Since $$$a_i+b_i \le a_{i+1}+b_{i+1}$$$, then $$$a_i+b_i$$$ can be equal to $$$a_{i+1}+b_{i+1}$$$.

Therefore, any permutation $$$b$$$ which satisfies $$$a_1+b_1=a_2+b_2=\ldots=a_n+b_n$$$ is a valid answer.

If $$$b_i=n+1-a_i$$$, then:

1. $$$b$$$ is a permutation;
2. $$$a_1+b_1=a_2+b_2=\ldots=a_n+b_n=n+1$$$

Consequently, $$$b=[n+1-a_1,n+1-a_2,\ldots,n+1-a_i,\ldots,n+1-a_n]$$$ is a valid answer.

Time complexity per test case : $$$O(N)$$$

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
void tc(){
    ll n;
    cin>>n;
    for(ll i=0;i<n;i++){
        ll x;
        cin>>x;
        cout<<n+1-x<<' ';
    }
    cout<<'\n';
}
int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    ll t; cin>>t; while(t--)
        tc();
    return 0;
}

Good problem

Ok problem

Bad problem

Didn't solve it, idk

Let $$$max_a(x)$$$ be the length of the longest subarray from $$$a$$$ containing only elements equal to $$$x$$$. If $$$x$$$ doesn't appear in $$$a$$$, then $$$max_a(x)=0$$$.

Similarly, let $$$max_b(x)$$$ be the length of the longest subarray from $$$b$$$ containing only elements equal to $$$x$$$. If $$$x$$$ doesn't appear in $$$b$$$, then $$$max_b(x)=0$$$.

$$$max_a$$$ and $$$max_b$$$ can be computed in $$$O(N)$$$ by scanning the array while updating current maximal subarray.

When merging two arrays, it is possible to force a particular subarray $$$[a_{l_1},a_{l_1+1},\ldots,a_{r_1}]$$$ to be adjacent to another particular subarray $$$[b_{l_2},b_{l_2+1},\ldots,b_{r_2}]$$$ in the merged array.

If $$$a_{l_1}=b_{l_2}$$$, then the merged array will have a subarray consisting of $$$(r_1-l_1+1)+(r_2-l_2+1)$$$ equal elements.

Therefore, the answer is equal to:

Time complexity per testcase: $$$O(N)$$$.

#include <bits/stdc++.h>

using namespace std;

int32_t main()
{
    cin.tie(nullptr)->sync_with_stdio(false);
    int q;
    cin >> q;
    while (q--)
    {
        int n;
        cin >> n;
        vector<int> a(n + 1);
        vector<int> b(n + 1);
        for (int i = 1; i <= n; ++i)
        {
            cin >> a[i];
        }
        for (int i = 1; i <= n; ++i)
        {
            cin >> b[i];
        }
        vector<int> fa(n + n + 1);
        vector<int> fb(n + n + 1);
        int p = 1;
        for (int i = 2; i <= n; ++i)
        {
            if (a[i] != a[i - 1])
            {
                fa[a[i - 1]] = max(fa[a[i - 1]], i - p);
                p = i;
            }
        }
        fa[a[n]] = max(fa[a[n]], n - p + 1);

        p = 1;
        for (int i = 2; i <= n; ++i)
        {
            if (b[i] != b[i - 1])
            {
                fb[b[i - 1]] = max(fb[b[i - 1]], i - p);
                p = i;
            }
        }
        fb[b[n]] = max(fb[b[n]], n - p + 1);

        int ans = 0;
        for (int i = 1; i <= n + n; ++i)
        {
            ans = max(ans, fa[i] + fb[i]);
        }

        cout << ans << '\n';
    }
}

Good problem

Ok problem

Bad problem

Didn't solve it, idk

This problem can be solved via dynamic programming. From here on out, step $$$1$$$ from the statement will be called a "scan".

Let $$$dp[i]$$$ be the number of scans needed to activate node $$$i$$$, and $$$id[i]$$$ be the index (in the order from the input) of the edge which activated node $$$i$$$.

Intially, since only $$$1$$$ is active, $$$dp[1]=1$$$ and $$$id[1]=0$$$.

We will perform a dfs traversal starting from node $$$1$$$. When we process an edge $$$(u,v)$$$, one of the following two cases can happen:

1. If $$$index((u,v)) \ge id[u]$$$, we can visit $$$v$$$ in the same scan as $$$u$$$:

1. If $$$index((u,v)) \lt id[u]$$$, $$$v$$$ will be visited in the next scan after $$$dp[u]$$$:

The answer is $$$max_{i=1}^n(dp[i])$$$.

Time complexity per test case: $$$O(n)$$$

#include<bits/stdc++.h>

using namespace std;
const int NMAX = 2e5+5;
int dp[NMAX], id[NMAX];
vector<pair<int,int>> edg[NMAX];
void dfs(int u){
    for(auto it : edg[u]){
        if(dp[it.first] == 0){
            dp[it.first] = dp[u] + (it.second <= id[u]);
            id[it.first] = it.second;
            dfs(it.first);
        }
    }
}
void tc(){
    int n;
    cin>>n;
    for(int i=1; i<=n; i++) dp[i] = id[i] = 0, edg[i].clear();
    for(int i=1; i<n; i++){
        int u,v;
        cin>>u>>v;
        edg[u].push_back({v,i});
        edg[v].push_back({u,i});
    }
    dp[1] = 1;
    dfs(1);
    int ans = 0;
    for(int i=1; i<=n; i++) ans=max(ans,dp[i]);
    cout<<ans<<'\n';
}
int main()
{
    ios_base::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--)
        tc();
    return 0;
}

#include <bits/stdc++.h>

const int NMAX = 3e5 + 5, INF = 1e9;

int n, f[NMAX], d[NMAX];
std :: vector < std :: pair < int, int > > G[NMAX];

void DFS(int node, int t) {
    f[node] = true;

    for (int i = 0; i < G[node].size(); ++ i) {
        int u = G[node][i].first, c = G[node][i].second;

        if (f[u] == false) {
            d[u] = (c < t) + d[node];
            DFS(u, c);
        }
    }

    return;
}

int main() {
    std :: ios_base :: sync_with_stdio(false);
    std :: cin.tie(nullptr);
    
    int tc;
    
    std :: cin >> tc;
    
    while (tc --) {
        std :: cin >> n;
    
        for (int i = 1, u, v; i < n; ++ i) {
            std :: cin >> u >> v;
    
            G[u].push_back({v, i});
            G[v].push_back({u, i});
        }
        
        for (int i = 1; i <= n; ++ i)
            f[i] = false;
    
        d[1] = 0;
        DFS(1, n);
    
        int Max = 0;
    
        for (int i = 1; i <= n; ++ i)
            Max = std :: max(Max, d[i]);
    
        std :: cout << Max << "\n";
        
        for (int i = 1; i <= n; ++ i)
            G[i].clear();
    }

    return 0;
}

Good problem

Ok problem

Bad problem

Didn't solve it, idk

Since $$$b_i \le n$$$ and $$$b_j \le n$$$, $$$b_i+b_j = a_i \cdot a_j \le 2 \cdot n$$$. Therefore, $$$\min(a_i,a_j) \le \sqrt{2 \cdot n}$$$.

Let $$$lim=\sqrt{2 \cdot n}$$$ and $$$fr[a_i][b_i]$$$ be the number of pairs $$$(a_i,b_i)$$$ from the input such that $$$a_i \le lim$$$.

A pair $$$(i,j)$$$ is good if it satisfies $$$a_i \cdot a_j=b_i+b_j$$$.

Firstly, we'll count the number of good pairs $$$(i,j)$$$ such that $$$a_i=a_j$$$. Since $$$min(a_i,a_j) \le lim$$$, we can see that $$$a_i=a_j \le lim$$$.

This sum can be written as:

The remaining good pairs will have $$$a_i \neq a_j$$$, and instead of counting the pairs which have $$$i \lt j$$$, we can count the pairs which have $$$a_i \gt a_j$$$.

Since $$$a_j=min(a_i,a_j)$$$, we can say that $$$a_j \le lim$$$.

Substituting $$$a_j$$$ for $$$j$$$, this second sum can be written as:

Since we've already established that $$$j \le lim = \sqrt{2 \cdot n}$$$, calculating this sum takes $$$O(n\sqrt{n})$$$ time.

Be especially careful when calculating these sums, as $$$a_i \cdot a_i - b_i$$$ and $$$a_i \cdot j-b_i$$$ can end up being either negative or greater than $$$n$$$.

Time complexity per testcase: $$$O(n \sqrt n)$$$

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const ll NMAX=2e5+5,SQRMAX=635,MOD=1e9+7;

int fr[SQRMAX][NMAX];
ll a[NMAX],b[NMAX];
void tc(){
    ll n,ans=0;
    cin>>n;
    ll lim=sqrt(n*2);
    for(ll i=0;i<n;i++){
        cin>>a[i];
    }
    for(ll i=0;i<n;i++){
        cin>>b[i];
        if(a[i]<=lim)
            fr[a[i]][b[i]]++;
    }
    for(ll i=0;i<n;i++){
        if(a[i]<=lim){
            if(a[i]*a[i]-b[i]>=1 && a[i]*a[i]-b[i]<=n)
                ans+=fr[a[i]][a[i]*a[i]-b[i]];
        }
    }
    for(ll i=2;i<=lim;i+=2)
        ans-=fr[i][i*i/2];
    ans/=2;

    for(ll i=0;i<n;i++){
        for(ll j=1;j<=lim && j<a[i] && j*a[i]<=2*n;j++){
            if(j*a[i]-b[i]>=1 && j*a[i]-b[i]<=n)
                ans+=fr[j][j*a[i]-b[i]];
        }
    }
    for(ll i=0;i<n;i++){
        if(a[i]<=lim)
            fr[a[i]][b[i]]=0;
    }
    cout<<ans<<'\n';
}
int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll t;
    cin>>t;
    while(t--)
        tc();
    return 0;
}

for _ in range(int(input())):
    n = int(input())
    
    ta = list(map(int, input().split()))
    tb = list(map(int, input().split()))
    
    a = [(x, y) for x, y in zip(ta, tb)]
    a.sort()
    
    cnt = [0] * (2 * n + 1)
    pr = 0
    ans = 0
    
    for i in range(n):
        if pr != a[i][0]:
            pr = a[i][0]
            
            if pr * pr > 2 * n:
                break
                
            cnt = [0] * (2 * n + 1)
            for j in range(i + 1, n):
                t = a[i][0] * a[j][0] - a[j][1]
                if t >= 0 and t <= 2 * n:
                    cnt[t] += 1
        
        ans += cnt[a[i][1]]
        
        if i + 1 < n:
            t = a[i][0] * a[i + 1][0] - a[i + 1][1]
            if t >= 0 and t <= 2 * n:
                cnt[t] -= 1
    
    print(ans)

Good problem

Ok problem

Bad problem

Didn't solve it, idk

First and foremost, the number of regular bracket strings of length $$$2 \cdot n$$$ is equal to $$$C_n=\frac{1}{n+1}\binom{2n}{n}$$$.

Secondly, for a bracket string $$$\overline{s_1s_2 \ldots s_k}$$$, let:

A bracket string $$$\overline{s_1s_2 \ldots s_k}$$$ is a regular bracket string if both of the following statements are true:

1. $$$\Delta_k=0$$$

2. $$$\Delta_i \ge 0, i=\overline{1,k}$$$

From now on we'll call a set of indices $$$i_1 < i_2 < \ldots < i_k$$$ a group if $$$\overline{s_{i_1}s_{i_2}\ldots s_{i_k}}$$$ must be an RBS.

There are two main cases to consider, both of which can be proven with the aforementioned conditions for a string to be an RBS:

By taking both of these cases into account, we can conclude that all indices $$$i_k$$$ covered by the same subset of intervals are part of the same group.

Finding the subset of intervals which cover a certain index $$$i$$$ can be implemented using difference arrays and xor hashing.

Time complexity: $$$O(maxn \cdot log(mod))$$$ for precomputing every $$$C_n$$$, and $$$O(k\cdot log(k))$$$ per test case.

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int NMAX=3e5+5, MOD=998244353;

mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<ll> rnd(0,LLONG_MAX);

ll fact[NMAX], invfact[NMAX], C[NMAX];

ll binPow(ll x, ll y){
    ll ans=1;
    for(;y ;y>>=1, x = x*x%MOD)
        if(y&1)
            ans = ans*x%MOD;
    return ans;
}
map<ll,ll> diff,freq;
void add_interval(ll l, ll r){
    ll Hash = rnd(gen);
    diff[l] ^= Hash, diff[r+1] ^= Hash;
}
void tc(){
    ll n,k;
    cin>>n>>k;
    diff.clear(), freq.clear();
    add_interval(1,n); /// the initial string must be an RBS

    for(ll i=0;i<k;i++){
        ll l,r;
        cin>>l>>r;
        add_interval(l,r);
    }
    ll Hash = diff[1];

    for(map<ll,ll> :: iterator it=next(diff.begin()); it!=diff.end(); it++){
        freq[Hash] += it->first - prev(it)->first;
        Hash ^= it->second;
    }

    ll ans=1;

    for(const auto& it : freq)
        ans=ans*C[it.second]%MOD;

    cout<<ans<<'\n';
}

int main()
{
    fact[0] = invfact[0] = 1;
    for(ll i=1; i<NMAX; i++){
        fact[i] = fact[i-1] * i % MOD;
        invfact[i] = binPow(fact[i], MOD - 2);
    }

    for(ll i=0; i*2<NMAX; i++) C[i*2] = fact[i*2] * invfact[i] % MOD * invfact[i+1] % MOD;

    ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll t;
    cin>>t;
    while(t--)
        tc();
    return 0;
}

Good problem

Ok problem

Bad problem

Didn't solve it, idk

Disclaimer: I ( tibinyte ) wanted to cut $$$O(n \sqrt{n})$$$ memory because I thought that setting 1024 MB memory limit would spoil the solution. So only blame me for this.

If we analyze how good a bipartite coloring of the given tree is we can note that it achieves a total cost of $$$n \cdot (n+1) - O(n)$$$.

Now suppose the tree has a connected component of size $$$k$$$. We can note that this decreses the maximum cost by $$$O(k^2)$$$.

By the $$$2$$$ claims above, we can note that in an optimal coloring, the maximum size of a connected component is $$$O(\sqrt{n})$$$.

Now let $$$dp_{i,j,color}$$$ be the minimum loss if we color the subtree of $$$i$$$ and the connected component of vertex $$$i$$$ has size $$$j$$$. We can calculate this in a knapsack style by adding subtrees successively. The computation takes $$$O(n \cdot \sqrt{n})$$$ if we use the $$$7$$$-th trick from this blog.

Now we are only left to optimize memory, since now it's $$$O(n \cdot \sqrt{n})$$$. We can directly apply this to get linear memory.

#include <bits/stdc++.h>
#define int long long

using namespace std;

const int lim = 893;
const int inf = 1e16;

struct dp_state
{
    vector<int> a;
    vector<int> b;
    void init()
    {
        a.resize(lim + 1, inf);
        b.resize(lim + 1, inf);
        a[1] = 1;
        b[1] = 2;
    }
};

int32_t main()
{
    cin.tie(nullptr)->sync_with_stdio(false);
    int q;
    cin >> q;
    while (q--)
    {
        int n;
        cin >> n;
        vector<vector<int>> g(n + 1);
        vector<int> sz(n + 1);

        for (int i = 1; i < n; ++i)
        {
            int x, y;
            cin >> x >> y;
            g[x].push_back(y);
            g[y].push_back(x);
        }

        function<int(int, int)> shuffle_kids = [&](int node, int parent)
        {
            int sz = 1;
            pair<int, int> best;
            for (int i = 0; i < (int)g[node].size(); ++i)
            {
                if (g[node][i] != parent)
                {
                    int sz2 = shuffle_kids(g[node][i], node);

                    best = max(best, {sz2, i});

                    sz += sz2;
                }
            }
            if (!g[node].empty())
            {
                swap(g[node][0], g[node][best.second]);
            }
            return sz;
        };
        shuffle_kids(1, 0);

        vector<vector<int>> merged(2 * lim + 1, vector<int>(2, inf));

        function<dp_state(int, int)> dfs = [&](int node, int parent)
        {
            dp_state dp;
            sz[node] = 1;
            bool hasinit = false;
            for (auto i : g[node])
            {
                if (i != parent)
                {
                    dp_state qui = dfs(i, node);
                    if (!hasinit)
                    {
                        dp.init();
                        hasinit = true;
                    }
                    for (int j = 0; j <= min(sz[node], lim) + min(sz[i], lim); ++j)
                    {
                        merged[j][0] = merged[j][1] = inf;
                    }
                    for (int k = 1; k <= min(sz[node], lim); ++k)
                    {
                        for (int l = 1; l <= min(sz[i], lim); ++l)
                        {
                            merged[k][0] = min(merged[k][0], dp.a[k] + qui.b[l]);
                            merged[k][1] = min(merged[k][1], dp.b[k] + qui.a[l]);
                            merged[k + l][0] = min(merged[k + l][0], dp.a[k] + qui.a[l] + k * l);
                            merged[k + l][1] = min(merged[k + l][1], dp.b[k] + qui.b[l] + k * l * 2);
                        }
                    }
                    sz[node] += sz[i];
                    for (int k = 1; k <= min(sz[node], lim); ++k)
                    {
                        dp.a[k] = merged[k][0];
                        dp.b[k] = merged[k][1];
                    }
                }
            }
            if (!hasinit)
            {
                dp.init();
                hasinit = true;
            }
            return dp;
        };
        dp_state dp = dfs(1, 0);
        int ans = inf;
        for (int i = 1; i <= lim; ++i)
        {
            ans = min(ans, dp.a[i]);
            ans = min(ans, dp.b[i]);
        }
        cout << n * (n + 1) - ans << '\n';
    }
}

Good problem

Ok problem

Bad problem

Didn't solve it, idk

First of all, we notice that if some element moves left, it will never move right.

Obviously, if an element moves right, it will never move left since it will continue to move right until it reaches its final position.

By the $$$2$$$ claims above, we conclude that left and right movers are distinct.

Now suppose we swap indicies $$$i$$$ and $$$j$$$ and note that such swap kills $$$2 \cdot (j-i) - 1$$$ inversions and the left mover $$$j$$$ moves $$$j-i$$$ steps.

Now the magic is that if we let $$$s$$$ be the sum over $$$2 \cdot (i-p_i)$$$ for all left movers we have $$$s - ans = inversions$$$, thus our answer is just $$$s-inversions$$$.

Now to handle the data structure part, we just need to be able to calculate inversions while being able to perform point updates. There are many ways to do this, for example for using a fenwick tree and a bitwise trie/ordered_set in $$$O(nlog^2n)$$$.

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace std;

using namespace __gnu_pbds;

#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>

typedef long long ll;

/*               .___                 __                 __           .__
  ____  ____   __| _/____     _______/  |______ ________/  |_  ______ |  |__   ___________   ____
_/ ___\/  _ \ / __ |/ __ \   /  ___/\   __\__  \\_  __ \   __\/  ___/ |  |  \_/ __ \_  __ \_/ __ \
\  \__(  <_> ) /_/ \  ___/   \___ \  |  |  / __ \|  | \/|  |  \___ \  |   Y  \  ___/|  | \/\  ___/
 \___  >____/\____ |\___  > /____  > |__| (____  /__|   |__| /____  > |___|  /\___  >__|    \___  >
     \/           \/    \/       \/            \/                 \/       \/     \/            \/
*/

struct bit
{
  int n;
  vector<ordered_set> lesgo;
  void resize(int _n)
  {
    n = _n;
    lesgo = vector<ordered_set>(n + 1);
  }
  void update(int pos, int val, int sign)
  {
    for (int i = pos; i <= n; i += i & (-i))
    {
      if (sign == -1)
      {
        lesgo[i].erase(val);
      }
      if (sign == 1)
      {
        lesgo[i].insert(val);
      }
    }
  }
  int query_smaller(int pos, int val)
  {
    int ans = 0;
    for (int i = pos; i; i -= i & (-i))
    {
      ans += lesgo[i].order_of_key(val);
    }
    return ans;
  }
  int query_greater(int pos, int val)
  {

    int ans = 0;
    for (int i = pos; i; i -= i & (-i))
    {
      int sz = lesgo[i].size();
      ans += sz - lesgo[i].order_of_key(val + 1);
    }
    return ans;
  }
  int query_smaller(int st, int dr, int val)
  {
    return query_smaller(dr, val) - query_smaller(st - 1, val);
  }
  int query_greater(int st, int dr, int val)
  {
    return query_greater(dr, val) - query_greater(st - 1, val);
  }
};

int32_t main()
{
  cin.tie(nullptr)->sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; ++i)
  {
    cin >> a[i];
  }
  bit lesgo;
  lesgo.resize(n);
  for (int i = 1; i <= n; ++i)
  {
    lesgo.update(i, a[i], 1);
  }

  ll inv = 0;

  for (int i = 1; i <= n; ++i)
  {
    inv += lesgo.query_greater(1, i, a[i]);
  }

  ll sum = 0;

  for (int i = 1; i <= n; ++i)
  {
    sum += max(a[i] - i, 0);
  }

  while (q--)
  {
    int x, y;
    cin >> x >> y;
    if (x != y)
    {
      lesgo.update(x, a[x], -1);
      lesgo.update(y, a[y], -1);
      sum -= max(a[x] - x, 0);
      sum -= max(a[y] - y, 0);
      if (x + 1 <= y - 1)
      {
        int lesgo1 = lesgo.query_smaller(x + 1, y - 1, a[x]);
        int lesgo2 = lesgo.query_greater(x + 1, y - 1, a[y]);
        inv += 2 * (y - x - 1 - lesgo1 - lesgo2);
      }

      if (a[x] > a[y])
      {
        inv--;
      }
      else
      {
        inv++;
      }

      swap(a[x], a[y]);
      sum += max(a[x] - x, 0);
      sum += max(a[y] - y, 0);
      lesgo.update(x, a[x], 1);
      lesgo.update(y, a[y], 1);
    }
    cout << sum + sum - inv << '\n';
  }
}

#include <bits/stdc++.h>

using namespace std;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int random(int st, int dr)
{
  uniform_int_distribution<int> dist(st, dr);
  return dist(rng);
}

typedef long long ll;

/*               .___                 __                 __           .__
  ____  ____   __| _/____     _______/  |______ ________/  |_  ______ |  |__   ___________   ____
_/ ___\/  _ \ / __ |/ __ \   /  ___/\   __\__  \\_  __ \   __\/  ___/ |  |  \_/ __ \_  __ \_/ __ \
\  \__(  <_> ) /_/ \  ___/   \___ \  |  |  / __ \|  | \/|  |  \___ \  |   Y  \  ___/|  | \/\  ___/
 \___  >____/\____ |\___  > /____  > |__| (____  /__|   |__| /____  > |___|  /\___  >__|    \___  >
     \/           \/    \/       \/            \/                 \/       \/     \/            \/
*/

struct decomp
{
  vector<int> a;
  vector<pair<int, int>> buckets;
  vector<int> sp;
  vector<int> spbucket;
  vector<int> wh;
  const int sz = 400;
  void resize(int n)
  {
    a = vector<int>(n + 1);
    sp = vector<int>(n + 1);
    wh = vector<int>(n + 1);
    for (int i = 1; i <= n; i += sz)
    {
      int st = i;
      int dr = min(n, st + sz - 1);
      for (int j = st; j <= dr; ++j)
      {
        wh[j] = (int)buckets.size();
      }
      buckets.push_back({st, dr});
      spbucket.push_back(0);
    }
  }
  void update(int pos, int x)
  {
    a[pos] += x;
    int qui = wh[pos];
    sp[buckets[qui].first] = a[buckets[qui].first];
    for (int i = buckets[qui].first + 1; i <= buckets[qui].second; ++i)
    {
      sp[i] = sp[i - 1] + a[i];
    }
    spbucket[0] = sp[buckets[0].second];
    for (int i = 1; i < (int)buckets.size(); ++i)
    {
      spbucket[i] = spbucket[i - 1] + sp[buckets[i].second];
    }
  }
  int query(int pos)
  {
    int qui = wh[pos];
    int ans = sp[pos];
    if (qui)
    {
      ans += spbucket[qui - 1];
    }
    return ans;
  }
};

struct geralt
{
  vector<int> a;
  vector<decomp> lesgo;
  vector<pair<int, int>> buckets;
  vector<int> wh;
  const int sz = 3000;
  void resize(int n)
  {
    a = vector<int>(n + 1);
    wh = vector<int>(n + 1);
    for (int i = 1; i <= n; i += sz)
    {
      int st = i;
      int dr = min(n, st + sz - 1);
      for (int j = st; j <= dr; ++j)
      {
        wh[j] = (int)buckets.size();
      }
      buckets.push_back({st, dr});
      lesgo.push_back({});
      lesgo.back().resize(n);
    }
  }
  void update(int pos, int val)
  {
    int qui = wh[pos];
    if (a[pos] != 0)
    {
      lesgo[qui].update(a[pos], -1);
    }
    a[pos] = val;
    lesgo[qui].update(a[pos], 1);
  }
  int query_brute(int st, int dr, int val)
  {
    int ans = 0;
    for (int i = st; i <= dr; ++i)
    {
      if (a[i] <= val)
      {
        ans++;
      }
    }
    return ans;
  }
  int query_smaller(int st, int dr, int val) // number of values <= val in st...dr
  {
    if (wh[st] == wh[dr])
    {
      return query_brute(st, dr, val);
    }
    int qui1 = wh[st], qui2 = wh[dr];
    int ans = query_brute(st, buckets[qui1].second, val) + query_brute(buckets[qui2].first, dr, val);
    qui1++;
    qui2--;
    for (int i = qui1; i <= qui2; ++i)
    {
      ans += lesgo[i].query(val);
    }
    return ans;
  }
  int query_greater(int st, int dr, int val) // number of values > val in st...dr
  {
    return (dr - st + 1) - query_smaller(st, dr, val);
  }
};

int32_t main()
{
  cin.tie(nullptr)->sync_with_stdio(false);
  int n, q;
  cin >> n >> q;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; ++i)
  {
    cin >> a[i];
  }

  geralt lesgo;
  lesgo.resize(n);

  ll inv = 0;

  for (int i = 1; i <= n; ++i)
  {
    lesgo.update(i, a[i]);
    inv += lesgo.query_greater(1, i, a[i]);
  }

  ll sum = 0;

  for (int i = 1; i <= n; ++i)
  {
    sum += max(a[i] - i, 0);
  }

  while (q--)
  {
    int x, y;
    cin >> x >> y;
    if (x != y)
    {
      sum -= max(a[x] - x, 0);
      sum -= max(a[y] - y, 0);
      if (x + 1 <= y - 1)
      {
        int lesgo1 = lesgo.query_smaller(x + 1, y - 1, a[x]);
        int lesgo2 = lesgo.query_greater(x + 1, y - 1, a[y]);
        inv += 2 * (y - x - 1 - lesgo1 - lesgo2);
      }

      if (a[x] > a[y])
      {
        inv--;
      }
      else
      {
        inv++;
      }

      swap(a[x], a[y]);
      sum += max(a[x] - x, 0);
      sum += max(a[y] - y, 0);
      lesgo.update(x, a[x]);
      lesgo.update(y, a[y]);
    }
    cout << sum + sum - inv << '\n';
  }
}

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

#define ll long long
#define ii pair<int,int>
#define i4 tuple<int,int,int,int>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << ": " << x << endl

#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound

#define rep(x,start,end) for(int x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--))
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());

struct FEN{ //we want to count the reverse
	int arr[500005];
	
	void update(int i,int k){
		i=500005-i;
		
		while (i<500005){
			arr[i]+=k;
			i+=i&-i;
		}
	}
	
	int query(int i){
		i=500005-i;
		
		int res=0;
		while (i){
			res+=arr[i];
			i-=i&-i;
		}
		return res;
	}
} fen;

int n,q;
int arr[500005];
long long ans[50005];

void dnc(int l,int r,vector<i4> upd,vector<i4> que){
	vector<i4> updl,updr;
	vector<i4> quel,quer;
	
	int m=l+r>>1;
	
	for (auto [a,b,c,d]:upd){
		if (c<=m) updl.pub({a,b,c,d});
		else updr.pub({a,b,c,d});
	}
	
	for (auto [a,b,c,d]:que){
		if (c<=m) quel.pub({a,b,c,d});
		else quer.pub({a,b,c,d});
	}
	
	int i=0;
	for (auto it:quer){
		while (i<sz(updl) && get<0>(updl[i])<get<0>(it)){
			fen.update(get<1>(updl[i]),get<3>(updl[i]));
			i++;
		}
		
		ans[get<2>(it)]+=fen.query(get<1>(it))*get<3>(it);
	}
	while (i){
		i--;
		fen.update(get<1>(updl[i]),-get<3>(updl[i]));
	}
	
	if (l!=m) dnc(l,m,updl,quel);
	if (m+1!=r) dnc(m+1,r,updr,quer);
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin.exceptions(ios::badbit | ios::failbit);
	
	cin>>n>>q;
	rep(x,1,n+1) cin>>arr[x];
	
	vector<i4> upd,que;
	
	rep(x,1,n+1) upd.pub({x,arr[x],-1,1});
	rep(x,1,n+1) que.pub({x,arr[x],0,-1});
	rep(x,1,n+1) ans[0]+=abs(x-arr[x]);
	
	rep(x,1,q+1){
		int a,b;
		cin>>a>>b;
		
		if (a==b) continue;
		if (arr[a]<arr[b]) ans[x]--;
		else ans[x]++;
		
		upd.pub({a,arr[a],x-1,-1});
		upd.pub({b,arr[b],x-1,-1});
		ans[x]-=abs(a-arr[a])+abs(b-arr[b]);
		
		swap(arr[a],arr[b]);
		
		upd.pub({a,arr[a],x,1});
		upd.pub({b,arr[b],x,1});
		ans[x]+=abs(a-arr[a])+abs(b-arr[b]);
		
		que.pub({b,arr[b],x,-2});
		que.pub({b,arr[a],x,2});
		que.pub({a,arr[b],x,2});
		que.pub({a,arr[a],x,-2});
		
	}
	
	sort(all(upd)),sort(all(que));
	
	dnc(-1,q,upd,que);
	
	rep(x,1,q+1) ans[x]+=ans[x-1];
	rep(x,1,q+1) cout<<ans[x]<<" "; cout<<endl;
}

Good problem

Ok problem

Bad problem

Didn't solve it, idk

Let $$$dp_i$$$ be the maximum sum of costs of activated points that are $$$< i$$$ over all states that has point $$$i$$$ activated.

When we transition from $$$dp_i$$$ to $$$dp_j$$$, we need to add the cost contributed by point $$$i$$$. The number of ranges where point $$$i$$$ is the largest coordinate within it are the ranges $$$[l,r]$$$ which satisfy $$$l \leq i \leq r < j$$$. So we have $$$dp_j = \max\limits_{i < j}(dp_i + p_i \cdot S_{i,j})$$$ where $$$S_{i,j}$$$ is the number of ranges $$$[l,r]$$$ which satisfy $$$l \leq i \leq r < j$$$. Now, we note that $$$S_{i,j} z$$$

With some work, we can get calculate this $$$dp$$$ in $$$O(n^2)$$$. But this will be too slow, let's try to speed this up.

Let us define $$$dp_{i,j} = \max\limits_{k \leq i}(dp_k + p_k \cdot S_{k,j})$$$. We have $$$dp_{j-1,j}=dp_j$$$. Our goal is to go from (implictly) storing $$$dp_{i,i+1},dp_{i,i+2},\ldots,dp_{i,n}$$$ to $$$dp_{i+1,i+2},dp_{i+1,i+3},\ldots,dp_{i+1,n}$$$. As $$$dp_i + p_i \cdot S_{i,j}$$$ looks like a linear function w.r.t. $$$S_{i,j}$$$, we can try to use convex hull data structures to maintain it.

Let's figure out how $$$S_{i+1,j}$$$ relates to $$$S_{i,j}$$$. We need to subtract the number of ranges with $$$r=i$$$ and add the number of ranges with $$$l=i+1$$$ and $$$r < j$$$. This corresponds to $$$S_{i,*}$$$ and $$$S_{i+1,*}$$$ differing by amortized $$$O(1)$$$ suffix increment updates. Also, note that $$$S_{i,*}$$$ is non-decreasing.

So we want to support the following data structure: Initially we have an arrays $$$A$$$ and $$$X$$$ that are both initially $$$0$$$. Handle the following updates:

• 1 m c $$$A_i \gets \max(A_i,m \cdot X_i + c)$$$ for all $$$i$$$
• 2 j k $$$X_i \gets X_i + k$$$ for all $$$j \leq i$$$. It is guaranteed that $$$X$$$ will always be non-decreasing.
• 3 i find $$$A_i$$$

This can be maintained in a lichao tree in $$$O(log ^2)$$$ time. In each lichao node, we need to store $$$s,m,e$$$, the start, middle and end indices and their corresponding $$$X$$$ values $$$X_s,X_m,X_e$$$ respectively. This way, we can support operations $$$1$$$ and $$$3$$$ already.

To support operation $$$2$$$, note that in a lichao tree, you can use $$$O(log^2)$$$ time to push down all lines that covers a certain point (refer to https://codeforces.com/blog/entry/86731). This way, all lines in the li chao tree are in $$$[1,j)$$$ or $$$[j,n]$$$, so you can do lazy updates on both the coordinates of the lines and the $$$X$$$ values.

The time complexity is $$$O(n \log^2)$$$.

There is another solution that works in $$$O(m \log^3)$$$ that we are unable to optimize further yet. Let us flip the array so that the condition is on the activated point with the smallest coordinate. Then we have $$$dp_j = \max\limits_{i < j}(dp_i + p_j \cdot S_{i,j})$$$ where $$$S_{i,j}$$$ counts the number of ranges $$$[l,r]$$$ such that $$$i< l \leq j \leq r$$$.

Now, we want to store the linear function $$$dp_i + x \cdot S_{i,j}$$$ in some sort of data structure so that we can evaluate the maximum value with $$$x=p_j$$$. Unfortunately, the value of $$$S_{i,j}$$$ can change by suffix additions, similar to above.

But since $$$S_{i,j}$$$ is non-increasing here, that means the optimal $$$i$$$ that maximizes $$$dp_i + x \cdot S_{i,j}$$$ decreases when $$$x$$$ increases. That is, for $$$l_1 \leq r_1< l_2 \leq r_2$$$ and we have two data structures that can get the maximum $$$f_j(x)=dp_i+x \cdot S_{i,j}$$$ for $$$l_j \leq i \leq r_j$$$ respectively. We can combine these $$$2$$$ data structures to make it for $$$[l_1,r_1] \cup [l_2,r_2]$$$ by binary searching the point $$$X$$$ where for all $$$x \leq X$$$, $$$f_1(x) \leq f_2(x)$$$ and for all $$$X \leq x,f_1(x) \geq f_2(x)$$$. Since querying this data structure takes $$$O(\log n)$$$, it takes $$$O(\log ^2)$$$ time to combine two such data structures. If we use a segment tree, we only need to rebuild $$$O(\log^3)$$$ different data structures (the rest can be handled using lazy tags), giving us a time complexity of $$$O(\log^3)$$$.

//もう布団の中から出たくない
//布団の外は寒すぎるから
//布団の中から出たくない
//布団の中はあたたかすぎるから

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

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

#define int long long
#define ll long long
#define ii pair<ll,ll>
#define iii pair<ii,ll>
#define fi first
#define se second
#define endl '\n'
#define debug(x) cout << #x << ": " << x << endl

#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define lb lower_bound
#define ub upper_bound

#define rep(x,start,end) for(int x=(start)-((start)>(end));x!=(end)-((start)>(end));((start)<(end)?x++:x--))
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());

const ll INF=1e18;

struct line{
	ll m,c;
	
	line (ll _m,ll _c){
		m=_m,c=_c;
	}
	
	ll get(ll x){
		return m*x+c;
	}
};

struct node{
	int s,e,m;
	int vs=0,ve=0,vm=0;
	line val={0,-INF};
	int lazy=0;
	node *l,*r;
	
	node(int _s,int _e){
		s=_s,e=_e,m=s+e>>1;
		
		if (s!=e){
			l=new node(s,m);
			r=new node(m+1,e);
		}
	}
	
	void propo(){
		if (lazy==0) return;
		
		val.c-=val.m*lazy;
		vs+=lazy;
		ve+=lazy;
		vm+=lazy;
		
		if (s!=e){
			l->lazy+=lazy;
			r->lazy+=lazy;
		}
		lazy=0;
	}
	
	void update(int i){
		if (s==i) lazy++;
		else{
			if (val.c!=-INF){
				l->propo(),r->propo();
				l->update(val),r->update(val);
				val={0,-INF};
			}
			
			if (m<i) r->update(i);
			else l->update(i),r->update(m+1);
			
			l->propo(),r->propo();
			vs=l->vs,vm=l->ve,ve=r->ve;
		}
	}
	
	void update(line i){
		bool lo=i.get(vs)>val.get(vs);
		bool mi=i.get(vm)>val.get(vm);
		bool hi=i.get(ve)>val.get(ve);
		
		if (mi) swap(i,val);
		
		if (s==e || i.c==-INF || lo==hi) return;
		
		l->propo(),r->propo();
		if (lo!=mi) l->update(i);
		else r->update(i);
	}
	
	ii query(ll i){
		propo();
		
        if (s==e) return {val.get(vs),vs};
		else if (i<=m){
			auto temp=l->query(i);
			return {max(temp.fi,val.get(temp.se)),temp.se};
		}
		else{
			auto temp=r->query(i);
			return {max(temp.fi,val.get(temp.se)),temp.se};
		}
	}
} *root;

const int maxn = 1e6;
const int maxm = 1e6;

int n,m;
ii arr[maxn+1];
ii brr[maxm+1];

int ans[maxm+1];

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	cin.exceptions(ios::badbit | ios::failbit);
	
	int TC;
	cin>>TC;
	while (TC--){
		cin>>n>>m;
		rep(x,1,n+1) cin>>arr[x].fi>>arr[x].se;
		rep(x,1,m+1){
            cin>>brr[x].se;
            brr[x].fi = x;
        }
    
		sort(arr+1,arr+n+1);
		
		root=new node(0,m+1);
		root->update({0,0});
		
		int curr=1;
		rep(x,0,m+2){
			while (curr<=n && arr[curr].fi==x) root->update(arr[curr++].se+1);
			auto temp=root->query(x);
			root->update({brr[x].se,temp.fi-brr[x].se*temp.se});
		}
		
		cout<<root->query(m+1).fi<<endl;
	}
}

#include <bits/stdc++.h>
#define all(x) (x).begin(),(x).end()
using namespace std;

using ll = long long;
using ld = long double;

//#define int ll
#define sz(x) ((int)(x).size())

using pii = pair<ll,ll>;
using tii = tuple<int,int,int>;

const int nmax = 5e5 + 5, qmax = 2e6 + 5;
const ll inf = 1e18 + 5; 

struct line {
  ll m = 0, b = -inf;
  ll operator()(const ll& x) const {
     return m * x + b;
  }
  bool operator ==(const line& x) const { return m == x.m && b == x.b; }
};

int M;

namespace AINT {
  struct Node {
    int vl = 0, vmid = 0, vr = 0;
    line val = line{0, -inf};
    Node() {;}
    void operator += (const int &x) { vl += x, vmid += x, vr += x; val.b -= val.m * x; }
  } aint[qmax * 4];
  int lazy[qmax * 4];
  
  void push(int node) {
    lazy[2 * node] += lazy[node];
    aint[2 * node] += lazy[node];
    lazy[2 * node + 1] += lazy[node];
    aint[2 * node + 1] += lazy[node];
    lazy[node] = 0;
  }
  void pushline(line x, int node = 1, int cl = 0, int cr = M) {
    bool l = x(aint[node].vl) > aint[node].val(aint[node].vl), m = x(aint[node].vmid) > aint[node].val(aint[node].vmid), r = x(aint[node].vr) > aint[node].val(aint[node].vr);
    if(m) swap(x, aint[node].val);
    if(cl == cr || x == line() || l == r) return;
    
    int mid = cl + cr >> 1;
    push(node);
    if(m != l) pushline(x, 2 * node, cl, mid);
    else pushline(x, 2 * node + 1, mid + 1, cr);
    return;
  }
  void add(int l, int node = 1, int cl = 0, int cr = M) {
    if(cl >= l) {
      lazy[node]++;
      aint[node] += 1;
      return;
    }
    if(cr < l) return;
    int mid = cl + cr >> 1;
    
    push(node);
    
    pushline(aint[node].val, 2 * node, cl, mid);
    pushline(aint[node].val, 2 * node + 1, mid + 1, cr);
    aint[node].val = line{0, -inf};
    
    add(l, 2 * node, cl, mid);
    add(l, 2 * node + 1, mid + 1, cr);
    aint[node].vl = aint[2 * node].vl;
    aint[node].vmid = aint[2 * node].vr;
    aint[node].vr = aint[2 * node + 1].vr;
  }
  pii query(ll p, int node = 1, int cl = 0, int cr = M) {
    if(cl == cr) return pii{aint[node].val(aint[node].vl), aint[node].vl};
    int mid = cl + cr >> 1;
    push(node);
    auto [mxv, atrv] = p <= mid? query(p, 2 * node, cl, mid) : query(p, 2 * node + 1, mid + 1, cr);
    return pii{max(mxv, aint[node].val(atrv)), atrv};
  } 
  
  void init(int node = 1, int cl = 0, int cr = M) {
    aint[node] = Node();
    lazy[node] = 0;
    if(cl == cr) return;
    int mid = cl + cr >> 1;
    init(2 * node, cl, mid);
    init(2 * node + 1, mid + 1, cr);
    return;
  }
};

vector<int> atraddp[qmax];
ll val[qmax];

void testcase() {
  int n, m;
  cin >> n >> m;
  M = m + 2;
  for(int i = 0, l, r; i < n; i++) {
    cin >> l >> r;
    atraddp[l].emplace_back(r + 1);
  }
  for(int i = 1; i <= m; i++)
    cin >> val[i];
  
  AINT::init();
  AINT::pushline({0, 0});
  
  for(int i = 0; i < m + 2; i++) {
    for(auto x : atraddp[i])
      AINT::add(x);
    auto [a, b] = AINT::query(i);
    //cerr << a << ' ' << b << '\n';
    AINT::pushline({val[i], a - val[i] * b});
  }
  
  for(int i = 0; i < m + 2; i++)
    atraddp[i].clear();
  
  cout << AINT::query(m + 1).first << '\n';
}

signed main() {
  cin.tie(0) -> sync_with_stdio(0);
  int t;
  cin >> t;
  while(t--) testcase();
  
}

/**
      Anul asta se da centroid.
-- Surse oficiale
*/

Good problem

Ok problem

Bad problem

Didn't solve it, idk

He didn't solve E because he realized it was NP-Hard and left the contest to spend his time on other problems. It turns out the tests weren't strong enough so some people upsolved it after the contest but he proved the problem is NP-Hard at the lesson after the contest.