(Analysis by Spencer Compton)

We begin by considering the structure of optimal solutions for a particular query $q$ where you must end at mana pool $e_q$ after $s_q$ seconds. Let us characterize a journey by an ordered list $(p_1, t_2), (p_2, t_2), \dots, (p_k, t_k)$, where $(p_i, t_i)$ represents Bessie last visiting pool $p_i$ at time $t_i$ and collecting all of its available mana. By definition, a valid journey for query $q$ must end at $e_q$ (meaning $p_k=e_q$) and end after $s_q$ seconds (meaning $t_k = s_q$). The mana collected in such a proposed journey is then $\sum_{i=1}^k m_{p_i} \cdot t_{i}$. Of course, some journeys we describe are infeasible. More formally, we will consider a journey feasible if and only if $t_1 \ge 0$ and $t_{i+1} \le t_i + D[p_i][p_{i+1}]$ for all $i \in \{1,\dots,k-1\}$, and where $D[a][b]$ represents the shortest path from pool $a$ to pool $b$ (note that $D$ can be pre-computed with the Floyd-Warshall algorithm in $O(N^3)$ time). With this in mind, what is the optimal feasible and valid journey for any fixed $p_1, \dots, p_k$? It can be shown with an inductive proof or exchange argument that it is optimal to wait as long as possible. More concretely, this means the optimal times $t^*$ are:

• $t_k^* = s_q$
• $t_{k-1}^* = s_q - D[p_{k-1}][p_{k}]$
• $t_{k-2}^* = s_q - D[p_{k-1}][p_{k}] - D[p_{k-2}][p_{k-1}]$
• $\dots$
• $t_1^* = s_q- \sum_{i=1}^{k-1} D[p_{i}][p_{i+1}]$

Note how if $t_1^* < 0$ then no journey with pools $p_1,\dots,p_k$ is feasible. For the first subtask, it is sufficient to find the optimal solution by exhaustively enumerating over all $O(N!)$ possible choices of $p_1,\dots,p_k$ and calculate the corresponding $t_1^*,\dots,t_k^*$ for each query in $O(N! \cdot (N+Q))$ total time.

For the second subtask, we can no longer afford to iterate over each query for each possible journey. We look for some structure of similarity that eliminates the need for this iteration. For simplicity of presentation, we will only consider journeys and queries ending at some fixed pool $e$, as we can handle other journey/queries separately. Based on our closed-form for $t^*$, the journey will be feasible and valid for some query if and only if $\sum_{i=1}^{k-1}D[p_{i}][p_{i+1}] \le s_q$. Moreover, the total mana gained will be equal to $\sum_{i=1}^k m_{p_i} \cdot t^*_{i} = \sum_{i=1}^{k} m_{p_{i}} \cdot \left( s_q- \sum_{j=i}^{k-1} D[p_{j}][p_{j+1}] \right) = \left( \sum_{i=1}^{k} m_{p_{i}} \right) \cdot s_q - \left( \sum_{i=1}^{k} \left(m_{p_{i}} \cdot \sum_{j=i}^{k-1} D[p_{j}][p_{j+1}]\right) \right)$.

We see how this expression takes the form of a function linear in $s_q$. In particular, we can determine that a journey $J$ is feasible if and only if $s_q \ge T_J$, and then journey $J$ gains mana $f_J(s_q) \triangleq slope_J \cdot s_q + add_J$ where:

• $T_J = \sum_{i=1}^{k-1}D[p_{i}][p_{i+1}]$
• $slope_J = \sum_{i=1}^{k} m_{p_{i}}$
• $add_J = \sum_{i=1}^{k} \left(m_{p_{i}} \cdot \sum_{j=i}^{k-1} D[p_{j}][p_{j+1}]\right)$

This immediately implies that the answer for a query $q$ is the maximum over a collection of linear functions $\max\limits_{J \textrm{ s.t. } T_J \le s_q} f_J(s_q)$. If not for the condition that $T_J \le s_q$, queries regarding the maximum of linear functions at a particular point can be answered efficiently with a data structure using Convex Hull Trick (CHT) or LineContainer. To handle the condition that $T_J \le s_q$, we can sort queries by $s_q$ and journeys by $T_J$, and incrementally add lines to our data structure such that when we answer query $q$, the journeys with $T_J \le s_q$ are exactly the set of journeys in our data structure. Accordingly, we can enumerate over all $O(N!)$ possible choices of $J = p_1,\dots,p_k$, calculate the corresponding $T_J$ and $f_J$, and then answer queries using our data structure in total time $O(N! \cdot N \cdot \log(N) + Q \log(Q))$ for the second subtask.

For the remaining subtasks, it is too slow to enumerate over all $O(N!)$ possibilities for $J = p_1, \dots, p_k$. We make two structural observations:

First, the condition that $T_J \le s_q$ is surprisingly unimportant, because $\max\limits_{J \textrm{ s.t. } T_J \le s_q} f_J(s_q) = \max\limits_{J} f_J(s_q)$. We will show this with a proof by contradiction. Suppose, for sake of contradiction, that some job $J'$ with $T_{J'} > s_q$ obtains the optimal amount of mana. Recall how $f_{J'}(s_q) = \sum_{i=1}^{k} m_{p_{i}} \cdot \left( s_q- \sum_{j=i}^{k-1} D[p_{j}][p_{j+1}] \right)$. If $T_{J'} > s_q$, then the summand corresponding to $p_1$ must be negative, and thus the journey corresponding to $p_2, \dots, p_k$ would be strictly better, and we have proven a contradiction. Accordingly, we can ignore the $T_J$ constraint and simply optimize for $\max\limits_{J} f_J(s_q)$.

Second, all journeys with the same unordered set of pools $p_1,\dots,p_k$ have the same slope (see the definition of $slope_J$ above). More concretely, let $\mathcal{P} \subseteq \{ 1, \dots, N \}$ denote some subset of pools to visit, and let $P_J$ denote the unordered set of pools that a journey $J$ visits. All journeys $J$ with $P_J = \mathcal{P}$ satisfy $slope_J = \sum_{p \in P} m_p$.

Our first observation shows how the only aspects of a line that matter are $slope_J$ and $add_J$ (not $T_J$). Our second observation notes that $slope_J$ is completely determined by the unordered set of pools $J$ visits. Combining our two observations, we conclude that we can optimize over subsets of pools to visit, and for each fixed subset $\mathcal{P}$ the only meaningful degree of freedom is $add_J$, so we can just consider the line with the best $add_J$ among those that visit $\mathcal{P}$. More formally: $\max\limits_{J \textrm{ s.t. } T_J \le s_q} f_J(s_q) = \max\limits_{J} f_J(s_q) = \max\limits_{J} slope_J \cdot s_q + add_J = \max\limits_{\mathcal{P}} \left( \sum_{p \in P} m_p \right) \cdot s_q + \left(\max\limits_{J \textrm{s.t. } P_J = \mathcal{P}} add_J \right)$.

This closed-form is also a function linear in $s_q$. We can then define our optimization as over linear functions for each subset as $g_{\mathcal{P}}(s_q) \triangleq slope_{\mathcal{P}} \cdot s_q + add_{\mathcal{P}}$. Where:

• $slope_{\mathcal{P}} = \sum_{p \in \mathcal{P}} m_{p}$
• $add_{\mathcal{P}} = \max\limits_{J \textrm{s.t. } P_J = \mathcal{P}} add_J$

Our hope is to optimize over $\max\limits_{\mathcal{P}} g_{\mathcal{P}}(s_q)$. All that remains is being able to compute each $add_{\mathcal{P}}$. We compute all $add_{\mathcal{P}}$ by a dynamic programming solution where the state $dp[now][rem]$ represents the set $rem$ of remaining nodes to visit (going backwards in the ordering), and the current location $now$. Recall how $add_J = \sum_{i=1}^{k} \left(m_{p_{i}} \cdot \sum_{j=i}^{k-1} D[p_{j}][p_{j+1}]\right)$. Accordingly, if we decide to transition to the next location $next \in rem$, then the summands in $add_J$ for each $p \in rem$ will increase by $m_{p} \cdot D[next][now]$. This formulation is sufficient to design a dynamic program with $O(2^N \cdot N)$ states, $O(N)$ transitions, and $O(2^N \cdot N^2)$ total running time.

Combining all steps, for each query we output $\max\limits_{\mathcal{P}} g_{\mathcal{P}}(s_q)$, where we compute the coefficients for the $O(2^N \cdot N)$ lines $g_{\mathcal{P}}$ by dynamic programming in $O(2^N \cdot N^2)$ time. We then use CHT to evaluate $\max\limits_{\mathcal{P}} g_{\mathcal{P}}(s_q)$ for each query. There is some flexibility in implementation. For example, one could sort the lines and queries and then use a vector/deque for a total runtime of $O(2^N \cdot N^2 + Q \log(Q))$ (as is done in the code below). Or, one could prepare the CHT data structure over the lines beforehand and then evaluate the queries without sorting them for a total runtime of $O(2^N \cdot N^2 + QN)$.

It is possible to get the third subtask by utilizing the above dynamic program but enumerating over all queries (instead of using CHT) for a $O(2^N \cdot (N^2 + Q))$ time algorithm. The fourth and fifth subtask of $N=16$ and $N=17$ exist to potentially give partial credit for slightly suboptimal solutions, or those needing constant factor optimization.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define f first
#define s second
vector<ll> rates;
ll dist[19][19];
ll inf = 1e9+1;
ll inf2 = 2e18;
int N, Q;
ll dp[18][1<<18];
vector<pair<ll, ll> > hull[18];
vector<ll> hull_pt;

ll go(int now, int rem){
// computing optimal add_J that ends at now, and visits everything remaining in rem
if(rem == 0){
return 0LL;
}
if(dp[now][rem] <= 0LL){
return dp[now][rem];
}
ll ret = -inf2;
for(int i = 0; i<N; i++){
if((rem&(1<<i))!=0){
ret = max(ret,go(i,rem-(1<<i)) - maskDeriv[rem] * dist[i][now]);
}
}
dp[now][rem] = ret;
return ret;
}

ll overcome(pair<ll, ll> a, pair<ll, ll> b){
// when does line "a" with slope a.f and add a.s, take over line "b" with slope b.f and add b.s
assert((a.f > b.f) && (a.s < b.s));
ll slopeDif = a.f - b.f;
ll addDif = b.s - a.s;
}

void ins(int loc, ll slope, ll add){
// insert a line to the hull
// corresponds to a journey ending at loc
if(hull[loc].size()>0 && (hull[loc].back().f == slope) && hull[loc].back().s >= add){
return;
}
while(true){
int sz = (int)(hull[loc].size());
if(sz==0){
break;
}
auto bef = hull[loc][sz-1];
// if the previous line has worse slope and add than the new line, delete it
hull[loc].pop_back();
continue;
}
if(sz == 1){
break;
}
auto grandBef = hull[loc][sz-2];
// then the second-newest is never on the hull, so we delete it
if(overcome(bef,grandBef) >= overcome(cur,bef)){
hull[loc].pop_back();
continue;
}
break;
}
hull[loc].push_back(cur);
}

ll query(int loc, ll t){
// query the hull for ending location loc at time t
int sz = (int)(hull[loc].size());
while(hull_pt[loc]+1<sz){
auto it = hull[loc].begin();
auto val1 = hull[loc][hull_pt[loc]];
it++;
auto val2 = hull[loc][hull_pt[loc]+1];
if(overcome(val2,val1) > t){
break;
}
hull_pt[loc]++;
}
auto val = (hull[loc][hull_pt[loc]]);
return t * val.f + val.s;
}

int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
int m;
cin >> N >> m;
hull_pt.resize(N);
for(int i = 0; i<N; i++){
ll x;
cin >> x;
rates.push_back(x);
}
for(int i = 0; i<=N; i++){
for(int j = 0; j<=N; j++){
dist[i][j] = inf;
}
}
for(int i = 0; i<m; i++){
int a, b;
ll t;
cin >> a >> b >> t;
a--;
b--;
dist[a][b] = t;
}
// compute shortest paths
for(int k = 0; k<N; k++){
for(int i = 0; i<N; i++){
for(int j = 0; j<N; j++){
dist[i][j] = min(dist[i][j],dist[i][k] + dist[k][j]);
}
}
}
// compute the total rate of all masks
for(int i = 0; i<N; i++){
break;
}
}
}
for(int i = 0; i<(1<<N); i++){
for(int j = 0; j<N; j++){
dp[j][i] = 1LL;
}
}
int all = (1<<N) - 1;
vector<pair<ll, pair<ll, int> > > lines;
for(int j = 0; j<N; j++){
// considers the line for a journey ending at location j
}

}
}
sort(lines.begin(),lines.end());
// insert lines into hull in order of slope
for(auto x: lines){
ins(x.s.s, x.f,x.s.f);
}
cin >> Q;
vector<ll> ans(Q);
vector<pair<ll, pair<int, int> > > queries;
for(int i = 0; i<Q; i++){
ll s;
int e;
cin >> s >> e;
e--;
// consider query i for ending at e after s seconds
queries.push_back(make_pair(s,make_pair(e,i)));
}
// consider queries in order of time
sort(queries.begin(),queries.end());
for(int i = 0; i<Q; i++){
ll t = queries[i].f;
int loc = queries[i].s.f;
int id = queries[i].s.s;
// query the relevant hull at the relevant time
ans[id] = query(loc,t);
}
for(int i = 0; i<Q; i++){
cout << ans[i] << "\n";
}
}