USACO 1352: Target Practice Analysis

USACO Silver 2023 Dec

Problem 3: Target Practice

Algorithmic Analysis & Solution

Topic: Simulation, Suffix Sums, Ad-hoc

1. Problem Summary

Input: $T$ target positions, $C$ commands string (L, R, F).
Actions:
- $L$: Move left ($pos \leftarrow pos - 1$)
- $R$: Move right ($pos \leftarrow pos + 1$)
- $F$: Fire (Hit target if one exists at current $pos$)
Constraint: You can change exactly one command in the string.
Goal: Maximize the number of distinct targets hit.

2. Challenges & Approach

Brute Force?

Iterate every command $i$, change it to the other 2 types, and re-simulate.

Complexity: $O(C^2)$. With $C \le 10^5$, this will Time Out (TLE).

Optimized Approach ($O(C)$):

Key Observation: Global Shift.

Changing command $i$ affects the absolute position of all steps $j > i$.
However, the relative movement between steps after $i$ remains unchanged.

3. Visualizing the Path Shift

Changing a move creates a constant offset for the entire future path.

Example: Changing $L$ (-1) to $R$ (+1) increases the position by 2 for all future steps.

4. The 5 Possible Offsets

We need to pre-calculate how many targets the suffix hits under 5 distinct shifts.

Original	New	Pos Change	Global Offset	Map Index
R (+1)	L (-1)	$-2$	Left 2	0
R (+1) / F (0)	F (0) / L (-1)	$-1$	Left 1	1
Any	Same	$0$	No Change	2
L (-1) / F (0)	F (0) / R (+1)	$+1$	Right 1	3
L (-1)	R (+1)	$+2$	Right 2	4

5. Algorithm Design

1. Offline Pre-calculation (Suffix):

We use 5 Maps to simulate the future:

map<int, int> right[5]
Key: Target Position. Value: Reference Count (how many times hit).
Stores hits assuming the path is shifted by $k \in \{-2, -1, 0, 1, 2\}$.

2. Simulation Scan (Prefix):

set<int> left: Unique targets hit before current index.
Iterate $i$ from $0$ to $C-1$. At each step, test changing the command.
Result = left.size() + right[shift_index].size().

6. The "Claiming" Mechanism

To count distinct targets, we must avoid double counting between the Prefix (left) and Suffix (right).


// If current command is 'F' and it hits a target P
if (targets.count(p) && left[p] == 0) {       
    // 1. Claim it for the "Left" (past) set
    left[p]++;
    
    // 2. Remove it from ALL "Right" (future) maps
    // Since we hit it now, future hits on this target add no value
    for (int j = 0; j < 5; j++) right[j].erase(p);
}

Greedy Removal: Once a target is claimed by the prefix, we erase it from the suffix maps entirely. This ensures left.size() + right.size() is always accurate.

7. Complete Solution (C++)


#include <bits/stdc++.h>
using namespace std;

int t, c;
char s[100005];
set<int> targets;
// left: Targets hit in the past (Prefix)
map<int,int> left_hits; 
// right[5]: Targets hit in the future (Suffix) under offsets -2 to +2
// Indexes: 0:(-2), 1:(-1), 2:(0), 3:(+1), 4:(+2)
map<int,int> right_hits[5];     
int ans;

int main() {
    scanf("%d %d", &t, &c);
    for (int i = 0; i < t; i++) {
        int p; scanf("%d", &p); targets.insert(p);
    }
    scanf("%s", s);
    int p = 0;
    
    // --- Step 1: Pre-calculate Suffixes ---
    for (int i = 0; i < c; i++) {
        if (s[i] == 'L') p--;
        if (s[i] == 'R') p++;
        if (s[i] == 'F') {
            // Record hits for all possible future offsets
            if (targets.count(p-2)) right_hits[0][p-2]++;
            if (targets.count(p-1)) right_hits[1][p-1]++;
            if (targets.count(p))   right_hits[2][p]++;
            if (targets.count(p+1)) right_hits[3][p+1]++;
            if (targets.count(p+2)) right_hits[4][p+2]++;
        }
    }

    // --- Step 2: Simulate Prefix & Try Changes ---
    p = 0;
    for (int i = 0; i < c; i++) {
        int current_score = left_hits.size();
        // Check if changing to F hits a *new* target right now
        int hit_if_change_to_F = (left_hits.count(p) == 0 && targets.count(p)) ? 1 : 0; 

        if (s[i] == 'L') {
            // Option A: Change L -> F (Global Shift +1)
            // Note: If right_hits has p, we rely on the map size.
            if (right_hits[3].count(p)) 
                ans = max(ans, current_score + (int)right_hits[3].size());
            else 
                ans = max(ans, current_score + hit_if_change_to_F + (int)right_hits[3].size());
                
            // Option B: Change L -> R (Global Shift +2)
            ans = max(ans, current_score + (int)right_hits[4].size());
            p--; // Execute actual move L
        } 
        else if (s[i] == 'R') {
            // Option A: Change R -> F (Global Shift -1)
            if (right_hits[1].count(p))
                ans = max(ans, current_score + (int)right_hits[1].size());
            else
                ans = max(ans, current_score + hit_if_change_to_F + (int)right_hits[1].size());

            // Option B: Change R -> L (Global Shift -2)
            ans = max(ans, current_score + (int)right_hits[0].size());
            p++; // Execute actual move R
        } 
        else { // s[i] == 'F'
            // We are passing this 'F', so remove its contribution from the FUTURE maps
            // Decrement counts; erase if count reaches 0
            for (int j = 0; j < 5; j++)
                if (right_hits[j].count(p+j-2)) {
                    if (--right_hits[j][p+j-2] == 0) right_hits[j].erase(p+j-2);
                }
            
            // Try changing F -> L (Shift -1) or F -> R (Shift +1)
            ans = max(ans, current_score + (int)right_hits[1].size()); 
            ans = max(ans, current_score + (int)right_hits[3].size()); 

            // Execute actual F: Claim target if valid
            if (targets.count(p) && left_hits.count(p) == 0) {
                left_hits[p]++;
                // Crucial: Remove claimed target from ALL future maps
                for (int j = 0; j < 5; j++) right_hits[j].erase(p);
            }
        }
    }
    // Finally, consider making NO changes
    ans = max(ans, (int)left_hits.size());

    printf("%d\n", ans);
    return 0;
}

8. Summary

Time Complexity:
- Pre-calculation: $O(C \log T)$ (due to Map ops).
- Simulation: $O(C \log T)$.
- Total: $O(C \log T)$.
Space Complexity: $O(T)$ to store the maps.
Key Takeaways:
1. Identify how local changes affect global state (Offsets).
2. Use Reference Counting for "Offline" suffix data.
3. Use Greedy "Claiming" to handle unique counts properly.