CF 2046C - Adventurers
Each city is a point on the plane. We choose a dividing point $(x0,y0)$, which splits the plane into four regions: $$begin{aligned} &x ge x0, y ge y0 &x < x0, y ge y0 &x ge x0, y < y0 &x < x0, y < y0 end{aligned}$$ Every city belongs to exactly one of these regions, according…
Rating: 2100
Tags: binary search, data structures, greedy, sortings, ternary search, two pointers
Solve time: 3m 19s
Verified: no
Solution
Problem Understanding
Each city is a point on the plane. We choose a dividing point $(x_0,y_0)$, which splits the plane into four regions:
$$\begin{aligned} &x \ge x_0,\ y \ge y_0 \ &x < x_0,\ y \ge y_0 \ &x \ge x_0,\ y < y_0 \ &x < x_0,\ y < y_0 \end{aligned}$$
Every city belongs to exactly one of these regions, according to the inequalities above.
For a chosen dividing point, let the four region counts be $c_1,c_2,c_3,c_4$. We want to maximize
$$\min(c_1,c_2,c_3,c_4).$$
We must output both the maximum achievable value and one dividing point that achieves it.
The total number of points over all test cases is at most $10^5$. Any solution that examines every pair of candidate $x_0$ and $y_0$ values would require $O(n^2)$ or worse work per test case, which is far too slow. We need something around $O(n \log^2 n)$.
A subtle point is that the inequalities are asymmetric. Cities with $x=x_0$ belong to the right half, and cities with $y=y_0$ belong to the upper half. A solution that accidentally treats the split as strict on both sides will produce incorrect quadrant counts.
Another tricky case occurs when many cities share the same $x$-coordinate. While sweeping possible vertical splits, all cities with the same $x$ must move together. Splitting inside such a group changes the meaning of the inequality and can create invalid states.
Approaches
A brute-force solution would try every possible $x_0$, every possible $y_0$, count the four quadrants, and keep the best answer. Even if we restrict candidates to coordinates appearing in the input, this still requires $O(n^2)$ candidate pairs and $O(n)$ work to evaluate each one, resulting in $O(n^3)$.
The key observation is that we do not need to search directly for the best minimum quadrant size. Instead, suppose we ask a simpler question:
Is it possible to make every quadrant contain at least $k$ cities?
If we can answer this efficiently, then the final answer can be found with binary search on $k$.
For a fixed $k$, imagine sweeping the vertical boundary from left to right. At any moment, cities are divided into two groups:
$$L = {x_i < x_0}, \qquad R = {x_i \ge x_0}.$$
For a chosen horizontal boundary $y_0$, we need
$$\begin{aligned} \text{left-below} &\ge k,\ \text{left-above} &\ge k,\ \text{right-below} &\ge k,\ \text{right-above} &\ge k. \end{aligned}$$
Let $P_L(y)$ and $P_R(y)$ be the numbers of cities in $L$ and $R$ whose $y$-coordinate is strictly smaller than $y$.
Then the conditions become
$$k \le P_L(y_0) \le |L|-k,$$
and
$$k \le P_R(y_0) \le |R|-k.$$
For each side, the valid values of $y_0$ form a contiguous interval because prefix counts are monotone. We only need to determine whether the interval for $L$ intersects the interval for $R$.
Fenwick trees allow us to maintain the prefix counts while sweeping $x$, and also to find the first position where a prefix reaches a given value.
| Approach | Time Complexity | Space Complexity | Verdict |
|---|---|---|---|
| Brute Force | $O(n^3)$ | $O(1)$ | Too slow |
| Binary Search + Sweep + Fenwick | $O(n \log^2 n)$ | $O(n)$ | Accepted |
Algorithm Walkthrough
Coordinate Compression
All operations depend only on the relative order of $y$-coordinates.
Compress all distinct $y$-values into indices $1 \ldots m$.
Feasibility Check for a Fixed $k$
- Put every point into the right structure $R$.
- The left structure $L$ starts empty.
- Sweep distinct $x$-coordinates in increasing order.
- Before processing a group with coordinate $x$, the split corresponds to $x_0=x$. Every point with smaller $x$ is already in $L$, and every point with $x\ge x_0$ remains in $R$.
- Let $tot_L$ and $tot_R$ be the current sizes of the two sides.
- If either side has fewer than $2k$ points, that side can never contribute at least $k$ points both above and below the horizontal line. Skip this sweep position.
- For $L$, find the interval of compressed $y$-positions where
$$k \le P_L \le tot_L-k.$$
Because prefix counts are monotone, this interval can be obtained with Fenwick order-statistics queries.
- Compute the corresponding interval for $R$.
- If the two intervals intersect, then a valid $y_0$ exists and the answer for this check is "possible".
- Move the entire current $x$-group from $R$ into $L$ and continue the sweep.
Binary Search
- The answer cannot exceed $\lfloor n/4 \rfloor$.
- Binary search $k$ in the range $[0,\lfloor n/4 \rfloor]$.
- Run the feasibility check.
- Keep the largest feasible $k$.
- Run the check once more for that value and store the corresponding $(x_0,y_0)$.
Why it works
For a fixed vertical split, every valid horizontal split must satisfy two independent constraints:
$$k \le P_L(y_0) \le |L|-k,$$
and
$$k \le P_R(y_0) \le |R|-k.$$
Since both prefix functions are monotone, each constraint defines a contiguous interval of possible $y_0$-positions. A horizontal split exists exactly when those intervals intersect.
The sweep enumerates every meaningful vertical split because the classification only changes when we pass a distinct $x$-coordinate. The binary search is valid because feasibility is monotone: if every quadrant can contain at least $k$ cities, then every smaller value is also achievable.
Python Solution
import sys
from bisect import bisect_left
input = sys.stdin.readline
class Fenwick:
def __init__(self, n):
self.n = n
self.bit = [0] * (n + 1)
def add(self, idx, val):
n = self.n
bit = self.bit
while idx <= n:
bit[idx] += val
idx += idx & -idx
def sum(self, idx):
res = 0
bit = self.bit
while idx:
res += bit[idx]
idx -= idx & -idx
return res
def kth(self, k):
idx = 0
bitmask = 1 << (self.n.bit_length() - 1)
while bitmask:
nxt = idx + bitmask
if nxt <= self.n and self.bit[nxt] < k:
k -= self.bit[nxt]
idx = nxt
bitmask >>= 1
return idx + 1
def solve():
t = int(input())
out = []
for _ in range(t):
n = int(input())
pts = [tuple(map(int, input().split())) for _ in range(n)]
ys = sorted(set(y for _, y in pts))
m = len(ys)
data = []
for x, y in pts:
data.append((x, bisect_left(ys, y) + 1))
data.sort()
groups = []
i = 0
while i < n:
j = i
while j < n and data[j][0] == data[i][0]:
j += 1
groups.append((data[i][0], data[i:j]))
i = j
def get_interval(ft, total, k):
if total < 2 * k:
return None
left = ft.kth(k)
limit = total - k
if limit == total:
right = m
else:
pos = ft.kth(limit + 1)
right = pos - 1
if left > right:
return None
return (left, right)
def check(k):
L = Fenwick(m)
R = Fenwick(m)
for _, y in data:
R.add(y, 1)
left_cnt = 0
right_cnt = n
for x0, grp in groups:
A = get_interval(L, left_cnt, k)
B = get_interval(R, right_cnt, k)
if A is not None and B is not None:
l = max(A[0], B[0])
r = min(A[1], B[1])
if l <= r:
y0 = ys[l - 1]
return True, x0, y0
for _, y in grp:
R.add(y, -1)
L.add(y, 1)
sz = len(grp)
left_cnt += sz
right_cnt -= sz
return False, 0, 0
lo, hi = 0, n // 4
best = 0
while lo <= hi:
mid = (lo + hi) // 2
ok, _, _ = check(mid)
if ok:
best = mid
lo = mid + 1
else:
hi = mid - 1
_, x0, y0 = check(best)
out.append(str(best))
out.append(f"{x0} {y0}")
sys.stdout.write("\n".join(out))
if __name__ == "__main__":
solve()
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | $O(n \log^2 n)$ | Binary search on $k$, each feasibility check performs a sweep with Fenwick order-statistic queries |
| Space | $O(n)$ | Compressed coordinates, grouped points, and Fenwick trees |
The total number of points across all test cases is at most $10^5$, so $O(n \log^2 n)$ easily fits within the limits. The memory usage is linear in the number of points.