## 1) Concise (abridged) problem statement

You are given an axis-aligned polyline (sequence of \(N\) vertices, \(1 \le N \le 4000\)) describing a railroad track. The train moves along the polyline in the given order.

A train of length \(L\) occupies a *contiguous* part of the track (between the head position and the tail position along the polyline). While moving, the train must **never self-intersect or self-touch** in the plane (except that the head and tail are allowed to coincide at one point).

Find the **maximum possible** train length \(L\).

---

## 2) Detailed editorial (explaining the provided solution)

### Key observation: when does the train become invalid?
At any moment, the train covers a continuous interval of arc-length positions along the polyline: \([s, s+L]\) (head at \(s+L\), tail at \(s\)).

The train becomes invalid exactly when there exist **two distinct positions along the polyline** within that interval that map to the **same point in the plane**, or when the covered curve touches itself. Because the polyline has no *overlapping* segments ("no self-covering"), the only way two different positions can coincide is at a **proper intersection of two non-adjacent segments** (or revisiting a previous vertex, which is also an intersection case).

So: the maximum valid \(L\) is limited by the **shortest "cycle length"** created by any self-intersection of the polyline.

### Turning an intersection into a limiting length
Suppose segment \(i\) and segment \(j\) (with \(j \ge i+2\), i.e., not adjacent) intersect at point \(P\).

Let:
- \(d_i\) = distance along the track from the start (vertex 0) to point \(P\) traveling along segment \(i\).
- \(d_j\) = distance along the track from the start to point \(P\) traveling along segment \(j\).

Assuming \(j > i\), then along the forward traversal of the polyline, you reach \(P\) first at \(d_i\) and later again at \(d_j\). The path between these two occurrences forms a closed loop of length:
\[
\text{cycleLen} = d_j - d_i
\]

If the train length \(L\) is **greater** than this cycle length, then at some moment the train will cover both occurrences of \(P\) simultaneously, forcing a self-touch/intersection. Therefore:
\[
L \le \min(\text{cycleLen over all intersections})
\]

If there are no intersections, the full track length is valid.

### Computing distances \(d_i\) and \(d_j\)
All segments are axis-aligned, so segment lengths are Manhattan distances.

Precompute prefix lengths:
- `prefix_len[k]` = total length of segments \(0..k-1\) (distance from start to vertex \(k\)).

For an intersection point \((px,py)\) that lies on segment \(i\), its distance from the start is:
\[
d_i = \text{prefix\_len}[i] + |px-x_i| + |py-y_i|
\]

### Detecting intersections
Two axis-aligned segments intersect properly only if:
- one is horizontal, the other vertical, and
- the vertical segment's x lies within horizontal segment's x-range, and
- the horizontal segment's y lies within vertical segment's y-range.

The solution checks all pairs \((i,j)\) with \(j \ge i+2\) (skip adjacent segments which share endpoints and are allowed).

### Complexity
The provided code is \(O(N^2)\) intersection checks, each \(O(1)\), which is fine for the intended constraints in this problem setting (and matches the author's comment).

---

## 3) C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<pair<int, int>> p;
vector<int64_t> prefix_len;

int64_t seg_len(int i) {
    return abs(p[i + 1].first - p[i].first) +
           abs(p[i + 1].second - p[i].second);
}

optional<pair<int, int>> get_intersection(int i, int j) {
    bool h1 = (p[i].second == p[i + 1].second);
    bool h2 = (p[j].second == p[j + 1].second);

    if(h1 == h2) {
        return nullopt;
    }

    if(!h1) {
        swap(i, j);
    }

    int hx1 = min(p[i].first, p[i + 1].first);
    int hx2 = max(p[i].first, p[i + 1].first);
    int hy = p[i].second;

    int vx = p[j].first;
    int vy1 = min(p[j].second, p[j + 1].second);
    int vy2 = max(p[j].second, p[j + 1].second);

    if(vx >= hx1 && vx <= hx2 && hy >= vy1 && hy <= vy2) {
        return make_pair(vx, hy);
    }

    return nullopt;
}

int64_t pos_on_track(int i, int px, int py) {
    return prefix_len[i] + abs(px - p[i].first) + abs(py - p[i].second);
}

void read() {
    cin >> n;
    p.resize(n);
    for(int i = 0; i < n; i++) {
        cin >> p[i].first >> p[i].second;
    }
}

void solve() {
    // The train occupies a contiguous portion of the track. As it moves,
    // different parts of the track are covered, meaning that the train
    // self-intersects when two different positions along the track map to the
    // same point P (or there is a cycle). We will anyways pass through every
    // cycle, meaning that we are bounded by the "shortest one". This means we
    // can simply try all possible pairs of segments, and if they intersect we
    // can directly figure out the length of the cycle: it's the length(start,
    // B) - length(start, A). Note that this might not be a simple cycle, but if
    // that's the case there is certainly a smaller one which we would consider
    // in this algorithm too. The complexity of this algorithm is O(N^2) which
    // is sufficient for the problem here.

    if(n <= 1) {
        cout << 0 << endl;
        return;
    }

    int m = n - 1;
    prefix_len.resize(m + 1);
    prefix_len[0] = 0;
    for(int i = 0; i < m; i++) {
        prefix_len[i + 1] = prefix_len[i] + seg_len(i);
    }

    int64_t ans = prefix_len[m];

    for(int i = 0; i < m; i++) {
        for(int j = i + 2; j < m; j++) {
            auto intersection = get_intersection(i, j);
            if(intersection) {
                auto [px, py] = *intersection;
                int64_t pos_i = pos_on_track(i, px, py);
                int64_t pos_j = pos_on_track(j, px, py);
                ans = min(ans, pos_j - pos_i);
            }
        }
    }

    cout << ans << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys
from typing import List, Tuple, Optional

def solve() -> None:
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n = int(next(it))
    p: List[Tuple[int, int]] = [(int(next(it)), int(next(it))) for _ in range(n)]

    # If there are no segments, track length is 0.
    if n <= 1:
        print(0)
        return

    m = n - 1  # number of segments

    # Segment length: since segments are axis-aligned, Manhattan distance equals segment length.
    def seg_len(i: int) -> int:
        x1, y1 = p[i]
        x2, y2 = p[i + 1]
        return abs(x2 - x1) + abs(y2 - y1)

    # Prefix lengths: prefix_len[k] = distance from start to vertex k
    prefix_len = [0] * (m + 1)
    for i in range(m):
        prefix_len[i + 1] = prefix_len[i] + seg_len(i)

    # Return intersection point of segment i and j, or None if no intersection.
    # Only handles proper crossing: one horizontal and one vertical.
    def get_intersection(i: int, j: int) -> Optional[Tuple[int, int]]:
        x1, y1 = p[i]
        x2, y2 = p[i + 1]
        x3, y3 = p[j]
        x4, y4 = p[j + 1]

        h1 = (y1 == y2)  # segment i horizontal?
        h2 = (y3 == y4)  # segment j horizontal?

        # Same orientation => ignore (no proper crossing; overlaps are disallowed).
        if h1 == h2:
            return None

        # Ensure i is horizontal, j is vertical (swap if needed).
        if not h1:
            i, j = j, i
            x1, y1 = p[i]
            x2, y2 = p[i + 1]
            x3, y3 = p[j]
            x4, y4 = p[j + 1]

        # Horizontal segment i spans x in [hx1, hx2] at y=hy
        hx1, hx2 = (x1, x2) if x1 <= x2 else (x2, x1)
        hy = y1

        # Vertical segment j spans y in [vy1, vy2] at x=vx
        vx = x3
        vy1, vy2 = (y3, y4) if y3 <= y4 else (y4, y3)

        # Check if they cross (including touching at endpoints)
        if hx1 <= vx <= hx2 and vy1 <= hy <= vy2:
            return (vx, hy)

        return None

    # Distance along the polyline to reach (px,py) while on segment i
    def pos_on_track(i: int, px: int, py: int) -> int:
        x, y = p[i]
        # Distance to segment start vertex + distance within segment
        return prefix_len[i] + abs(px - x) + abs(py - y)

    # Start with total track length (valid if no intersections exist)
    ans = prefix_len[m]

    # Check all non-adjacent pairs of segments:
    # j starts at i+2 to avoid adjacent segments sharing a vertex.
    for i in range(m):
        for j in range(i + 2, m):
            inter = get_intersection(i, j)
            if inter is None:
                continue
            px, py = inter

            # Arc-length positions of the same geometric point along the path
            pos_i = pos_on_track(i, px, py)
            pos_j = pos_on_track(j, px, py)

            # The loop length between the two occurrences
            cycle_len = pos_j - pos_i
            if cycle_len < ans:
                ans = cycle_len

    print(ans)

if __name__ == "__main__":
    solve()
```

---

## 5) Compressed editorial

- The train covers a contiguous arc-length interval of the polyline.
- A self-touch/intersection happens iff two different arc-length positions map to the same plane point, which (given "no self-covering") occurs only at intersections of a horizontal and a vertical segment (including revisiting a point).
- For any intersecting segment pair \((i,j)\) with \(j \ge i+2\), compute intersection point \(P\).
- Let \(d_i, d_j\) be arc-length distances from the start to \(P\) along segments \(i\) and \(j\), using prefix sums.
- Then \(d_j - d_i\) is a loop length that upper-bounds the train length. The answer is the minimum such loop length over all intersections, or total track length if none.
