Target Practice II

USACO 2024 Feb Silver

Solution & Algorithm Analysis

Difficulty: Silver / Hard

1. Problem Overview

  • Given $N$ rectangular targets ($X_1, x_2[i], y_1[i], y_2[i]$).
  • Goal: Position 4N cows, each with its slope ($s[i]$), on the Y-axis. Each cow shoots at one corner of a target.
  • Constraint: Shots must not pass through the interior of any target.
  • Objective: Minimize $\max(y_{cow}) - \min(y_{cow})$.
  • $N < 4 \times 10^4$.
Target

2. Geometric Formulation

Cow at $y_c$, target point $(x, y)$, slope $s$.

Linear Equation: $$y_c = y - s \cdot x$$

Key Observation:
The failure condition (hitting the interior) is determined solely by the right-side corners of the rectangle.
  • If $s < 0$ (Shooting Down): To avoid the interior, the shot must hit top right corner (TR).
  • If $s > 0$ (Shooting Up): The shot must hit bottom right corner (BR).

Constraint Visualization

Target Interior TR (x2, y2) BR (x2, y1) s < 0: Must satisfy constraints of Top-Right s > 0: Must satisfy constraints of Bottom-Right

3. The Splitting Strategy

The problem can be decoupled into two independent parts:

  • Negative Group ($s < 0$): Determines the lower bound of the upper cow range (Minimum $y$ for negative slopes).
  • Positive Group ($s > 0$): Determines the upper bound of the lower cow range (Maximum $y$ for positive slopes).

The Challenge: What about the points on the left edge $(x_1)$?

$\rightarrow$ Assign them greedily!

Assigning Left-Edge Points

To minimize the total range ($\max - \min$), we want the Positive Group to start as high as possible and the Negative Group as low as possible (conceptually separating them).

  1. Collect all $2N$ left-edge Y-coordinates.
  2. Sort them.
  3. Lower Half: Assign to Negative Slope group.
  4. Upper Half: Assign to Positive Slope group.
Sorted Left Ys To Pos Group To Neg Group

4. Solving for One Group

Focus on the Positive Slope Group:

  • Set of points $P$: Includes all Target BR corners + Upper half of Left Edge points.
  • Set of slopes $S$: All available positive slopes.
  • Goal: Find the maximum possible $y_{cow}$ such that every point $p \in P$ can be matched to a unique slope $s \in S$ satisfying: $$y_{cow} \le y_p - s \cdot x_p$$
Rearranged: $s \le \frac{y_p - y_{cow}}{x_p}$

Algorithm: Binary Search on Answer

Since slopes and coordinates are integers, we can binary search for $y_{cow}$.

  1. Binary Search for $y_{mid}$ (Potential cow position).
  2. Check($y_{mid}$):
    • For each point $p$, calculate max allowed slope: $s_{req} = \lfloor (y_p - y_{mid}) / x_p \rfloor$.
    • Greedy Logic: To save "easier" slopes for harder points, the current point should consume the largest possible slope $\le s_{req}$.
    • Implementation: Use `multiset` and `upper_bound` to find and erase the optimal slope.

Time Complexity: $O(N \log N \cdot \log(\text{Range}))$.

5. The Symmetry Trick

We don't need to write separate logic for negative slopes.

$$y_c = y - s \cdot x \iff (-y_c) = (-y) - (-s) \cdot x$$

Steps for Negative Group:
  1. Negate all target Y-coordinates: $y' = -y$.
  2. Negate all negative slopes to make them positive: $s' = -s$.
  3. Reuse the solve() function used for the positive group.
  4. Negate the result back: $ans = -solve()$.

Final Answer: $Ans_{total} = (-Ans_{neg}) - Ans_{pos}$

Code: The Check Function


bool check(vector<pi> &pts, vector<int> &slopes, long long y) {
    multiset<int> ss;
    for (auto s: slopes) ss.insert(s);
    
    for (auto p: pts) {
        // Calculate max allowable slope for this point given cow at y
        long long s0 = (p.second - y) / p.first;
        
        // Find largest slope in set <= s0
        auto it = ss.lower_bound(s0);
        if (it == ss.end() || *it > s0) {
            if (it == ss.begin()) return false; // No valid slope found
            it--;
        }
        ss.erase(it); // Use this slope
    }
    return true;
}

Code: Binary Search

// Maximize the minimum y for a set of points and positive slopes
ll solve(vector<pi> &pts, vector<int> &slopes) {
    long long l = -2e14, r = 2e14; // Sufficiently large range
    long long ans = -2e14;
    
    while (l <= r) {
        long long mid = (l + r) / 2;
        // Adjust for negative division behavior in C++ if necessary
        if (l+r < 0 && (l+r)%2 != 0) mid = (l+r-1)/2; 
        
        if (check(pts, slopes, mid)) {
            ans = mid;
            l = mid + 1; // Try to go higher
        } else {
            r = mid - 1;
        }
    }
    return ans;
}

Code: Main Logic

// 1. Separate Slopes
for (int i=0; i<4*n; i++) {
    if (s[i] > 0) slopes[0].push_back(s[i]);
    else slopes[1].push_back(-s[i]); // Store as positive for reuse
}

// 2. Assign Corner Points (Top-Right vs Bottom-Right)
for (int i=0; i<n; i++) {
    pts[0].push_back({x2[i], y1[i]});      // To Pos Group
    pts[1].push_back({x2[i], -y2[i]});     // To Neg Group (Negated)
}

// 3. Greedy Assignment of Left Edge
vector<int> y_all;
for (int i=0; i<n; i++) { y_all.push_back(y1[i]); y_all.push_back(y2[i]); }
sort(y_all.begin(), y_all.end());

for (auto y: y_all) {
    if (pts[1].size() < 2*n) pts[1].push_back({x1, -y}); // Lower halves
    else pts[0].push_back({x1, y}); // Upper halves
}

Summary

  • Key Insight: The problem splits into "Upper Constraints" vs "Lower Constraints".
  • Greedy 1 (Data): Sort left-edge points; small ones go to the negative slope group, large ones to the positive slope group.
  • Greedy 2 (Matching): In the check function, always use the largest valid slope to preserve smaller slopes for stricter constraints.
  • Technique: Binary Search on the Answer converts the optimization problem into a validity check.
Result = (-solve(neg_part)) - solve(pos_part)