Solve for $$$k = 1$$$.

Solve for $$$k = n$$$.

What other $$$k$$$ work for a given $$$n$$$?

Read the hints.

For $$$k=1$$$, the construction $$$1,2,3\dots n$$$ will always work because in any other cyclic shift, $$$n$$$ will be before $$$1$$$.

Now, consider what happens if we want to find an array with two or more cyclic shifts that are sorted. If we consider two of these cyclic shifts, notice they are actually shifts of each other as well.

So, there should be a sorted array, and a shift of it which is also sorted. What this means is that some of the first elements in the array move to the back and stays sorted, which can only happen if all values in the array are equal.

If the array has all equal values, this gives us our construction for $$$k=n$$$. As we have seen, for $$$k \gt 1$$$, only $$$k=n$$$ is possible. Thus, for any other $$$k$$$ not equal to $$$1$$$ or $$$n$$$, we can print $$$-1$$$.

#include <iostream>
using namespace std;
int main(){
	int t; cin >> t;
	while(t--){
		int n, k; cin >> n >> k;
		if(k == 1) for(int i = 0; i < n; i++) cout << i + 1 << " ";
		else if(k == n) for(int i = 0; i < n; i++) cout << 1 << " ";
		else cout << -1;
		cout << "\n";
	}
}

We will construct the solution forward.

Separate the $$$a_i$$$'s given into negative and positive cases. What does this tell us about the $$$\texttt{MEX}$$$?

We can find $$$p_1, p_2, ... , p_n$$$ in order, looking at positive and negative cases. Note that $$$a_i \neq 0$$$ because $$$p_i$$$ would equal $$$\texttt{MEX}$$$($$$p_1 \dots p_i$$$), which can never happen.

• If $$$a_i \gt 0$$$, then $$$\texttt{MEX}$$$($$$p_1, p_2, ... p_i$$$) must increase from $$$\texttt{MEX}$$$($$$p_1, p_2, ... p_{i-1}$$$), so we know $$$p_i$$$ must equal $$$\texttt{MEX}$$$($$$p_1, p_2, ... p_{i-1}$$$).
• Otherwise, the $$$\texttt{MEX}$$$ stays the same, so $$$p_i$$$ is just simply $$$\texttt{MEX}$$$($$$p_1, p_2, ... p_{i-1}$$$) — $$$a_i$$$.

Thus, we can just maintain the $$$\texttt{MEX}$$$ and find each $$$p_i$$$ as we go forward.

Note that this is also a way to show $$$p$$$ is always unique.

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

void solve(){
	int n; cin >> n;
	vector<int> a(n);
	for(int& i: a) cin >> i;
	vector<int> p(n), has(n + 1);
	int mex = 0;
	for(int i = 0; i < n; i++){
		if(a[i] >= 0){
			p[i] = mex;
		}
		else{
			p[i] = mex - a[i];
		}
		has[p[i]] = true;
		while(has[mex]) mex++;
	}
	for(int i: p) cout << i << " ";
	cout << "\n";
}

int main(){
	cin.tie(0) -> sync_with_stdio(0);
	int T = 1;
	cin >> T;
	while(T--) solve();
}

/*   /\_/\
*   (= ._.)
*   / >  \>
*/

We will construct the solution backwards.

The $$$\texttt{MEX}$$$ is determined at the last index, since all of $$$0, 1 \dots n-1$$$ appear in $$$p$$$.

Read the hints.

Since we know the $$$\texttt{MEX}$$$ of the last position is $$$n$$$, then $$$n - p_n = a_n$$$. From this equation, we can find that $$$p_n = n - a_n$$$.

Now, because we know $$$p_n$$$, we can determine the $$$\texttt{MEX}$$$ of the first $$$n-1$$$ numbers. Like how we found $$$p_n$$$, we can do a similar process for finding $$$p_{n-1}$$$. Doing this for $$$i = n, n - 1 \dots 1$$$ will get us a valid answer $$$p$$$.

Note that this is also a way to show $$$p$$$ is always unique.

Ignoring $$$n$$$ for now, let's just focus on the $$$x$$$ chosen vertices. Sort the $$$x$$$ vertices and connect adjacent vertices to form its own smaller polygon. By drawing out some cases or if you're familiar with triangulation (video that proves by induction), you can form $$$x - 2$$$ triangles by drawing diagonals in a polygon with $$$x$$$ vertices. If you don't know it, one possible construction that always work is fixing one vertex $$$v$$$ and drawing diagonals to every other vertex not adjacent to $$$v$$$. Now, let's consider the original $$$n$$$-sided polygon. In additional to the aforementioned construction, to close the $$$x$$$-sided polygon up: for every non-adjacent vertex in the chosen vertices based on the bigger $$$n$$$-sided polygon, draw a diagonal between them. Through this construction, we can always guarantee $$$x - 2$$$ triangles.

However, this doesn't account for all triangles, as some triangles can form sides with the bigger $$$n$$$-sided polygon. These triangles occur exactly when two adjacent vertex in $$$x$$$ have exactly one vertex not in $$$x$$$ between them but part of the bigger polygon. This is because one side is from the diagonals we drew, and the other two sides form from the $$$n$$$-sided polygon.

Therefore, the answer is $$$x - 2$$$ + (number of adjacent chosen vertices $$$2$$$ apart). Note that the first chosen vertex is also adjacent to last chosen vertex in $$$x$$$.

Read the solution to the easy version first.

We can now reduce the problem to the following: For every vertex we choose, we can make one more triangle. For every chosen vertices two apart, we can make one more triangle. Let's focus on the latter condition. To maximize the effect of our $$$y$$$ vertices, we want to prioritize vertices $$$v$$$ that satisfy the following: the vertex is not included in $$$x$$$ and there is a chosen vertex two apart. Let's denote such $$$v$$$ as good.

Let vertex $$$a$$$ and $$$b$$$ $$$(a \lt b)$$$ be adjacent chosen vertices in the $$$x$$$-sided polygon. There are $$$g = b - a - 1$$$ vertices for us to choose between them. There are two cases: if $$$g$$$ is odd, then we can choose $$$\lfloor \frac{g}{2} \rfloor$$$ vertices to form $$$\lfloor \frac{g}{2}\rfloor + 1$$$ extra triangles. This is because all of the vertices we choose will be good. if $$$g$$$ is even, then we can choose $$$\frac{g}{2}$$$ vertices to make $$$\frac{g}{2}$$$ extra triangles. This is because one of the vertices we choose will not be good. This shows that it is always optimal to process smaller and odd $$$g$$$ first.

After processing these adjacent gaps, if we have any leftover vertices, we can just ignore them. This is because since we have maximized the number of good vertices, even though any further vertex we place will increase the answer by $$$1$$$, it will break two good vertices which will decrease the answer by $$$1$$$.

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

void solve(){
	int n, x, y; cin >> n >> x >> y;
	int initial_y = y;
	vector<int> chosen(x);
	for(int& i: chosen) cin >> i;
	sort(chosen.begin(), chosen.end());
	int ans = x - 2;
	int triangles_from_even_g = 0;
	vector<int> odd_g;
	auto process_gap = [&](int g) -> void{
		if(g <= 1){
			// already two apart
			ans += g;
		}
		else if(g % 2 == 1){
			odd_g.push_back(g / 2);
		}
		else{
			triangles_from_even_g += g / 2;
		}
	};
	for(int i = 0; i < x - 1; i++){
		process_gap(chosen[i + 1] - chosen[i] - 1);
	}
	process_gap((chosen[0] + n) - chosen[x - 1] - 1);
	sort(odd_g.begin(), odd_g.end());
	for(int g: odd_g){
		if(y >= g){
			// all vertices are good, so we can make g + 1 triangles
			ans += g + 1;
			y -= g;
		}
		else{
			ans += y;
			y = 0;
			break;
		}
	}
	int even_triangles = min(triangles_from_even_g, y);
	y -= even_triangles;
	ans += even_triangles;
	int used_vertices = initial_y - y;
	ans += used_vertices;
	cout << ans << "\n";
}

int main(){
	cin.tie(0) -> sync_with_stdio(0);
	int T = 1;
	cin >> T;
	while(T--) solve();
}

/*   /\_/\
*   (= ._.)
*   / >  \>
*/

We can solve this with dynamic programming. Let $$$\texttt{dp}[i]$$$ store a list of all $$$\min(k, 2^i)$$$ highest beauties of a painting in non-increasing order if you only paint cells $$$1,2,\ldots ,i$$$.

Transitioning boils down to finding the $$$k$$$ largest elements from $$$n$$$ non-increasing lists. Try to do this in $$$\mathcal{O}((n + k) \log n)$$$ time.

Let $$$\texttt{dp}[i]$$$ store a list of all $$$\min(k, 2^i)$$$ highest beauties of a painting in non-increasing order if you only paint cells $$$1,2,\ldots ,i$$$.

Let's define merging two lists as creating a new list that stores the $$$k$$$ highest values from the two lists.

First, let's look at a slow method of transitioning. We iterate over the left endpoint $$$l$$$ such that $$$l \ldots i$$$ is a maximal painted interval. At each $$$l$$$, we merge $$$\texttt{dp}[l - 2]$$$, with $$$a_{l,i}$$$ added to each value, into $$$\texttt{dp}[i]$$$. We also merge $$$\texttt{dp}[i - 1]$$$ into $$$\texttt{dp}[i]$$$ to handle the case in which we do not paint cell $$$i$$$. If implemented well, transitioning takes $$$\mathcal{O}(nk)$$$ time leading to an $$$\mathcal{O}(n^2k)$$$ solution which is too slow.

We can speed this up. This merging process boils down to finding the $$$k$$$ largest elements from $$$n$$$ non-increasing lists in $$$\mathcal{O}((n + k) \log n)$$$ time. We can use a priority queue that stores ($$$\texttt{element}, \texttt{index}$$$) for each list. We can do the following $$$k$$$ times: add the top of the priority queue to our answer, advance its index, and update the priority queue accordingly. This allows us to transition in $$$\mathcal{O}((n + k) \log n)$$$ time which leads to an $$$\mathcal{O}((n^2 + nk) \log n)$$$ solution.

Bonus: find an $$$\mathcal{O}(n^2 + k)$$$ solution.

#include <bits/stdc++.h>
using namespace std;
 
void solve(){
    int n, k;
    cin >> n >> k;
 
    int A[n + 1][n + 1];
 
    for (int i = 1; i <= n; i++)
        for (int j = i; j <= n; j++)
            cin >> A[i][j];
    
    // dp[i] = Answer if we consider 1...i
    vector<int> dp[n + 1];
    dp[0] = {0};
 
    for (int i = 1; i <= n; i++){
        priority_queue<array<int, 3>> pq;
 
        // Don't create an interval
        pq.push({dp[i - 1][0], i - 1, 0});
 
        // Create interval j+2...i (transition from j)
        for (int j = i - 2; j >= -1; j--){
            pq.push({(j < 0 ? 0 : dp[j][0]) + A[j + 2][i], j, 0});
        }
 
        while (pq.size() and dp[i].size() < k){
            auto [val, j, num] = pq.top(); pq.pop();
            dp[i].push_back(val);
            
            if (j < 0 or num + 1 >= dp[j].size())
                continue;
            
            // Don't create interval
            if (j == i - 1)
                pq.push({dp[i - 1][num + 1], i - 1, num + 1});
            
            // Create interval j+2...i (transition from j)
            else 
                pq.push({dp[j][num + 1] + A[j + 2][i], j, num + 1});
        }
    }
    for (int i : dp[n])
        cout << i << " ";
    cout << "\n";
}
 
int main(){
    ios_base::sync_with_stdio(0); cin.tie(0); 
    int tc; 
    cin >> tc;
 
    while (tc--) 
        solve();
}

Think about the gaps between $$$(a_1,b_1), (a_2,b_2), ... ,(a_n,b_n)$$$. Let the gaps be $$$g_1,g_2,...,g_n$$$. What effect do the moves have on the gaps?

WLOG let the first cow be FJ's cow. FN wins if all $$$g_i$$$ are even. Otherwise FJ wins. Note that the game will always end.

Try to count the number of ways FN wins. We can iterate over the sum of $$$g_i$$$ and use stars and bars as well as other counting techniques.

Consider the gaps between $$$(a_1,b_1), (a_2,b_2), ... ,(a_n,b_n)$$$. In the example below, '$$$\texttt{a}$$$' represents FJ's cow, '$$$\texttt{b}$$$' represents FN's cow, and '$$$\texttt{.}$$$' represents an empty space. The gaps are $$$[2,3]$$$.

Consider the gaps $$$g_1,g_2,...,g_n$$$. When a farmer moves, they choose some non-empty subset of non-zero $$$g_i$$$ and add or subtract $$$1$$$ from every element in the subset (so long such move is possible). The game ends when some farmer cannot move (which implies that all $$$g_i$$$ must be $$$0$$$ when the game ends).

WLOG let the first cow be FJ's cow. The winner of the game is determined as follows:

1) If all $$$g_i$$$ are even, then FN wins. If FJ pushes the cows corresponding to gaps $$$i_1, i_2, \ldots, i_k$$$ some direction in a move, then $$$g_{i_1}, g_{i_2}, \ldots g_{i_k}$$$ will change by $$$1$$$ and become odd. FN can then push the cows corresponding to gaps $$$i_1, i_2, \ldots, i_k$$$ left so that $$$g_{i_1}, g_{i_2}, \ldots g_{i_k}$$$ will become even again. Therefore, all $$$g_i$$$ will be even when it's FJ's turn and at least one $$$g_i$$$ will be odd when it's FN's turn. Since the game ends when all $$$g_i$$$ are $$$0$$$ (which is even), then FJ will lose. Note that the game will always end. This is since the sum of positions of FN's cows will always be decreasing since FN will always push some cow left.

2) Otherwise, FJ wins since he can subtract $$$1$$$ to all odd $$$g_i$$$ in his first move to make all elements even. Then it's FN's turn and all $$$g_i$$$ are even so by the reasoning above, FJ wins.

Let's use complementary counting and determine the number of ways FN wins (which is when all $$$g_i$$$ are even). We can iterate over the sum, $$$s$$$ ($$$s$$$ is even), of all $$$g_i$$$. At each $$$s$$$, we can use stars and bars to find how many such $$$g_1,g_2,...,g_n$$$ (all $$$g_i$$$ are even) sum to $$$s$$$. Specifically, the number of ways are $$$\frac{s}{2} + n - 1 \choose n - 1$$$. The number of ways to place the cows in the available positions given the gaps will be $$$2 \cdot {l - s - n \choose n}$$$. We multiply by $$$2$$$ since we can alternate starting with either FJ's cow or FN's cow.

Finally, we subtract this result from the total number of configurations, $$$2 \cdot {l \choose 2 \cdot n}$$$, to get the number of ways FJ wins. This runs in $$$\mathcal{O}(l)$$$ time.

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

#define ll long long

const int MAX = 2e6 + 5, MOD = 998244353;

ll fact[MAX], ifact[MAX];

ll bpow(ll a, int p){
    ll ans = 1;

    for (;p; p /= 2, a = (a * a) % MOD) 
        if (p & 1) 
            ans = (ans * a) % MOD;

    return ans;
}
ll ncr(int n, int r){
    if (n < 0)
        return 0;
    if (r > n)
        return 0;

    return fact[n] * ifact[r] % MOD * ifact[n - r] % MOD;
}
void solve(){
    int l, n;
    cin >> l >> n;
 
    ll all_even = 0;
 
    for (int s = 0; s <= l; s += 2){
        all_even += 2 * ncr(s / 2 + n - 1, n - 1) % MOD * ncr(l - s - n, n) % MOD;
        all_even %= MOD;
    }
    cout << (2 * ncr(l, 2 * n) % MOD - all_even + MOD) % MOD << "\n";
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    
    for (int i = 0; i < MAX; i++)
        fact[i] = !i ? 1 : fact[i - 1] * i % MOD;
    
    for (int i = MAX - 1; i >= 0; i--)
        ifact[i] = (i == MAX - 1) ? bpow(fact[MAX - 1], MOD - 2) : ifact[i + 1] * (i + 1) % MOD;

    int tc;
    cin >> tc;

    while (tc--)
        solve();
}

Consider the case where all $$$f(i)$$$ are integers. Decreasing any $$$a_i$$$ will decrease $$$\lfloor f(n) \rfloor$$$. So solutions that iterate over the last few elements will not work.

If $$$n \ge 6$$$, changing $$$a_1$$$ will only affect $$$\lfloor f(n) \rfloor$$$ by at most $$$1$$$.

We can take the floor each time we square root. Specifically, we can define $$$f$$$ as:

• $$$f(1) = \lfloor \sqrt{a_1} \rfloor$$$
• For all $$$i \gt 1$$$, $$$f(i) = \lfloor \sqrt{f(i-1)+a_i} \rfloor$$$

This allows us to work with integers.

We can divide our array into blocks of size $$$b \ge 6$$$ (note that the value of $$$b$$$ depends on the solution). How can we use hint $$$2$$$? What should we store for each block?

Let's first look at the impact of earlier numbers on the final result. Earlier numbers can still influence $$$\lfloor f(n) \rfloor$$$ when $$$n$$$ is large. Suppose that all $$$f(i)$$$ are integers. Then decreasing any $$$a_i$$$ will decrease $$$\lfloor f(n) \rfloor$$$.

However, it is clear that the impact of earlier number on $$$\lfloor f(n) \rfloor$$$ is still extremely small. A key observation is that when $$$n \ge 6$$$, changing $$$a_1$$$ will only affect $$$\lfloor f(n) \rfloor$$$ by at most $$$1$$$.

Another observation is that we can take the floor each time we square root. Specifically, we can define $$$f$$$ as:

• $$$f(1) = \lfloor \sqrt{a_1} \rfloor$$$
• For all $$$i \gt 1$$$, $$$f(i) = \lfloor \sqrt{f(i-1)+a_i} \rfloor$$$

This allows us to work with integers. From here on, we work with this new definition of $$$f$$$.

Let's divide our array into blocks of size $$$b=6$$$. We can append zeroes to the front of the array to make $$$n$$$ a multiple of $$$b$$$.

Consider some block representing range $$$[l,r]$$$. Let's consider the subarray $$$a_l,a_{l+1},\ldots ,a_r$$$. Let $$$v$$$ be the value of $$$f(r)$$$ if we let $$$f(l - 1)=0$$$. Using our first observation, we know that $$$f(r)$$$ will be either $$$v$$$ or $$$v+1$$$ depending on $$$f(l - 1)$$$. So let's find the smallest value $$$c$$$ such that if $$$f(l - 1)=c$$$, then $$$f(r)=v+1$$$. This can be done by iterating over the elements of the block backwards.

For each block, we store its corresponding ($$$v,c$$$). We can build a segment tree over the blocks for an $$$\mathcal{O}((n + q) \log n)$$$ solution.

Alternatively, we can do square root decomposition by having $$$b=\sqrt n$$$ which leads to an $$$\mathcal{O}(n + q \sqrt n)$$$ solution (in practice, we should set $$$b$$$ to something small like $$$100$$$).

"Draw $$$1$$$" cards do not matter because we can immediately play them when we draw them.

Treat "draw $$$2$$$" as $$$+1$$$ and "draw $$$0$$$" as $$$-1$$$. We start with a balance of $$$+5$$$. We draw all the cards up to the first prefix where the balance dips to $$$0$$$. We are interested in the number of ways where the special cards lie in this prefix.

Read the hints.

Treat the special cards as "draw $$$0$$$" cards and multiply by the appropriate binomial coefficient at the end. Also treat cards of the same type as indistinguishable and multiply by the appropriate factorials at the end.

Enumerate the length of the prefix where the balance first hits $$$0$$$. This dictates how many $$$+1$$$ and $$$-1$$$ we have. Now we want to count the number of ways of arranging $$$+1$$$ and $$$-1$$$ such that the balance never dips to $$$0$$$, assuming we start with $$$+5$$$.

To solve this subproblem, we can draw inspiration from the path counting interpretation for counting balanced bracket sequences. We have $$$n$$$ open and $$$n$$$ close brackets and want to count the number of balanced bracket sequences starting with (((( (note that there are only $$$4$$$ instead of $$$5$$$ open brackets to shift from all balances being strictly positive to all balances being non-negative). The number of sequences ignoring the balance constraint is $$$\binom{2n-4}{n}$$$. Any bracket sequence where some balance dips negative corresponds to a path that walks above the line $$$y = x$$$. For those paths, we reflect over the line $$$y = x + 1$$$ at the first point where it steps over the line. So there is a bijection between these paths and paths that reach the point $$$(n-1,n+1)$$$, of which there are $$$\binom{2n-4}{n+1}$$$. So the total number of ways is $$$\binom{2n-4}{n} - \binom{2n-4}{n+1}$$$.

Our final answer is summing up the number of ways over all prefix lengths. Make sure to handle the case where you draw the entire deck correctly. The complexity is $$$\mathcal O(\min(a,c))$$$.

Model this as a flow problem where growth events are an input of some flow to a node and harvest events and the necessary amount of fruits are an output of flow to the sink. We will define the $$$a_i$$$ edge to be the input edge of every node, $$$b_i$$$ to be the output edge for necessary fruits, and $$$c_i$$$ to be the output edge for growth events. The connections for each individual node will look like:

$$$(\text{source}, a_i, \text{# of growth events})$$$

$$$(a_i, \text{temporary node}, \text{INF})$$$

$$$(\text{temporary node}, b_i, \text{INF})$$$

$$$(\text{temporary node}, c_i, \text{INF})$$$

$$$(b_i, \text{sink}, \text{# of required peaches})$$$

$$$(c_i, \text{sink}, \text{# of harvest events})$$$

To simulate growth events affecting the subtree, every temporary node has infinite capacity directed edges down to the temporary nodes of its children. To simulate growth events affecting the parent, every temporary node has an infinite capacity directed edge to its parent's $$$b_i$$$. To simulate harvest events affecting subtree, every node has infinite capacity directed edges to all its ancestors' $$$c_i$$$ nodes. Lets try to find the min cut of the graph. If we don't put cut some node's $$$a_i$$$, then we must cut both $$$b_i$$$ and $$$c_i$$$ for all of its descendants. If we don't cut some node's $$$a_i$$$, then we also must cut $$$c_i$$$ for all its ancestors and $$$b_i$$$ for its parent. Thus, we can model a $$$dp[i][0/1/2]$$$ to know if we are currently cutting the a_i input edge, cutting the b_i and c_i output edge, or cutting the a_i edge but there is some node in the subtree where we did not cut the a_i edge, in which case we need to cut the c_i edge as well. Transitions look like:

$$$dp[i][0] = a_i + \sum_j dp[j][0]$$$

$$$dp[i][1] = b_i + c_i + \sum_j dp[j][1]$$$

$$$dp[i][2] = a_i + c_i + \min(\sum_j \min(dp[j][0], dp[j][2]), b_i + \sum_j \min(dp[j][0], dp[j][1], dp[j][2]))$$$ where at least one child is a $$$dp[j][2]$$$ or $$$dp[j][1]$$$

$$$dp[i][2]$$$ can be rewritten as:

$$$dp[i][2] = a_i + c_i + \min(\min(dp[j][2] - \min(dp[j][0], dp[j][2])) + \sum_j \min(dp[j][0], dp[j][2]),$$$ $$$\min(dp[j][1] - \min(dp[j][0], dp[j][1], dp[j][2])) + b_i + \sum_j \min(dp[j][0], dp[j][1], dp[j][2])))$$$

Use dynamic dp by representing dp transitions as a matrix where all the light children dp values are known and the heavy child dp values are variable. Then, we can accelerate each chain in the heavy light decomposition of the tree. To get the matrix representation of each transition to the heavy child, there are 5 cases when considering a $$$dp[j][0/1/2]$$$ -> $$$dp[j][2]$$$ transiton where $$$dp[j][0]$$$ is the min, $$$dp[j][1]$$$ is the min, and $$$dp[j][2]$$$ is the min, $$$dp[j][2] - \min(dp[j][0], dp[j][2])$$$ is the optimal added value, or $$$dp[j][0] - min(dp[j][0], dp[j][1], dp[j][2])$$$, is the optimal added value. For each cell in the matrix, put the min of the $$$5$$$ cases.

Since you will only ever change hld chains $$$\log n$$$ times, the total complexity will be $$$n \log^2 n \cdot 3^3$$$. If we use balanced hld or other $$$n \log n$$$ hld techniques, then the complexity becomes $$$n \log^2 n$$$. Both solutions should pass under the given constraints.

    #ifndef LOCAL
    #pragma GCC optimize("O3,unroll-loops")
    #pragma GCC target("avx,avx2,fma")
    #pragma GCC target("sse4,popcnt,abm,mmx,tune=native")
    #endif
    #include <bits/stdc++.h>
    #include <ext/pb_ds/assoc_container.hpp>
    #include <ext/pb_ds/tree_policy.hpp>
    using namespace __gnu_pbds;
    using namespace std;
    
    #define pb push_back
    #define ff first
    #define ss second
    
    typedef long long ll;
    typedef long double ld;
    typedef pair<int, int> pii;
    typedef pair<ll, ll> pll;
    typedef pair<ld, ld> pld;
    
    const int INF = 1e9;
    const ll LLINF = 1e18;
    const int MOD = 1e9 + 7;
    
    template<class K> using sset =  tree<K, null_type, less<K>, rb_tree_tag, tree_order_statistics_node_update>;
    
    inline ll ceil0(ll a, ll b) {
        return a / b + ((a ^ b) > 0 && a % b);
    }
    
    void setIO() {
        ios_base::sync_with_stdio(0); cin.tie(0);
    }
    
    struct Matrix {
    
        array<array<ll, 3>, 3> mat;
    
        array<ll, 3> &operator[](int x){
            return mat[x];
        }
    
        const array<ll, 3> &operator[](int x) const {
            return mat[x];
        }
    
        Matrix(){
            for(int i = 0; i < 3; i++) for(int j = 0; j < 3; j++) mat[i][j] = 0;
        }
    
        Matrix operator*(const Matrix &x){
            Matrix ret;
            for(int i = 0; i < 3; i++) for(int j = 0; j < 3; j++) ret[i][j] = LLINF;
            for(int i = 0; i < 3; i++){
                for(int k = 0; k < 3; k++){
                    for(int j = 0; j < 3; j++){
                        ret[i][j] = min(ret[i][j], mat[i][k] + x[k][j]);
                    }
                }
            }
            return ret;
        }
    
        Matrix &operator*=(const Matrix &x){
            (*this) = (*this) * x;
            return *this;
        }
    };
    
    struct fastset {
    
        priority_queue<ll, vector<ll>, greater<ll>> q, rem;
    
        void insert(ll x){
            q.push(x);
        }
    
        void erase(ll x){
            rem.push(x);
        }
    
        ll query(){
            while(rem.size() && q.top() == rem.top()){
                rem.pop();
                q.pop();
            }
            return q.top();
        }
    
        void clear(){
            while(q.size()) q.pop();
            while(rem.size()) rem.pop();
        }
    };
    
    vector<int> g[500005];
    int sub[500005];
    int head[500005];
    int par[500005];
    int tim = 0;
    vector<int> chain[500005];
    int weight[500005];
    
    void dfs1(int x){
        sub[x] = 1;
        for(int &i : g[x]){
            dfs1(i);
            sub[x] += sub[i];
            if(sub[i] > sub[g[x][0]]) swap(i, g[x][0]);
        }
    }
    
    int in[500005], out[500005];
    int dir[500005];
    
    void dfs2(int x, int p){
        chain[head[x]].pb(x);
        par[x] = p;
        weight[x] = 1;
        in[x] = tim++;
        for(int i : g[x]){
            head[i] = (i == g[x][0] ? head[x] : i);
            dfs2(i, x);
            if(i != g[x][0]) weight[x] += sub[i];
        }
        out[x] = tim - 1;
    }
    
    Matrix dp[500005]; 
    Matrix cum[500005];
    int rt[500005];
    int left0[500005], right0[500005], ppar[500005];
    
    int dnq(int l, int r, int p, const vector<int> &v){
        if(l > r) return 0;
        int sum = 0;
        for(int i = l; i <= r; i++) sum += weight[v[i]];
        int cur = 0;
        int mid = -1;
        for(int i = l; i <= r; i++){
            cur += weight[v[i]];
            if(cur >= sum/2){
                mid = i;
                break;
            }
        }
        if(p == -1) ppar[v[mid]] = v[mid];
        else ppar[v[mid]] = p;
        left0[v[mid]] = dnq(l, mid - 1, v[mid], v);
        right0[v[mid]] = dnq(mid + 1, r, v[mid], v);
        return v[mid];
    }
    
    void merge(int x){
        if(right0[x] && left0[x]){
            cum[x] = cum[left0[x]]*dp[x]*cum[right0[x]];
        } else if(left0[x] && !right0[x]){
            cum[x] = cum[left0[x]]*dp[x];
        } else if(right0[x] && !left0[x]){
            cum[x] = dp[x]*cum[right0[x]];
        } else {
            cum[x] = dp[x];
        }
    }
    
    void pull(int x){
        while(x != rt[x]){
            merge(x); 
            x = ppar[x];
        }
        merge(x);
    }
    
    ll a[500005], b[500005], c[500005];
    ll dp0[500005][3];
    
    void dfs3(int x){
        dp0[x][0] = a[x];
        dp0[x][2] = b[x] + c[x];
        ll deg1 = a[x] + c[x], deg2 = b[x] + a[x] + c[x];
        ll mn1 = LLINF, mn2 = LLINF;
        for(int i : g[x]){
            dfs3(i);
            dp0[x][0] += dp0[i][0];
            dp0[x][2] += dp0[i][2];
            deg1 += min(dp0[i][0], dp0[i][1]);
            deg2 += min({dp0[i][0], dp0[i][1], dp0[i][2]});
            mn1 = min(mn1, max((ll)0, dp0[i][1] - dp0[i][0]));
            mn2 = min(mn2, max((ll)0, min(dp0[i][1], dp0[i][2]) - dp0[i][0]));
        }
        dp0[x][1] = min(deg1 + mn1, deg2 + mn2);
    }
    
    fastset deg1[500005], deg2[500005];
    ll sum0[500005], sum2[500005], sum01[500005], sum012[500005];
    
    void rebuild(int x){
        dp[x][0][0] = a[x] + sum0[x];
        dp[x][0][1] = LLINF;
        dp[x][0][2] = LLINF;
        if(g[x].size()){
            dp[x][1][0] = min(a[x] + b[x] + c[x] + deg2[x].query() + sum012[x], a[x] + c[x] + deg1[x].query() + sum01[x]);
            dp[x][1][1] = min(a[x] + c[x] + sum01[x], a[x] + b[x] + c[x] + sum012[x]);
            dp[x][1][2] = a[x] + b[x] + c[x] + sum012[x];
        } else {
            dp[x][1][0] = dp[x][1][1] = dp[x][1][2] = LLINF;
        }
        dp[x][2][0] = LLINF;
        dp[x][2][1] = LLINF;
        dp[x][2][2] = b[x] + c[x] + sum2[x];
    }
    
    ll eval(int x, int ind){
        return min({cum[rt[x]][ind][0], cum[rt[x]][ind][1], cum[rt[x]][ind][2]});
    }
    
    void rem(int x){
        while(head[x] != 1){
            if(x == head[x]){
                ll v0 = eval(x, 0), v1 = eval(x, 1), v2 = eval(x, 2);
                deg1[par[x]].erase(max((ll)0, v1 - v0));
                deg2[par[x]].erase(max((ll)0, min(v1, v2) - v0));
                sum012[par[x]] -= min({v0, v1, v2});
                sum01[par[x]] -= min(v0, v1);
                sum0[par[x]] -= v0;
                sum2[par[x]] -= v2;
                x = par[x];
            }
            x = head[x];
        }
    }
    
    void add(int x){
        while(head[x] != 1){
            if(x == head[x]){
                ll v0 = eval(x, 0), v1 = eval(x, 1), v2 = eval(x, 2);
                deg1[par[x]].insert(max((ll)0, v1 - v0));
                deg2[par[x]].insert(max((ll)0, min(v1, v2) - v0));
                sum012[par[x]] += min({v0, v1, v2});
                sum01[par[x]] += min(v0, v1);
                sum0[par[x]] += v0;
                sum2[par[x]] += v2;
                rebuild(par[x]);
                pull(par[x]);
                x = par[x];
            }
            x = head[x];
        }
    }
    
    int main(){
        setIO();
        int t;
        cin >> t;
        while(t--){
            int n, q;
            cin >> n >> q;
            for(int i = 2; i <= n; i++){
                int x;
                cin >> x;
                g[x].pb(i);
            }
            ll sum = 0;
            for(int i = 1; i <= n; i++) cin >> b[i], sum += b[i];
            dfs1(1);
            head[1] = 1;
            dfs2(1, 1);
            dfs3(1);
            for(int i = 1; i <= n; i++){
                if(chain[i].size()){
                    int p = dnq(0, chain[i].size() - 1, -1, chain[i]);
                    for(int j : chain[i]) rt[j] = p;
                }
            }
            for(int i = 1; i <= n; i++){
                deg1[i].insert(LLINF);
                deg2[i].insert(LLINF);
                for(int j : g[i]){
                    if(head[j] == head[i]) continue;
                    deg1[i].insert(max((ll)0, dp0[j][1] - dp0[j][0]));
                    deg2[i].insert(max((ll)0, min(dp0[j][1], dp0[j][2]) - dp0[j][0]));
                    sum012[i] += min({dp0[j][0], dp0[j][1], dp0[j][2]});
                    sum01[i] += min(dp0[j][0], dp0[j][1]);
                    sum0[i] += dp0[j][0];
                    sum2[i] += dp0[j][2];
                }
                rebuild(i);
                pull(i);
            } 
            while(q--){
                int t, x, v;
                cin >> t >> x >> v;
                rem(x);
                if(t == 1) a[x] += v;
                else c[x] += v, sum += v;
                rebuild(x);
                pull(x);
                add(x);
                cout << (min({eval(1, 0), eval(1, 1), eval(1, 2)}) == sum ? "YES" : "NO") << "\n";
            }
            tim = 0;
            for(int i = 1; i <= n; i++){
                a[i] = b[i] = c[i] = 0;
                g[i].clear();
                chain[i].clear();
                deg1[i].clear();
                deg2[i].clear();
                sum0[i] = sum2[i] = sum01[i] = sum012[i] = 0;
            }
        }
    }

There also exists a solution without flows. Thanks to rainboy for discovering it during testing!

Lets greedily assign the growth events. We will always try to use the growth events to satisfy harvest events or required peaches in its subtree, and if there is any excess then we give it to the parent's required peaches. Similarly, we will always use harvest events on growth events in its subtree. Lets define a $$$dp[u]$$$ for each node $$$u$$$. If $$$dp[x] \gt 0$$$ then it means we can give $$$dp[u]$$$ growths to $$$u$$$'s parent. Otherwise, it means $$$dp[u]$$$ requires $$$-dp[u]$$$ growths from its ancestors. Let $$$pos_u$$$ denote the sum of positive $$$dp[v]$$$ where $$$v$$$ is a child a of $$$u$$$, and $$$neg_u$$$ denote the sum of negative $$$dp[v]$$$. First we used the excess growth to satisfy the required peaches, so we can set $$$b_u = \max(0, b_u - pos_u)$$$. Then we distribute the growth events at node $$$u$$$ to satisfy the requirements in the subtree. Define $$$a_i$$$ as the number of growth events on node $$$i$$$ and $$$c_i$$$ as the number of harvest events on node $$$i$$$. To account for the harvest events at node $$$i$$$, we can define $$$bal_u = \text{sum of } a_i \text{ in u's subtree} - \text{sum of } b_i \text{ in u's subtree} - \text{sum of } c_i \text{ in u's subtree}$$$. We know that $$$dp[u] \le bal_u$$$, and as it turns out, as long as we keep $$$bal_u$$$ as an upper bound on $$$dp[u]$$$, the harvest events will always be satisfied. Why? $$$a_u + neg_u - \max(0, b_u - pos_u)$$$ is the assignment of growth events without considering harvest events. The amount of unused growth events in the subtree of $$$u$$$ is given by $$$bal_u - (a_u + neg_u - \max(0, b_u - pos_u)) + c_u$$$. If $$$a_u + neg_u - \max(0, b_u - pos_u) \le bal_u$$$, then it means we can use all of our harvest events on unused growth events in the subtree. If $$$a_u + neg_u - \max(0, b_u - pos_u) \gt bal_u$$$, then it means we will use all of our harvest events but still have $$$bal_u - (a_u + neg_u - \max(0, b_u - pos_u))$$$ harvest events remaining. Thus, we can formulate our dp as $$$dp[u] = \min(a_u + neg_u - \max(0, b_u - pos_u), bal_u)$$$. Let $$$v$$$ be the heavy child of $$$u$$$ and $$$pos'_u$$$ be the sum of positive dp excluding $$$v$$$ and $$$neg'_u$$$ be the sum of positive dp excluding $$$v$$$. Then, we can rewrite our dp as $$$dp[u] = \min(dp[v] + a_u + neg'_u - \max(0, b_u - pos'_u), a_u + neg'_u, bal_u)$$$. Define $$$x_u = a_u + neg'_u - \max(0, b_u - pos'_u)$$$ and $$$y_u = a_u + neg'_u$$$. Then $$$dp[u] = \min(dp[v] + x_u, y_u, bal_u)$$$. This can be accelerated by representing the transition as a matrix and accelerating it using dynamic dp. However, we can also notice that the dp can be represented as the smallest $$$x_1 + x_2 + \ldots + x_{i - 1} + \min(y_i, bal_i)$$$ over all prefixes of each heavy light decomposition chain. Thus, we can maintain the dp by using a range add and prefix min segment tree.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define N	300000
#define N_	(1 << 19)	/* N_ = pow2(ceil(log2(N))) */
#define INF	0x3f3f3f3f3f3f3f3fLL

long long min(long long a, long long b) { return a < b ? a : b; }
long long max(long long a, long long b) { return a > b ? a : b; }

int *ej[N], eo[N], n;

void append(int i, int j) {
	int o = eo[i]++;

	if (o >= 2 && (o & o - 1) == 0)
		ej[i] = (int *) realloc(ej[i], o * 2 * sizeof *ej[i]);
	ej[i][o] = j;
}

int dd[N], pp[N], qq[N], jj[N], ta[N], tb[N], qu[N];
int bb[N]; long long aa[N], zz[N], zzp[N], zzn[N];

int dfs1(int i, int d) {
	int o, s, k_, j_;

	dd[i] = d;
	s = 1, k_ = 0, j_ = -1;
	zz[i] = -bb[i];
	for (o = eo[i]; o--; ) {
		int j = ej[i][o], k = dfs1(j, d + 1);

		s += k;
		if (k_ < k)
			k_ = k, j_ = j;
		zz[i] += zz[j];
	}
	qq[i] = j_, jj[i] = j_ == -1 ? i : jj[j_];
	return s;
}

int t_;

void dfs2(int i, int q) {
	int o, j_;

	qu[ta[i] = t_++] = i;
	j_ = qq[i], qq[i] = q;
	if (j_ != -1)
		dfs2(j_, q);
	for (o = eo[i]; o--; ) {
		int j = ej[i][o];

		if (j != j_)
			dfs2(j, j);
	}
	tb[i] = t_;
}

long long stx[N_ * 2], sty[N_ * 2], stz[N_ * 2], lz[N_]; int h_, n_;

void put(int i, long long x) {
	stz[i] += x;
	if (i < n_)
		lz[i] += x;
}

void pus(int i) {
	if (lz[i])
		put(i << 1 | 0, lz[i]), put(i << 1 | 1, lz[i]), lz[i] = 0;
}

void pul(int i) {
	if (!lz[i]) {
		int l = i << 1, r = l | 1;

		stx[i] = stx[l] + stx[r];
		sty[i] = min(sty[l], stx[l] + sty[r]);
		stz[i] = min(stz[l], stx[l] + stz[r]);
	}
}

void push(int i) {
	int h;

	for (h = h_; h > 0; h--)
		pus(i >> h);
}

void pull(int i) {
	while (i > 1)
		pul(i >>= 1);
}

void build() {
	int i, j, a, o;
        h_ = 0;

	while (1 << h_ < n)
		h_++;
	n_ = 1 << h_;
	memset(stx, 0, n_ * 2 * sizeof *stx);
	memset(sty, 0, n_ * 2 * sizeof *sty);
	memset(stz, 0, n_ * 2 * sizeof *stz);
	memset(lz, 0, n_ * sizeof *lz);
	memset(aa, 0, n * sizeof *aa);
	memset(zzp, 0, n * sizeof *zzp);
	memset(zzn, 0, n * sizeof *zzn);
	for (i = 0; i < n; i++) {
		a = ta[i];
		stz[n_ + a] = zz[i];
		for (o = eo[i]; o--; ) {
			j = ej[i][o];
			if (qq[j] == j)
				zzn[i] += -zz[j];
		}
		stx[n_ + a] = -zzn[i] - bb[i];
		sty[n_ + a] = -zzn[i];
	}
	for (i = n_ - 1; i > 0; i--)
		pul(i);
}

void upd(int i) {
	int i_ = n_ + ta[i];

	push(i_);
	stx[i_] = aa[i] - zzn[i] - max(bb[i] - zzp[i], 0);
	sty[i_] = aa[i] - zzn[i];
	pull(i_);
}

void update_z(int l, int r, long long x) {
	int l_ = l += n_, r_ = r += n_;

	push(l_), push(r_);
	for ( ; l <= r; l >>= 1, r >>= 1) {
		if ((l & 1) == 1)
			put(l++, x);
		if ((r & 1) == 0)
			put(r--, x);
	}
	pull(l_), pull(r_);
}

long long query(int l, int r) {
	long long xl, yl, zl, xr, yr, zr;

	push(l += n_), push(r += n_);
	xl = 0, yl = INF, zl = INF, xr = 0, yr = INF, zr = INF;
	for ( ; l <= r; l >>= 1, r >>= 1) {
		if ((l & 1) == 1) {
			yl = min(yl, xl + sty[l]), zl = min(zl, xl + stz[l]), xl += stx[l];
			l++;
		}
		if ((r & 1) == 0) {
			yr = min(sty[r], yr == INF ? INF : stx[r] + yr), zr = min(stz[r], zr == INF ? INF : stx[r] + zr), xr += stx[r];
			r--;
		}
	}
	return min(min(xl + xr, min(yl, yr == INF ? INF : xl + yr)), min(zl, zr == INF ? INF : xl + zr));
}

int main() {
	int t;

	scanf("%d", &t);
	while (t--) {
		int q_, i, p, q;

		scanf("%d%d", &n, &q_);
		for (i = 0; i < n; i++)
			ej[i] = (int *) malloc(2 * sizeof *ej[i]), eo[i] = 0;
		pp[0] = -1;
		for (i = 1; i < n; i++) {
			scanf("%d", &pp[i]), pp[i]--;
			append(pp[i], i);
		}
		for (i = 0; i < n; i++)
			scanf("%d", &bb[i]);
		dfs1(0, 0);
		t_ = 0, dfs2(0, 0);
		build();
		while (q_--) {
			int t, x;

			scanf("%d%d%d", &t, &i, &x), i--;
			if (t == 2)
				x = -x;
			if (t == 1)
				aa[i] += x, upd(i);
			while (i >= 0) {
				q = qq[i], p = pp[q];
				if (p >= 0) {
					if (zz[q] > 0)
						zzp[p] -= zz[q];
					else
						zzn[p] -= -zz[q];
				}
				update_z(ta[q], ta[i], x);
				zz[q] = query(ta[q], ta[jj[q]]);
				if (p >= 0) {
					if (zz[q] > 0)
						zzp[p] += zz[q];
					else
						zzn[p] += -zz[q];
					upd(p);
				}
				i = p;
			}
			printf(zz[0] >= 0 ? "YES\n" : "NO\n");
		}
		for (i = 0; i < n; i++)
			free(ej[i]);
	}
	return 0;
}