Longest Trip
Case D = 3: complete graph Every pair is connected, so any permutation of all N nodes is a Hamiltonian path. Case D = 2 Any three nodes span at least 2 edges. A Hamiltonian path always exists. Case D = 1: general When...
Problem Statement
Rendered from the "Problem Statement" section in the LaTeX write-up.
$N$ landmarks are connected by roads. The graph satisfies a density condition: for any triple of nodes, at least $D$ of the three possible edges exist ($D \in \{1, 2, 3\}$). Using an interactive function $\texttt{are\_connected}(A, B)$ (which tests whether any edge exists between sets $A$ and $B$), find the longest simple path.
Editorial
The solution write-up is rendered from the LaTeX source, with equations kept live through MathJax.
Solution
Case $D = 3$: complete graph
Every pair is connected, so any permutation of all $N$ nodes is a Hamiltonian path.
Case $D = 2$
Any three nodes span at least 2 edges. A Hamiltonian path always exists.
Case $D = 1$: general
Lemma.
lem:merge When adding node $i$ that is not adjacent to any endpoint of path $P$ or path $Q$, the density condition forces $P$'s last node and $Q$'s last node to be adjacent. Hence we can concatenate $P$ and $Q$.
Proof.
Consider the triple ($i$, $P.\mathrm{back}$, $Q.\mathrm{back}$). We tested that $i$--$P.\mathrm{back}$ and $i$--$Q.\mathrm{back}$ are non-edges. Since at least 1 edge must exist among the three, the edge $P.\mathrm{back}$--$Q.\mathrm{back}$ exists.
Algorithm.
Maintain two paths $P$ and $Q$ (as deques). For each new node $i$:
If $i$ connects to an endpoint of $P$, extend $P$.
Else if $i$ connects to an endpoint of $Q$, extend $Q$.
Else by Lemma lem:merge, merge $P$ and $Q$ (appending $Q$ reversed to $P$), then start a new $Q = \{i\}$.
After all nodes are processed, try to merge $P$ and $Q$ by testing all four endpoint combinations. If no endpoint connection exists, use binary search to find an internal connection point (costing $O(\log N)$ additional queries) and splice the paths together.
Complexity
Queries during construction: $O(N)$ --- at most 4 queries per node for endpoint checks.
Queries for merging: $O(\log N)$ for binary search on each path. Total: $O(\log^2 N)$.
Overall: $O(N + \log^2 N) = O(N)$ queries.
Space: $O(N)$.
Code
C++ solution used for this page.
#include <bits/stdc++.h>
using namespace std;
// IOI 2023 - Longest Trip (Interactive)
// Graph with N nodes, density D: any triplet has >= D edges among 3 possible.
// Find longest simple path using are_connected(A, B) queries.
//
// Algorithm:
// 1. Maintain two paths P and Q.
// 2. For each new node, try to attach to an endpoint of P or Q.
// 3. If neither works, by density >= 1, P.back()-Q.back() must be an edge,
// so merge P and Q and start Q fresh with the new node.
// 4. After processing all nodes, merge P and Q via endpoint checks,
// or binary search for an internal edge and splice the longer halves.
bool are_connected(vector<int> A, vector<int> B); // grader function
vector<int> longest_trip(int N, int D) {
if (N == 1) return {0};
deque<int> P, Q;
P.push_back(0);
if (N > 1) Q.push_back(1);
// Phase 1: Build two paths by attaching each node to an endpoint.
for (int i = 2; i < N; i++) {
if (are_connected({i}, {(int)P.back()})) {
P.push_back(i);
} else if (are_connected({i}, {(int)P.front()})) {
P.push_front(i);
} else if (!Q.empty() && are_connected({i}, {(int)Q.back()})) {
Q.push_back(i);
} else if (!Q.empty() && are_connected({i}, {(int)Q.front()})) {
Q.push_front(i);
} else {
// D >= 1 on triplet {i, P.back(), Q.back()}: i-P.back() = false,
// i-Q.back() = false => P.back()-Q.back() must be an edge.
while (!Q.empty()) {
P.push_back(Q.back());
Q.pop_back();
}
Q.push_back(i);
}
}
if (Q.empty())
return vector<int>(P.begin(), P.end());
// Phase 2: Merge P and Q via endpoint connections.
if (are_connected({(int)P.back()}, {(int)Q.front()})) {
for (auto x : Q) P.push_back(x);
return vector<int>(P.begin(), P.end());
}
if (are_connected({(int)P.back()}, {(int)Q.back()})) {
while (!Q.empty()) { P.push_back(Q.back()); Q.pop_back(); }
return vector<int>(P.begin(), P.end());
}
if (are_connected({(int)P.front()}, {(int)Q.front()})) {
vector<int> result;
while (!P.empty()) { result.push_back(P.back()); P.pop_back(); }
for (auto x : Q) result.push_back(x);
return result;
}
if (are_connected({(int)P.front()}, {(int)Q.back()})) {
while (!Q.empty()) { P.push_front(Q.back()); Q.pop_back(); }
return vector<int>(P.begin(), P.end());
}
// Phase 3: No endpoint-endpoint edge. Binary search for an internal edge.
vector<int> pv(P.begin(), P.end()), qv(Q.begin(), Q.end());
if (!are_connected(pv, qv)) {
// Disconnected components; return the longer path.
return (pv.size() >= qv.size()) ? pv : qv;
}
// Binary search on P to find p_idx where pv[p_idx] connects to some node in Q.
int lo = 0, hi = (int)pv.size() - 1;
while (lo < hi) {
int mid = (lo + hi) / 2;
vector<int> sub(pv.begin() + lo, pv.begin() + mid + 1);
if (are_connected(sub, qv))
hi = mid;
else
lo = mid + 1;
}
int p_idx = lo;
// Binary search on Q to find q_idx where qv[q_idx] connects to pv[p_idx].
lo = 0; hi = (int)qv.size() - 1;
while (lo < hi) {
int mid = (lo + hi) / 2;
vector<int> sub(qv.begin() + lo, qv.begin() + mid + 1);
if (are_connected({pv[p_idx]}, sub))
hi = mid;
else
lo = mid + 1;
}
int q_idx = lo;
// We have edge pv[p_idx] -- qv[q_idx].
// P is a path: pv[0]-...-pv[end]. Splitting at p_idx gives two sub-paths:
// P_left = pv[0..p_idx] (size p_idx + 1)
// P_right = pv[p_idx..end] (size |P| - p_idx)
// Similarly Q splits at q_idx:
// Q_left = qv[0..q_idx] (size q_idx + 1)
// Q_right = qv[q_idx..end] (size |Q| - q_idx)
//
// We can form 4 valid paths through the bridge edge pv[p_idx]-qv[q_idx]:
// 1. P_left fwd + Q_right fwd: pv[0]...pv[p_idx]-qv[q_idx]...qv[end]
// 2. P_left fwd + Q_left rev: pv[0]...pv[p_idx]-qv[q_idx]...qv[0]
// 3. P_right rev + Q_right fwd: pv[end]...pv[p_idx]-qv[q_idx]...qv[end]
// 4. P_right rev + Q_left rev: pv[end]...pv[p_idx]-qv[q_idx]...qv[0]
//
// Sizes: pick the longer P half + the longer Q half.
// Best = max(p_idx+1, |P|-p_idx) + max(q_idx+1, |Q|-q_idx)
int pn = (int)pv.size(), qn = (int)qv.size();
bool use_p_left = (p_idx + 1 >= pn - p_idx); // true => use P_left, false => P_right
bool use_q_right = (qn - q_idx >= q_idx + 1); // true => use Q_right, false => Q_left
vector<int> result;
if (use_p_left) {
// P_left forward: pv[0]...pv[p_idx]
for (int i = 0; i <= p_idx; i++) result.push_back(pv[i]);
} else {
// P_right reversed: pv[end]...pv[p_idx]
for (int i = pn - 1; i >= p_idx; i--) result.push_back(pv[i]);
}
if (use_q_right) {
// Q_right forward: qv[q_idx]...qv[end]
for (int i = q_idx; i < qn; i++) result.push_back(qv[i]);
} else {
// Q_left reversed: qv[q_idx]...qv[0]
for (int i = q_idx; i >= 0; i--) result.push_back(qv[i]);
}
return result;
}
// The grader provides are_connected() and calls longest_trip().
// No main() is needed for submission.
#ifdef LOCAL
int main() {
return 0;
}
#endif
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