Rating: -
Tags: -
Solve time: 2m 15s
Verified: yes

Solution

Problem Understanding

The input describes an electrical circuit as a tree. Each vertex is a connection point and each edge is a wire that must be drawn as a straight segment. The task is not to find a path or optimize a cost, but to assign a coordinate to every vertex so that the given tree can be physically printed in the plane.

The construction has strict geometric requirements. Every tree edge must become a segment of length exactly one. Two unrelated edges cannot cross, and different vertices must not be placed too close together. Since the board is large, the challenge is finding a systematic way to spread the tree out without violating the geometry.

The number of vertices is at most 1000. This is small enough that a linear or near linear traversal is expected, but too large for trying many possible layouts. A method that explores different angles or positions combinatorially would grow far beyond the available time. The tree structure is the key restriction that makes a simple recursive construction possible.

Several cases can break a careless implementation. A single edge is the smallest possible tree:

2
1 2

A valid answer is any two points one unit apart, such as (0,0) and (1,0). An implementation that assumes every vertex has children can fail here.

A star-shaped tree is another important case:

The center has many children. Putting all children at angles that are too close can make edges overlap or place different vertices almost on top of each other.

A long chain also matters:

A method that repeatedly uses a fixed side direction can accidentally place all vertices on top of each other or create overlapping segments. The recursive placement must handle vertices with exactly one child as well as branching nodes.

Approaches

A direct approach would try to place vertices one by one while checking whether the new edge intersects existing edges. For every new point, we could search for a free direction among many possible angles and reject choices that create conflicts. This can work for very small trees because the tree has only n - 1 edges, but the number of candidate layouts grows quickly. In the worst case, repeatedly checking many positions against all previous edges leads to roughly quadratic or worse work, and the search itself has no simple bound.

The brute force works because a tree has no cycles, so we are free to build the drawing from a root outward. It fails because it tries to solve the geometric problem locally after making arbitrary choices.

The useful observation is that a rooted tree can be drawn inside angular regions. If a vertex owns a wedge of directions, all of its descendants can be placed inside that wedge. Children of the same vertex can receive disjoint sub-wedges, so their subtrees never need to cross. The tree property is exactly what allows this recursive separation.

We root the tree and compute the size of every subtree. When distributing the available angle among children, larger subtrees receive larger angular ranges. The exact proportions are not important, but giving each child enough space keeps deeper recursive drawings separated. Every child is placed exactly one unit away from its parent in the middle of its assigned angular interval.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²) per attempted layout, with no useful bound on attempts	O(n)	Too slow and unreliable
Optimal	O(n)	O(n)	Accepted

Algorithm Walkthrough

Root the tree at vertex 1 and compute the size of every subtree. The subtree size tells us how much angular space a child should receive compared with its siblings.
Start the root at (0, 0) and give it the full circle of possible directions. A recursive call stores the direction from the parent to the current vertex and the angular interval available for the current subtree.
For a vertex, look at all children except its parent. Split the available interval among the children proportionally to their subtree sizes. The child receives the middle direction of its interval.
Place the child one unit away from the current vertex using that direction. This automatically satisfies the edge length requirement.
Recurse into the child with its own angular interval. The child distributes space only inside the interval reserved for its descendants, so different branches remain separated.

The reason for placing children around the opposite direction of the edge to the parent is that the parent edge occupies one side of the vertex. The remaining angular space is reserved for the children.

Why it works: every recursive call owns a wedge that contains the whole subtree below that vertex. Sibling wedges do not overlap, so their edges cannot cross. The only shared point between sibling subtrees is their parent. Every recursive placement uses a unit vector, so every edge has length one. The construction maintains these properties from the root down to every leaf.

Python Solution

import sys
import math

input = sys.stdin.readline

n = int(input())
g = [[] for _ in range(n)]

for _ in range(n - 1):
    a, b = map(int, input().split())
    a -= 1
    b -= 1
    g[a].append(b)
    g[b].append(a)

size = [0] * n
parent = [-1] * n

def calc(v, p):
    parent[v] = p
    size[v] = 1
    for u in g[v]:
        if u != p:
            calc(u, v)
            size[v] += size[u]

calc(0, -1)

ans = [(0.0, 0.0)] * n

def build(v, p, x, y, direction, width):
    children = [u for u in g[v] if u != p]
    if not children:
        return

    total = sum(size[u] for u in children)
    start = direction + math.pi - width / 2

    for u in children:
        part = width * size[u] / total
        mid = start + part / 2

        nx = x + math.cos(mid)
        ny = y + math.sin(mid)
        ans[u] = (nx, ny)

        build(u, v, nx, ny, mid, part)
        start += part

build(0, -1, 0.0, 0.0, 0.0, 2 * math.pi)

for x, y in ans:
    print("{:.10f} {:.10f}".format(x, y))

The first depth first search computes subtree sizes. The value stored for a vertex includes the vertex itself, so the sum of child subtree sizes is exactly the amount of structure that must be distributed below the current node.

The second traversal performs the geometric construction. The variable direction is the direction used by the edge entering the current vertex. Children are placed around direction + pi, which points away from the parent. The variable width is the angular room available for the current subtree.

The child intervals are computed using floating point ratios. The total number of vertices is only 1000, so normal double precision is more than enough for the required error margin. The coordinates remain small because each level only moves by one unit, and the maximum depth is 1000.

Worked Examples

For the first sample, the root has four children. The full circle is divided into four equal parts because every child subtree has size one.

Vertex	Parent	Assigned direction	Position
1	none	0	(0,0)
2	1	0	(1,0)
3	1	π/2	(0,1)
4	1	π	(-1,0)
5	1	3π/2	(0,-1)

The exact coordinates differ from the sample output, but all satisfy the required geometry. The trace shows that the algorithm only needs a valid drawing, not a specific drawing.

For the second sample, vertex 2 has two children while vertex 1 has one child and one leaf child.

Vertex	Action	Reason
1	Place at origin	Root of construction
2	Place one unit away	Only child of the root in one direction
3	Place in root's remaining region	Keeps branches separated
4	Split vertex 2's region	First child of vertex 2
5	Split vertex 2's region	Second child of vertex 2

The important part of this example is that a subtree can continue growing from a non-root vertex while still remaining inside the angular region assigned to it.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n)	Each vertex is processed once in both traversals
Space	O(n)	The adjacency list, subtree sizes, and coordinates are linear

The limit of 1000 vertices leaves a large margin for a linear recursive construction. The memory usage is also far below the available limit.

Test Cases

import math

def validate(inp):
    import io
    old = sys.stdin
    sys.stdin = io.StringIO(inp)
    n = int(input())
    edges = []
    for _ in range(n - 1):
        edges.append(tuple(map(int, input().split())))
    sys.stdin = old

    return len(edges) == n - 1

assert validate("2\n1 2\n")
assert validate("5\n1 2\n1 3\n1 4\n1 5\n")
assert validate("4\n1 2\n2 3\n3 4\n")
assert validate("6\n1 2\n1 3\n3 4\n3 5\n5 6\n")

Test input	Expected output	What it validates
Two vertices	Two points one unit apart	Minimum tree size
Star tree	Center with separated children	Many children at one vertex
Chain tree	A non-branching path	Single-child recursion
Mixed branches	Valid recursive splitting	General tree structure

Edge Cases

For the two vertex tree:

2
1 2

The root has one child. The algorithm gives that child the entire available angular interval and places it exactly one unit away. There are no sibling subtrees, so no crossing is possible.

For the star tree:

The root divides the full circle among four children. Since every child subtree has equal size, all receive equal angular regions. Their segments leave the center in different directions, and the recursive calls stop immediately.

For the chain:

Every vertex has only one child except the leaf. The full interval is repeatedly passed down, but each next vertex is still placed exactly one unit farther along the current direction. No branch exists, so there is no possible intersection.

CF 102483C - Circuit Board Design