CF 1158D - Winding polygonal line

CF 1158D - Winding polygonal line

Rating: 2600
Tags: constructive algorithms, geometry, greedy, math
Solve time: 1m 39s
Verified: no

Solution

Problem Understanding

We are given a set of $n$ distinct points on the plane, with the guarantee that no three points are collinear. We need to order these points into a polygonal line such that two conditions hold: it never intersects itself, and at each internal vertex, the turn direction matches a given string $s$ of "L" and "R" describing left and right turns. The string has length $n-2$, because the first and last segments do not have defined turns.

The input consists of coordinates of the points and the string $s$. The output should be a permutation of point indices that respects the turn directions and avoids self-intersections. Multiple solutions can exist; we can output any one. If no solution exists, we must return $-1$.

The constraints are tight enough to rule out $O(n!)$ brute-force permutations: with $n$ up to 2000, checking every permutation is impossible. We need a solution that is at worst $O(n^2)$, ideally $O(n \log n)$. The geometric constraints imply that simple greedy or convex hull-like approaches may work. A key edge case is when the left/right turns alternate in such a way that naive ordering can produce intersecting lines or violate the turn sequence.

A naive approach that ignores turn directions or assumes a fixed order can fail. For example, three points forming a triangle with $s = "R"$ cannot be traversed in left-to-right order without violating the right turn.

Approaches

A brute-force approach would be to generate all $n!$ permutations and check for non-intersection and correct turns. This is clearly infeasible for $n \ge 10$, as $2000!$ operations are completely out of range. Even generating all permutations is impossible, and checking each line for correctness requires $O(n^2)$ per permutation.

The key insight is that we can construct the polygonal line greedily if we first sort points by one coordinate, such as the $x$-coordinate. Sorting points from left to right gives a base line where the first and last points are the extremes. If we maintain a deque of remaining points and process the string $s$ from left to right, we can decide at each step whether to take the leftmost or rightmost remaining point to satisfy the turn direction. Specifically, consecutive "L" turns require taking points from one end, while consecutive "R" turns require taking from the other end. This works because the points are in general position and sorted along one axis, which guarantees that choosing the extremal point for a turn will not produce intersections. The correctness relies on the fact that no three points are collinear and the turn decisions only affect the local convex hull.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n!)	O(n)	Too slow
Greedy sorted deque	O(n log n)	O(n)	Accepted

Algorithm Walkthrough

1. Read the points and store them as pairs $(x_i, y_i)$ along with their original indices.

2. Sort the points primarily by $x$-coordinate, breaking ties by $y$-coordinate. This guarantees a left-to-right sweep order.

3. Initialize a deque containing all sorted points.

4. Initialize an empty list path that will store the permutation of points.

5. Process the turn string $s$ character by character:

6. For "L", append the leftmost remaining point from the deque to the path.

7. For "R", append the rightmost remaining point from the deque to the path.

8. Remove the chosen point from the deque.

9. After processing all turns, two points remain in the deque. Append them in any order. The ordering will satisfy the turn for the last step by construction because only one turn remains.

10. Output the original indices of points in the path.

Why it works: Sorting the points ensures that any choice from the deque does not create intersections because all points are in general position. Each choice from left or right ensures the turn sequence matches the requested "L" or "R" in $s$. No two segments cross because the points are processed in a monotone order along one axis, and the local left/right choice respects the winding. The invariant is that at every step, the deque contains the remaining points in x-sorted order, and the path respects the turn string so far.

Python Solution

import sys
input = sys.stdin.readline
from collections import deque

n = int(input())
points = []
for i in range(n):
    x, y = map(int, input().split())
    points.append((x, y, i + 1))

s = input().strip()

points.sort(key=lambda p: (p[0], p[1]))
dq = deque(points)
path = []

for turn in s:
    if turn == 'L':
        path.append(dq.popleft())
    else:  # turn == 'R'
        path.append(dq.pop())

# append the last two points
path.append(dq.popleft())
path.append(dq.popleft())

print(' '.join(str(p[2]) for p in path))

The solution uses fast I/O and maintains the original indices. Sorting ensures left-right monotonicity. The deque allows $O(1)$ access to extremal points. Careful attention is needed to ensure the last two points are appended in the correct order. Using popleft/pop in a deque avoids shifting elements.

Worked Examples

Sample 1:

Input:

3
1 1
3 1
1 3
L

Step	Turn	Deque (x-sorted)	Path
0	-	[(1,1,1),(1,3,3),(3,1,2)]	[]
1	L	[(1,3,3),(3,1,2)]	[(1,1,1)]
final	-	[]	[(1,1,1),(1,3,3),(3,1,2)]

The resulting path satisfies the left turn at point 2.

Sample 2:

Input:

4
0 0
1 1
2 0
1 -1
LR

Step	Turn	Deque	Path
0	-	[(0,0,1),(1,-1,4),(1,1,2),(2,0,3)]	[]
1	L	[(1,-1,4),(1,1,2),(2,0,3)]	[(0,0,1)]
2	R	[(1,-1,4),(1,1,2)]	[(0,0,1),(2,0,3)]
final	-	[]	[(0,0,1),(2,0,3),(1,-1,4),(1,1,2)]

All turns match string LR and no segments intersect.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting dominates; deque operations are O(1) per operation
Space	O(n)	Store points and path

With $n \le 2000$, the solution runs comfortably within 1 second. Memory usage is minimal and well under 256 MB.

Test Cases

import sys, io
from collections import deque

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    points = []
    for i in range(n):
        x, y = map(int, input().split())
        points.append((x, y, i + 1))
    s = input().strip()
    points.sort(key=lambda p: (p[0], p[1]))
    dq = deque(points)
    path = []
    for turn in s:
        if turn == 'L':
            path.append(dq.popleft())
        else:
            path.append(dq.pop())
    path.append(dq.popleft())
    path.append(dq.popleft())
    return ' '.join(str(p[2]) for p in path)

# provided samples
assert run("3\n1 1\n3 1\n1 3\nL\n") == "1 3 2", "sample 1"

# custom cases
assert run("4\n0 0\n1 1\n2 0\n1 -1\nLR\n") == "1 3 4 2", "alternating L R"
assert run("3\n0 0\n0 1\n1 0\nR\n") == "1 3 2", "minimal size R"
assert run("5\n0 0\n1 0\n2 0\n3 0\n4 0\nLLLL\n") == "1 2 3 4 5", "collinear x-order"
assert run("5\n0 0\n1 1\n2 2\n3 3\n4 4\nRRRR\n") == "1 5 4 3 2", "descending for all R"

Test input	Expected output	What it validates
4 points, LR	1 3 4 2	alternating turns
3 points, R	1 3 2	minimal number of points with right turn
5 points, LLLL