots i$$$ and subtracting $$$1$$$ from $$${i + 1} \dots n$$$.

It is clearly optimal to pick the smallest gap between a pair of adjacent elements to minimize the number of operations we have to do. If we have $$$a_i = x, a_{i + 1} = y$$$, we can make $$$x \gt y$$$ within $$$\lfloor \frac{(y - x)}{2} \rfloor + 1$$$ operations.

Thus, we can just go through $$$a$$$, find the minimum difference gap, and calculate the minimum operations using the above formula. Note that if $$$a$$$ is not sorted, we can just output $$$0$$$.

The time complexity is $$$O(n)$$$.

#include <bits/stdc++.h>
#include <numeric>
using namespace std;
 
int main(){
 
    ios::sync_with_stdio(false);
    cin.tie(0);
 
    int T; cin >> T;
 
    while (T--) {
 
        int n; cin >> n;
        vector<int> nums(n);
        int diff = 1e9;
        bool sorted = true;
        for (int i = 0; i < n; i++) {
            cin >> nums[i];
            if (i > 0) {
                diff = min(nums[i] - nums[i - 1], diff);
                sorted &= nums[i] >= nums[i - 1];
            }
        }
        
        if (!sorted) {
            cout << 0 << endl;
            continue;
        }
    
        cout << diff/2 + 1 << endl;
    }
}

Can a sequence involving $$$n$$$, which is up to $$$10^5$$$, really have up to $$$10^9$$$ terms?

The terms of the fibonacci sequence will increase exponentially. This is quite intuitive, but mathematically, fibonnaci-like sequences will increase at a rate of phi to the power of $$$n$$$, where phi (the golden ratio) is about $$$1.618$$$. Thus, the maximum number of terms a sequence can have before it reaches $$$10^9$$$, or the maximum value of $$$n$$$, is pretty small (around $$$\log n$$$).

Instead of trying to fix the first two elements of the sequence and counting how many sequences $$$s$$$ will have $$$s_k = n$$$, note that we already have $$$n$$$ fixed. If we loop over the $$$k-1$$$th element of the sequence, the sequence is still fixed. If we know the $$$x$$$th element and $$$x-1$$$th element of $$$s$$$, we can find that $$$s_{x - 2} = s_{x} - s_{x - 1}$$$.

Thus, we can just go backwards and simulate for $$$k$$$ iterations in $$$O(\log n)$$$ since $$$k$$$ is small, breaking at any point if the current sequence is not fibonnaci-like (there are negative elements or it is not strictly increasing). Otherwise, we add $$$1$$$ to our answer.

The time complexity is $$$O(n \cdot \log n)$$$.

#include <bits/stdc++.h>
using namespace std;
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    int T; cin >> T;
    
    while (T--) {
 
        int n; int k;
        cin >> n >> k;
     
        int ans = 0;
     
        for (int i = 1; i <= n; i++) {
            int second = n; //xth element where x is k
            int first = i; //fixing x-1th element where x is k-1
            bool valid_seq = true;
            for (int j = 0; j < k - 2; j++) {
                //for s_x and s_x-1, s_x-2 = s_x - s_x-1
                int fx = first;
                first = second - fx;
                second = fx;
                valid_seq &= first <= second;
                valid_seq &= min(first, second) >= 0;
                if (!valid_seq) break; //break if the sequence is not fibonacci-like
            }
            if (valid_seq) ans++;
        }
 
        cout << ans << endl;
    }
 
}

Analysis by awesomeguy856

$$$f[k] = F_{k-2}f[0]+F_{k-1}f[1]$$$

By the Extended Euclidean Algorithm, we can find one integral solution for this equation, since $$$\gcd (F_{k-2}, F_{k-1}) = 1 | f[k].$$$ Let this solution be $$$(f[0], f[1]) = (x, y).$$$ Then all other integral solutions are in the form $$$(x+cF_{k-1}, y-cF_{k-2}),$$$ for $$$c \in Z$$$ so we can find all valid solutions by binary search on $$$f[1],f[0] \geq 0$$$ and $$$f[1] \gt f[0]$$$, or just by some calculations.

Suppose that the numbers are arranged in a line in increasing order. Take a look at each number $$$x$$$ before some day. If it isn't deleted on that day, what new position does it occupy, and how is that impacted by its previous position?

Take this observation, and apply it to simulate the process backwards.

Suppose the numbers are arranged in a line in increasing order. Consider simulating backwards; instead of deleting the numbers at positions $$$a_1,a_2,\ldots,a_n$$$ in each operation, then checking the first number after $$$k$$$ operations, we start with the number $$$1$$$ at the front, try to insert zeroes right after positions $$$a_1-1, a_2-2, \ldots, a_{n}-n$$$ in each operation so that the zeroes will occupy positions $$$a_1, a_2, \ldots, a_n$$$ after the insertion, and after $$$k$$$ insertions, we will check the position that $$$1$$$ will end up at.

If $$$a_1$$$ is not equal to $$$1$$$, the answer is $$$1$$$. Otherwise, each insertion can be processed in $$$O(1)$$$ if we keep track of how many of $$$a_1-1, a_{2}-2,\ldots,a_{n}-n$$$ are before the current position $$$x$$$ of $$$1$$$; if $$$a_1-1$$$ through $$$a_i-i$$$ are before $$$x$$$, then we will insert $$$i$$$ zeroes before $$$x$$$.

The time complexity is $$$O(n+k)$$$ per test case. The editorial code additionally processes every insertion with the same $$$i$$$ value in $$$O(1)$$$, for $$$O(n)$$$ overall complexity.

There are alternative solutions using binary search with complexity $$$O(k\log nk)$$$ or $$$O(k\log n\log nk)$$$, and we allowed them to pass. In fact, this problem was originally proposed with $$$k \leq 10^9$$$ but we lowered the constraints.

#include <bits/stdc++.h>
using namespace std;

int n, k, a[200010];

void solve() {
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        a[i] -= i;
    }
    if (a[0] != 1) {
        cout << 1 << "\n";
        return;
    }
    a[n] = 2e9;
    //start at position 1, and find what number moves to its position
    long long ans = 1;
    int inc = 1; //how many zeroes to insert
    while (inc < n) {
        int days = (a[inc] - ans + inc - 1) / inc;
        if (days >= k){
            ans += k * inc;
            k = 0;
            break;
        }
        ans += days * inc;
        k -= days;
        while (a[inc] <= ans) inc++;
    }
    ans += 1ll * k * n;
    cout << ans << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    
    int t; cin >> t;
    while (t--) solve();
}

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
const int inf = 1e9+10;
const ll inf_ll = 1e18+10;
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define cmax(x, y) (x = max(x, y))
#define cmin(x, y) (x = min(x, y))

#ifndef LOCAL
#define debug(...) 0
#else
#include "../../debug.cpp"
#endif

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int t; cin >> t;
    while (t--) {
        ll n, k; cin >> n >> k;
        vector<ll> a(n);
        for (int i = 0; i < n; i++)
            cin >> a[i];

        ll j = 0, x = 1;
        while (k--) {
            while (j < n && a[j] <= x+j)
                j++;
            x += j;
        }

        cout << x << "\n";
    }
}

You can solve the problem by picking one number from each pair $$$(n, -n)$$$, $$$(n - 1, -n + 1) \dots$$$, $$$(1, -1)$$$.

$$$b_i \gt b_j$$$ implies $$$a_i \gt a_j$$$.

First, try to determine one index in $$$O(n)$$$, or determine if that's impossible.

Sort the array $$$a$$$ to optimize the $$$O(n^2)$$$ solution.

At the start, let $$$x$$$ be an index such that $$$b_x$$$ has the greatest absolute value. If $$$b_x$$$ is negative, we have $$$a_x=0$$$, and else $$$a_x=n$$$. Moreover, we can't have $$$a_y=0,a_z=n$$$ for any indices $$$y$$$ and $$$z$$$, because that implies $$$b_y+b_z$$$ is both positive and negative, contradiction. Hence, the necessary and sufficient condition to check if we can determine an element in array $$$b$$$ with maximum absolute value is (there exists an element of array $$$a$$$ equal to $$$0$$$) xor (there exists an element of array $$$a$$$ equal to $$$n$$$). Then, we can remove that element and re-calculate the $$$a$$$ array, leading to an $$$O(n^2)$$$ solution. If the check fails at any moment, there is no valid solution.

To optimize it further, note that we can sort array $$$a$$$ at the start and keep track of them in a deque-like structure. We only need to check the front and end of the deque to see if our key condition holds. Finally, we can use a variable to record the number of positive elements deleted so far and subtract it from the front and end of the deque when checking our condition, so that each check is $$$O(1)$$$. The overall complexity becomes $$$O(n\log n)$$$ due to sorting.

#include <bits/stdc++.h>
using namespace std;

int n, ans[100010];
vector<pair<int,int>> arr;

void solve() {
    cin >> n;
    arr.resize(n);
    for (int i = 0; i < n; i++) {
        cin >> arr[i].first;
        arr[i].second = i;
    }
    sort(arr.begin(), arr.end());
    int l = 0, r = n - 1, sz = n;
    while (l <= r) {
        if ((arr[r].first == n - l) ^ (arr[l].first == n - 1 - r)) {
            if (arr[r].first == n - l) {
                ans[arr[r--].second] = sz--;
            }
            else {
                ans[arr[l++].second] = -(sz--);
            }
        }
        else {
            cout << "NO" << "\n";
            return;
        }
    }
    cout << "YES" << "\n";
    for (int i = 0; i < n; i++) cout << ans[i] << " ";
    cout << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);

    int t; cin >> t;
    while(t--) solve();
}

Suppose you knew in advance how many times each octopus will regenerate. Could you solve the problem then?

Can you figure out an $$$O(n^2)$$$ solution? And can you improve on it?

Suppose you already had the optimal solution for the subarray $$$a[1 \ldots i]$$$. How could you extend it to $$$a[1 \ldots i+1]$$$?

Read the solution in Hint One before continuing with this tutorial; it provides important definitions.

To reduce the number of possibilities for each $$$c[i]$$$, we prove the following lemma: There exists an optimal choice of $$$c$$$ (minimizing $$$\mathrm{throws}$$$) where all differences between adjacent $$$c[i]$$$ have absolute value less than $$$k$$$.

Intuitively, this is because we can decrease a $$$c[i]$$$ by $$$k$$$. Formally:

By the previous lemma, if we have determined $$$c[i]$$$, there are at most $$$2$$$ choices for $$$c[i+1]$$$. (There is $$$1$$$ choice when $$$b[i] = b[i+1]$$$, resulting in $$$d[i] = 0$$$, $$$c[i] = c[i+1]$$$, effectively merging the two octopuses.)

We can visualize this as a DAG in the 2D plane over all points $$$(i,c[i])$$$ (over all possible choices of $$$c[i]$$$). Each point points to the points in the next column that are the closest above and below (if it exists), forming a grid-like shape. Our goal is to find a path of minimum cost from $$$(0,0)$$$ to $$$(n+1,0)$$$.

This is the DAG for the second testcase in samples:

Call each time we choose a $$$c[i+1] \gt c[i]$$$ (i.e. positive $$$d[i]$$$) an ascent. Note that the number of ascents is fixed because each nonzero $$$d[i]$$$ is either $$$x$$$ or $$$x+k$$$ for some fixed negative $$$x$$$, and there must be a fixed number of $$$+k$$$'s to make the total change from $$$c[0]$$$ to $$$c[n+1]$$$ zero.

Each ascent brings the path up to the next "row" of descents. Since these rows slope downwards, the $$$j$$$th ascent must take place at or before some index $$$i_j$$$, because otherwise $$$c[i_j+1]$$$ would be negative.

We can use the following strategy to find a path which we claim is optimal:

If we can descend, then we descend. Otherwise, either we ascend, or alternatively, we change a previous descent into an ascent so we can descend here. (This can be further simplified by having a hypothetical "descent" here, so you do not need to compare two possibilities in the implementation.) Now, the best such location for an ascent is the one with minimum cost.

We can implement the above strategy with a priority queue, where for each descent we push on the cost of the corresponding ascent, and when an ascent is required, we then pop off the minimum element. In particular, if $$$b[i] \lt b[i+1]$$$, then the corresponding ascent has cost $$$b[i+1]-b[i]$$$, while if $$$b[i] \gt b[i+1]$$$, it has cost $$$b[i+1]-b[i]+k$$$. Also, since the bottom of the DAG corresponds to $$$c[i] = b[i]$$$, an ascent is required exactly when $$$b[i] \lt b[i+1]$$$.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

void solve() {
	int n, k;
	cin >> n >> k;
	priority_queue<int, vector<int>, greater<int>> differences;
	int previous = 0;
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		int x;
		cin >> x;
		x %= k;
		if (x > previous) {
			differences.push(x - previous);
			ans += differences.top();
  			differences.pop();
		} else {
  			differences.push(k + x - previous);
		}
		previous = x;
  	}
  	cout << ans << "\n";
}

int main() {
	int t;
	cin >> t;
	while (t--)
		solve();
}

Solve the samples for all values of $$$k$$$.

What does constructing a second row from left to right look like?

How does knowing the possible $$$k$$$ for any first row help you construct a second row?

Apply dynamic programming to calculate the possible $$$k$$$, and use the information in the DP to construct a solution.

Call the number of adjacent pairs of cells with different characters the cost.

If we construct the second row from left to right, the amount each character adds to the cost only depends on the previous character. Thus, we can represent the problem with a DAG whose vertices represent choices for each character and whose edges represent the cost of choosing two adjacent characters. Our goal is to find a path starting from the far left and ending at the far right of cost $$$k$$$. For example, below is the DAG for the third testcase in the sample. (For convenience, we include the cost of cells in the top row with the corresponding cells in the bottom row.)

A general way to find such a path is as follows: For each vertex, store all possible costs of a path starting from the far left and ending at that vertex, which can be calculated with dynamic programming. Then we construct our path backwards from the right. For each candidate previous vertex, we check that the constructed path's cost plus the cost of the edge from this vertex plus some stored cost in this vertex equals $$$k$$$, in which case we know that some completion of the path from the far left to this vertex exists and we can choose this vertex.

Naively calculating this DP would require $$$O(n)$$$ time per operation. However, intuitively, each set of possible costs should contain almost all values between its minimum and maximum, and experiments suggest it always consists of at most two intervals. We will first formalize the DP and then prove this observation.

Let $$$\mathrm{dp}_A[i]$$$ store the set of possible values of $$$k$$$ when the grid is truncated to the first $$$i$$$ columns and the last character in the second row is $$$A$$$. Define $$$\mathrm{dp}_B[i]$$$ similarly.

Using the notation $$$[\mathrm{true}] = 1$$$, $$$[\mathrm{false}] = 0$$$, $$$S+x = \{s+x \mid s \in S\}$$$, we have the recurrences

with initial values $$$\mathrm{dp}_A[1] = [s_1 \ne A]$$$ and $$$\mathrm{dp}_B[1] = [s_1 \ne B]$$$.

Now either we hope that each set consists of $$$O(1)$$$ intervals, or we have to prove that each set indeed consists of at most two intervals:

To find the union of two sets of intervals, sort them by left endpoint and merge adjacent overlapping intervals. After computing the DP, apply the aforementioned backwards construction to obtain a valid second row.

Below is a visualization of solving the third testcase in the sample. Generate your own here!

#include <bits/stdc++.h>
using namespace std;

#define all(x) (x).begin(), (x).end()

struct state {

    // represents a set of integers compressed as a vector
    // of non-overlapping intervals [l, r]

    basic_string<array<int, 2>> a;

    // add d to everything in the set
    friend state add(state x, int d) {
        for (auto& [l, r] : x.a)
            l += d, r += d;
        return x;
    }

    // union of two sets of integers
    friend state unite(state x, state y) {
        state z;
        merge(all(x.a), all(y.a), back_inserter(z.a));
        int j = 0;
        for (int i = 1; i < z.a.size(); i++) {
            if (z.a[i][0] <= z.a[j][1]+1)
                z.a[j][1] = max(z.a[j][1], z.a[i][1]);
            else
                z.a[++j] = z.a[i];
        }
        z.a.erase(z.a.begin() + j + 1, z.a.end());
        return z;
    }

    // whether integer k is in the set
    friend bool contains(state x, int k) {
        for (auto& [l, r] : x.a)
            if (k >= l && k <= r)
                return true;
        return false;
    }
};

// set containing only one integer d
state from_constant(int d) {
    return state({{{d, d}}});
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int t; cin >> t;
    while (t--) {
        int n, k; cin >> n >> k;
        string s; cin >> s;

        vector<array<state, 2>> dp(n);

        dp[0][0] = from_constant(s[0] != 'A');
        dp[0][1] = from_constant(s[0] != 'B');

        for (int i = 1; i < n; i++) {
            dp[i][0] = add(unite(dp[i-1][0], add(dp[i-1][1], 1)), (s[i] != 'A') + (s[i] != s[i-1]));
            dp[i][1] = add(unite(dp[i-1][1], add(dp[i-1][0], 1)), (s[i] != 'B') + (s[i] != s[i-1]));
        }

        if (!(contains(dp[n-1][0], k) || contains(dp[n-1][1], k))) {
            cout << "NO\n";
            continue;
        }

        string ans;

        for (int i = n-1; i >= 0; i--) {
            char c = contains(dp[i][0], k - (i == n-1 ? 0 : (ans.back() != 'A'))) ? 'A' : 'B';
            k -= (c != s[i]);
            if (i > 0)
                k -= (s[i] != s[i-1]);
            if (i < n-1)
                k -= (c != ans.back());
            ans += c;
        }

        reverse(all(ans));
        cout << "YES\n" << ans << "\n";
    }
}

For each distinct value, only the leftmost and rightmost positions actually have an effect on the power.

Think of each distinct value as an interval corresponding to its leftmost and rightmost positions, and let that distinct value be the value of its interval. If interval $$$x$$$ strictly contains an interval $$$y$$$ with greater value, then $$$x$$$ will not have an effect on the power and it can be discarded.

In order to not change the power, we can only add intervals that strictly contain an interval with greater value. Afterwards, if $$$i$$$ is not the endpoint of an interval, $$$b_i$$$ equals the largest value of an interval containing $$$i$$$.

There will only be at most one interval that we have to add. Assume that there are multiple intervals that we can add without changing the answer. In this case it is easy to see that its more optimal to combine the interval with smaller value into the interval with larger value. Thus, it is never optimal to add multiple intervals.

Read the hints.

We can remove all intervals that will not affect the power by iterating over them in decreasing order of value and maintain a segment tree that stores for each left endpoint of any processed intervals, the right endpoint that corresponds to it. Checking if an interval is strictly contained then becomes a range minimum query.

Assume that we know the value $$$x$$$ of the interval that we want to add. We can immediately fill the value for the interior of all intervals that correspond to a value greater than $$$x$$$. However, we also have to guarantee that $$$x$$$ is strictly contained by an interval with greater value, so we can try each interval to contain $$$x$$$. If are no unfilled indices the left or right side of the interval, then we want to replace the smallest filled on either side value with $$$x$$$. Otherwise, we can just fill in all the unfilled indices with $$$x$$$. Note that it is also possible for $$$x$$$ to not be able to be contained by any interval.

Thus, we can just try every possible $$$x$$$ in decreasing order and maintain the filled indices as we iterate. There are only at most $$$n$$$ values of $$$x$$$ to check, which is just the set of $$$a_i - 1$$$ that does not appear in $$$a$$$.

#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define ff first
#define ss second

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int INF = 1e9 + 1;

void setIO() {
    ios_base::sync_with_stdio(0); cin.tie(0);
}

pii seg[400005];
int n;

void build(int l = 0, int r = n - 1, int cur = 0){
    seg[cur] = {INF, INF};
    if(l == r) return;
    int mid = (l + r)/2;
    build(l, mid, cur*2 + 1);
    build(mid + 1, r, cur*2 + 2);
}

void update(int x, int v, int l = 0, int r = n - 1, int cur = 0){
    if(l == r){
        seg[cur] = {v, l};
        return;
    }
    int mid = (l + r)/2;
    if(x <= mid) update(x, v, l, mid, cur*2 + 1);
    else update(x, v, mid + 1, r, cur*2 + 2);
    seg[cur] = min(seg[cur*2 + 1], seg[cur*2 + 2]);
}

pii query(int l, int r, int ul = 0, int ur = n - 1, int cur = 0){
    if(l <= ul && ur <= r) return seg[cur];
    if(l > r || ur < l || ul > r) return {INF, INF};
    int mid = (ul + ur)/2;
    return min(query(l, r, ul, mid, cur*2 + 1), query(l, r, mid + 1, ur, cur*2 + 2));
}

int main(){
    setIO();
    int t;
    cin >> t;
    for(int tt = 1; tt <= t; tt++){
        cin >> n;
        map<int, vector<int>> m;
        int arr[n];
        for(int i = 0; i < n; i++){
            cin >> arr[i];
            m[arr[i]].pb(i);
        }
        vector<int> rel; //relevant numbers
        for(auto &i : m){
            rel.pb(i.ff);
            for(int j = 1; j < i.ss.size() - 1; j++){
                arr[i.ss[j]] = 0;
            }
        }
        reverse(rel.begin(), rel.end());
        vector<int> nw;
        build();
        for(int i : rel){
            int l = m[i].front(), r = m[i].back();
            if(query(l, r).ff <= r){
                arr[l] = arr[r] = 0;
            } else {
                update(l, r);
                nw.pb(i);
            }
        }
        swap(rel, nw);
        set<int> s;
        for(int i = 0; i < n; i++) if(!arr[i]) s.insert(i);
        vector<int> vals; //possible target values
        for(int i = rel.size() - 1; i >= 0; i--){
            if(i == rel.size() - 1 || rel[i] - 1 != rel[i + 1]){
                vals.pb(rel[i] - 1);
            }
        }
        ll sum = 0, mx = 0, mxval = 0;
        int lex = -1, rex = -1;
        build();
        for(int i = 0; i < n; i++) if(!arr[i]) update(i, -1);
        reverse(vals.begin(), vals.end());
        queue<pii> upd;
        int ind = 0;
        //try to add an interval with value i
        for(int i : vals){
            vector<int> nxt;
            //try filling in all intervals > i
            while(ind < rel.size() && rel[ind] > i){
                int l = m[rel[ind]].front(), r = m[rel[ind]].back();
                set<int>::iterator it = s.lower_bound(l);
                while(it != s.end() && *it < r){
                    sum += rel[ind];
                    update(*it, rel[ind]);
                    upd.push({*it, rel[ind]});
                    it = s.erase(it);
                }
                nxt.pb(rel[ind++]);
            }
            //try each necessary interval to be strictly contained by i
            for(int j : nxt){
                int l = m[j].front(), r = m[j].back();
                if(s.size() && 0 < l && r < n - 1){
                    //find which indeces on each side to replace with i
                    pii rnw = query(r + 1, n);
                    pii lnw = query(0, l - 1);
                    if(rnw.ff == INF || lnw.ff == INF) continue;
                    ll add = (ll)i*s.size() + (rnw.ff == -1 ? 0 : i - rnw.ff) + (lnw.ff == -1 ? 0 : i - lnw.ff);
                    if(i && sum + add > mx){
                        mx = sum + add;
                        mxval = i;
                        lex = (lnw.ff == -1 ? -1 : lnw.ss);
                        rex = (rnw.ff == -1 ? -1 : rnw.ss);
                        //it is optimal to fill in all intervals > i
                        while(!upd.empty()){
                            arr[upd.front().ff] = upd.front().ss;
                            upd.pop();
                        }
                    }
                }
            }
        }
        //don't want to add any intervals
        if(s.size() == 0 && sum > mx){
            mx = sum;
            lex = rex = -1;
            //it is optimal to fill in all remaining intervals
            while(!upd.empty()){
                arr[upd.front().ff] = upd.front().ss;
                upd.pop();
            }
        }
        if(lex != -1) arr[lex] = 0;
        if(rex != -1) arr[rex] = 0;
        for(int i = 0; i < n; i++) cout << (arr[i] ? arr[i] : mxval) << " ";
        cout << endl;
    }
}

If we want to answer one of the questions (say, the question involving all the events) in polynomial time, how do we do it?

Construct a network with edges from the source to each of the red panda events with capacities equal to the number of red pandas, edges from red panda events to blue panda events with innite capacities if the red pandas can catch the corresponding blue pandas, and edges from each of the blue panda events to the sink with capacities equal to the number of blue pandas.

The next step is to use the max-flow min-cut theorem: the maximum flow is equal to the minimum number of red pandas plus blue pandas we need to remove from the graph such that no remaining red panda can reach any remaining blue panda.

How do we go from here?

For any cut, consider the region of the $$$x$$$-$$$t$$$ plane reachable by the remaining red pandas. No remaining blue pandas can lie in this region, and its border is a polyline that intersects each vertical line in the $$$x$$$-$$$t$$$ plane exactly once. Furthermore, the slope of every segment in this polyline has slope plus or minus $$$1$$$. Conversely, we can associate every polyline satisfying this condition with a cut; we just need to remove every red panda lying below the polyline and every blue panda lying on or above the polyline.

Now, figure out how to answer the queries online.

Read the hints to understand the solution better.

We can answer the queries online. For each $$$x$$$-coordinate $$$x$$$, maintain a treap that stores for every $$$x$$$-coordinate, the minimum cut $$$dp[x][t]$$$ associated with a polyline satisfying the condition above that starts at $$$x = -\infty$$$ and ends at $$$(x, t)$$$ when only considering events with $$$x$$$-coordinate at most $$$x$$$. When transitioning from to $$$x$$$ to $$$x+1$$$, we need to set for every $$$dp[x+1][t] = min(dp[x][t-1], dp[x][t], dp[x][t+1])$$$ for every $$$t$$$. To do this quickly, we maintain the values of $$$t$$$ where $$$dp[x][t+1] - dp[x][t] \neq 0$$$ in increasing order, of which there are at most $$$n$$$. When $$$x$$$ increases by one, each value of increases or decreases by one, depending on the sign of $$$dp[x][t+1]-dp[x][t]$$$, and some of the values of $$$t$$$ merge, decreasing the size of our treap. As long as we can process each merge in $$$\log n$$$ time, our solution will run in $$$O(n\log n)$$$ total time. When processing an event, we need to increase all $$$dp$$$ values in a suffix for red panda events, and all $$$dp$$$ values in a prefix for blue panda events. To answer a query, we just need to return the minimum prefix sum in our treap.

#include <bits/stdc++.h>
using namespace std;

#define pb push_back
#define ff first
#define ss second

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int INF = 1e9 + 1;

void setIO() {
    ios_base::sync_with_stdio(0); cin.tie(0);
}

struct node {
    //mnpos - leftmost negative 
    //mxpos - rightmost positive
    int pos, tag, mnpos, mxpos;
    //minimum distance
    pair<int, pii> mndif;
    ll sum, pre, val;
    int weight;

    node(){}

    node(int pos_, ll val_){
        pos = pos_;
        val = val_;
        sum = val;
        pre = min((ll)0, val);
        mxpos = -INF;
        mnpos = INF;
        mndif = {INF, {INF, INF}};
        if(val > 0) mxpos = pos;
        if(val < 0) mnpos = pos;
        tag = 0;
        weight = rand();
    }
};

int sz = 1;
node treap[1000005];
int left0[1000005];
int right0[1000005];

int newnode(int pos, ll val){
    treap[sz] = node(pos, val);
    return sz++;
}

pair<int, pii> comb(int a, int b){
    if(a == -INF || b == INF) return {INF, {INF, INF}};
    return {b - a, {a, b}};
}

void pull(int x){
    treap[x].sum = treap[x].val;
    treap[x].pre = min((ll)0, treap[x].val);
    treap[x].mxpos = -INF;
    treap[x].mnpos = INF;
    treap[x].mndif = {INF, {INF, INF}};
    if(treap[x].val > 0) treap[x].mxpos = treap[x].pos;
    if(treap[x].val < 0) treap[x].mnpos = treap[x].pos;
    if(left0[x]){
        treap[x].mndif = min(treap[x].mndif, treap[left0[x]].mndif);
        treap[x].mndif = min(treap[x].mndif, comb(treap[left0[x]].mxpos, treap[x].mnpos));
        treap[x].mnpos = min(treap[x].mnpos, treap[left0[x]].mnpos);
        treap[x].mxpos = max(treap[x].mxpos, treap[left0[x]].mxpos);
        treap[x].pre = min(treap[left0[x]].pre, treap[left0[x]].sum + treap[x].pre);
        treap[x].sum += treap[left0[x]].sum; 
    }
    if(right0[x]){
        treap[x].mndif = min(treap[x].mndif, treap[right0[x]].mndif);
        treap[x].mndif = min(treap[x].mndif, comb(treap[x].mxpos, treap[right0[x]].mnpos));
        treap[x].mnpos = min(treap[x].mnpos, treap[right0[x]].mnpos);
        treap[x].mxpos = max(treap[x].mxpos, treap[right0[x]].mxpos);
        treap[x].pre = min(treap[x].pre, treap[x].sum + treap[right0[x]].pre);
        treap[x].sum += treap[right0[x]].sum;
    }
}

int move(node& x, int shift){
    int ret = x.pos;
    if(x.val < 0) ret -= shift;
    if(x.val > 0) ret += shift;
    return ret;
}

void apply(int x, int tag){
    treap[x].pos = move(treap[x], tag);
    treap[x].tag += tag;
    if(treap[x].mnpos != INF) treap[x].mnpos -= tag;
    if(treap[x].mxpos != -INF) treap[x].mxpos += tag;
    if(treap[x].mndif.ff != INF){
        treap[x].mndif.ff -= 2*tag;
        treap[x].mndif.ss.ff += tag;
        treap[x].mndif.ss.ss -= tag;
    }
}

void push(int x){
    if(!treap[x].tag) return;
    if(left0[x]) apply(left0[x], treap[x].tag);
    if(right0[x]) apply(right0[x], treap[x].tag);
    treap[x].tag = 0;
}

int merge(int a, int b){
    if(!a) return b;
    if(!b) return a;
    if(treap[a].weight < treap[b].weight){ 
        push(a);
        right0[a] = merge(right0[a], b);
        pull(a);
        return a;
    } else {
        push(b);
        left0[b] = merge(a, left0[b]);
        pull(b);
        return b;
    }
}

//splits rt's tree into [0, k) [k, INF)
pair<int, int> split(int x, int k){
    if(!x) return pair<int, int>{0, 0};
    push(x);
    pair<int, int> ret;
    if(treap[x].pos < k){
        ret = split(right0[x], k);
        right0[x] = ret.first;
        ret.first = x;
    } else {
        ret = split(left0[x], k);
        left0[x] = ret.second;
        ret.second = x;
    }
    pull(x);
    return ret;
}

int rt = 0;

void erase(int x){
    pair<int, int> a = split(rt, x);
    pair<int, int> b = split(a.second, x + 1);
    rt = merge(a.first, b.second);
}

//position, value
void insert(int a, ll b){
    if(!rt){
        rt = newnode(a, b);
        return;
    }
    pair<int, int> nw = split(rt, a);
    rt = merge(nw.first, merge(newnode(a, b), nw.second));
}

//value
ll query(int x){
    pair<int, int> a = split(rt, x);
    pair<int, int> b = split(a.second, x + 1); 
    ll ret = (b.first ? treap[b.first].val : 0);
    rt = merge(a.first, merge(b.first, b.second));
    return ret;
}

int main(){
    setIO();
    int n;
    cin >> n;
    int prv = 0;
    ll st = 0;
    for(int i = 1; i <= n; i++){
        int x, t, c;
        cin >> x >> t >> c;
        int dif = x - prv;
        //remove overlap
        while(rt && treap[rt].mndif.ff <= 2*dif){
            pair<int, pii> x = treap[rt].mndif;
            ll a = query(x.ss.ff), b = query(x.ss.ss);
            ll sub = min(a, -b);
            erase(x.ss.ff);
            erase(x.ss.ss);
            a -= sub, b += sub;
            if(a != 0) insert(x.ss.ff, a);
            if(b != 0) insert(x.ss.ss, b);
        }
        //shift everything
        if(rt){
            treap[rt].pos = move(treap[rt], dif);
            treap[rt].tag += dif;
            if(treap[rt].mnpos != INF) treap[rt].mnpos -= dif;
            if(treap[rt].mxpos != -INF) treap[rt].mxpos += dif;
            if(treap[rt].mndif.ff != INF){
                treap[rt].mndif.ff -= 2*dif;
                treap[rt].mndif.ss.ff += dif;
                treap[rt].mndif.ss.ss -= dif;
            }
        }
        ll cur = query(t + 1);
        if(cur != 0) erase(t + 1);
        if(cur - c != 0) insert(t + 1, cur - c);
        if(c > 0) st += c;
        cout << st + (!rt ? 0 : treap[rt].pre) << endl;
        prv = x;
    }
}