Overview

By modeling the debris as vertices and jumps between debris within distance \(D\) as edges, the minimum number of jumps can be found using Breadth-First Search (BFS). Instead of examining all edges, we use a spatial partitioning tree similar to a KD-tree to manage debris by region and efficiently find reachable unvisited debris.

Analysis

1. The Minimum Number of Jumps Can Be Found Using BFS

Every jump costs uniformly \(1\) jump. Therefore, by searching in the order of debris reachable from the mothership in \(1\) jump, debris reachable in \(2\) jumps, and so on, we can determine the minimum number of jumps.

However, moving directly from the mothership to the station is forbidden. Thus, we handle it as follows:

• Use debris directly reachable from the mothership as the starting points of the search with distance \(1\).
• Pre-determine whether each piece of debris can reach the station.
• When a piece of debris that can reach the station is discovered for the first time, add \(1\) to the number of jumps taken to reach it.

This ensures that at least one piece of debris is visited.

2. Examining All Pairs of Debris Is Too Slow

Naively, one could compute the distance between all pairs of debris and add an edge between pairs where movement is possible, then run BFS.

However, the number of pairs is \(O(N^2)\). With \(N \leq 10^5\), there is neither enough time nor enough memory to store the edges.

Therefore, instead of building the edges of the graph explicitly, we only search for unvisited debris within distance \(D\) from the current debris.

Furthermore, in BFS, the shortest distance is determined the first time a vertex is discovered, so already-discovered debris can be removed from future searches.

3. Managing Sets of Points with Rectangles

Instead of inspecting points one by one, we consider bounding rectangles that enclose multiple points.

Relative to the circle of radius \(D\) centered at the current position, there are three possible geometric relationships for a rectangle:

• The entire rectangle lies outside the circle
  Points inside cannot be reached, so we can ignore all of them at once.
• The entire rectangle is contained within the circle
  All points inside can be reached, so we discover all of them at once.
• Otherwise
  Subdivide the rectangle into smaller regions and examine them.

By performing this check using a tree structure, we reduce unnecessary distance calculations.

Algorithm

1. Determine Distances to the Mothership and Station

For each piece of debris \((x_i,y_i)\), check the following conditions:

• Reachable from the mothership: \(x_i^2+y_i^2 \leq D^2\)
• Can reach the station: \(x_i^2+(W-y_i)^2 \leq D^2\)

If there is any debris that satisfies both conditions, we can reach the station in \(2\) jumps by going from the mothership via that debris. Since direct movement is forbidden, this is optimal.

Otherwise, add the debris reachable from the mothership to frontier. Store the remaining debris along with a flag indicating whether it can reach the station.

At this point, if either of the following holds, the answer is -1:

• No debris is reachable from the mothership.
• No debris can reach the station.

2. Build a Spatial Partitioning Tree from Unvisited Debris

Each node maintains the bounding rectangle

\([x_{\min},x_{\max}] \times [y_{\min},y_{\max}]\)

enclosing the points it is responsible for. It also maintains whether any debris among its points can reach the station.

The tree is constructed as follows:

1. Compute the bounding rectangle enclosing the set of points.
2. If the number of points is \(12\) or fewer, make this node a leaf that directly stores the points.
3. Otherwise, choose the dimension (width or height) along which the rectangle is larger.
4. Sort the points by the coordinates in that dimension and split them into two halves of nearly equal size.
5. Recursively build the tree for each half.

Because we split evenly by the number of points, the height of the tree is \(O(\log N)\).

3. Search for Reachable Unvisited Debris from the Current Location

Let \((x,y)\) be the current location, and traverse the tree starting from the root.

Pruning by Minimum Distance to the Rectangle

Let the minimum distance from the current location to the rectangle along each axis be

\(\delta_x=\max(x_{\min}-x,\ 0,\ x-x_{\max})\)

\(\delta_y=\max(y_{\min}-y,\ 0,\ y-y_{\max})\).

If

\(\delta_x^2+\delta_y^2>D^2\),

there are no reachable points in the rectangle. We can skip exploring this node’s subtree.

Batch Processing When the Entire Rectangle is Reachable

Let the maximum distance along each axis be

\(\Delta_x=\max(|x-x_{\min}|,\ |x-x_{\max}|)\)

\(\Delta_y=\max(|y-y_{\min}|,\ |y-y_{\max}|)\).

If

\(\Delta_x^2+\Delta_y^2\leq D^2\),

all points in the rectangle are reachable.

• If any point that can reach the station is included, terminate the search immediately.
• If not, add all points to nxt as candidates for the next round of search.
• Delete this entire node from the tree.

Otherwise

If the node is a leaf, directly compute the distance to each stored point. Add the reachable points to nxt and remove them from the leaf.

If it is an internal node, recursively explore both the left and right children.

After deletion, remove any empty parts and update the bounding rectangle to match the remaining points. This helps maintain effective pruning in subsequent searches.

4. Advance BFS Layer by Layer

Let distance be the minimum number of jumps to reach the debris currently in frontier. Its initial value is \(1\).

Search from each point in frontier, collecting newly discovered unvisited debris into nxt. All of them can be reached from the mothership in distance + 1 jumps.

If a piece of debris that can reach the station is discovered here, the answer is

\(\text{distance}+2\).

This is because we take \(1\) jump from the current debris to the new debris, and \(1\) more jump from there to the station.

If none is found, replace frontier with nxt and proceed to the next layer. If the station is not reached and no more points remain to explore, output -1.

Correctness

BFS explores debris in increasing order of minimum jumps from the mothership. Therefore, the moment an unvisited piece of debris is first discovered, its minimum jump count is determined.

In the spatial partitioning tree, exploration is pruned only when the entire rectangle is unreachable, and points are discovered in batch only when the entire rectangle is reachable. Thus, all reachable unvisited debris are discovered without omissions or false positives.

Consequently, the number of jumps computed when a piece of debris capable of reaching the station is first discovered is the global minimum.

Complexity

• Time Complexity: \(O(N^2)\) in the worst case; tree construction takes \(O(N\log^2 N)\).
• Space Complexity: \(O(N)\).

During construction, points are sorted at each node, which takes \(O(N\log^2 N)\) overall.

The search time depends on point distribution and how effectively pruning works. Although each piece of debris is added and deleted at most once, internal nodes may be visited multiple times, so the total search cannot be guaranteed to run in \(O(N\log N)\). While the worst-case time complexity is \(O(N^2)\), pruning via bounding rectangles and batch deletions significantly reduces the actual number of inspected nodes.

Implementation Notes

• Do not compute square roots
  Comparing squared distances with \(D^2\) avoids floating-point inaccuracies and allows exact comparisons.

• Remove debris from search candidates upon discovery
  This prevents the same debris from being added multiple times to nxt from different search sources.

• No need to update the station-reachability flag after deletion
  In this implementation, the algorithm terminates immediately upon discovering any debris that can reach the station. As long as the search continues, only debris that cannot reach the station is deleted, so the node flags remain valid.

• Use exceptions to exit early from recursion
  Reached is used to immediately propagate the discovery of a debris that can reach the station all the way out of the nested recursive calls.

  Source Code

import sys


def main():
    input = sys.stdin.buffer.readline
    N, W, D = map(int, input().split())
    D2 = D * D

    frontier = []
    remaining = []
    has_goal = False

    for _ in range(N):
        x, y = map(int, input().split())
        goal = x * x + (W - y) * (W - y) <= D2
        if x * x + y * y <= D2:
            if goal:
                print(2)
                return
            frontier.append((x, y, False))
        else:
            remaining.append((x, y, goal))
            has_goal |= goal

    if not frontier or not has_goal:
        print(-1)
        return

    def key_x(p):
        return p[0]

    def key_y(p):
        return p[1]

    def build(points):
        xmin = xmax = points[0][0]
        ymin = ymax = points[0][1]
        for x, y, _ in points:
            if x < xmin:
                xmin = x
            if x > xmax:
                xmax = x
            if y < ymin:
                ymin = y
            if y > ymax:
                ymax = y

        if len(points) <= 12:
            goal = any(p[2] for p in points)
            return [xmin, xmax, ymin, ymax, None, None, points, goal]

        if xmax - xmin >= ymax - ymin:
            points.sort(key=key_x)
        else:
            points.sort(key=key_y)

        mid = len(points) // 2
        left = build(points[:mid])
        right = build(points[mid:])
        return [
            xmin, xmax, ymin, ymax,
            left, right, None, left[7] or right[7]
        ]

    root = build(remaining)
    del remaining

    class Reached(Exception):
        pass

    nxt = []

    def search(node, x, y):
        xmin, xmax, ymin, ymax = node[0], node[1], node[2], node[3]

        if x < xmin:
            dx = xmin - x
        elif x > xmax:
            dx = x - xmax
        else:
            dx = 0

        if y < ymin:
            dy = ymin - y
        elif y > ymax:
            dy = y - ymax
        else:
            dy = 0

        if dx * dx + dy * dy > D2:
            return node

        dx = x - xmin
        t = xmax - x
        if t > dx:
            dx = t
        dy = y - ymin
        t = ymax - y
        if t > dy:
            dy = t

        if dx * dx + dy * dy <= D2:
            if node[7]:
                raise Reached
            stack = [node]
            while stack:
                current = stack.pop()
                points = current[6]
                if points is not None:
                    nxt.extend(points)
                else:
                    stack.append(current[4])
                    stack.append(current[5])
            return None

        points = node[6]
        if points is not None:
            size = len(points)
            write = 0
            for i in range(size):
                p = points[i]
                dx = p[0] - x
                dy = p[1] - y
                if dx * dx + dy * dy <= D2:
                    if p[2]:
                        raise Reached
                    nxt.append(p)
                else:
                    points[write] = p
                    write += 1

            if write == 0:
                return None
            if write == size:
                return node

            del points[write:]
            xmin = xmax = points[0][0]
            ymin = ymax = points[0][1]
            for px, py, _ in points:
                if px < xmin:
                    xmin = px
                if px > xmax:
                    xmax = px
                if py < ymin:
                    ymin = py
                if py > ymax:
                    ymax = py

            node[0] = xmin
            node[1] = xmax
            node[2] = ymin
            node[3] = ymax
            return node

        left = search(node[4], x, y)
        right = search(node[5], x, y)

        if left is None:
            return right
        if right is None:
            return left

        node[4] = left
        node[5] = right
        node[0] = left[0] if left[0] < right[0] else right[0]
        node[1] = left[1] if left[1] > right[1] else right[1]
        node[2] = left[2] if left[2] < right[2] else right[2]
        node[3] = left[3] if left[3] > right[3] else right[3]
        return node

    distance = 1
    try:
        while frontier and root is not None:
            nxt = []
            for x, y, _ in frontier:
                root = search(root, x, y)
                if root is None:
                    break
            frontier = nxt
            distance += 1
    except Reached:
        print(distance + 2)
        return

    print(-1)


if __name__ == "__main__":
    main()

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This editorial was generated by gpt-6-astra-medium.