1. Abridged Problem Statement  
Given n fence boards of width 1 and integer heights h₁…hₙ. The perimeter of the fence (joining the rectangles) is  
  P = 2 + Σ_{i=1 to n–1}(2 + |hᵢ – hᵢ₊₁|) + h₁ + hₙ.  
You may steal (remove) some boards and reassemble the remaining in order. Let P′ be the perimeter of the new (smaller) fence computed the same way. You must ensure P′ ≥ ⌈P/2⌉. Under this constraint, maximize the total height of stolen boards. Output that maximum sum, the count of stolen boards, and their indices (1-based).

2. Detailed Editorial  

Overview  
We want to choose a subset S of boards to steal so that the perimeter of the remaining boards is not less than half the original. Equivalently, we choose which boards to keep and track two quantities:  
  • The sum of heights stolen (we want to maximize this).  
  • The perimeter contributed by the kept boards (we want this ≥ ⌈P/2⌉).  

Compute the original perimeter P in O(n). Let T = ⌈P/2⌉.

Dynamic Programming  
Define DP[pos][last_h][per] = maximum total stolen height after processing the first pos boards, where:  
  – last_h = height of the last kept board (or 0 if none kept yet),  
  – per = perimeter accumulated so far from kept boards, *excluding* the rightmost vertical side.  
Initialize DP[0][0][0] = 0, all others = –∞ (or –1 to denote impossible).

Transition for board at position pos (0-based, height = h = heights[pos]):  
  1. **Steal it**:  
     DP[pos+1][last_h][per] = max(DP[pos+1][last_h][per], DP[pos][last_h][per] + h)  
     (stolen sum increases by h, perimeter unchanged)
  2. **Keep it**:  
     new_per = per + 2 + |last_h – h|  
     DP[pos+1][h][new_per] = max(DP[pos+1][h][new_per], DP[pos][last_h][per])  
     (no change to stolen sum)

After filling up to pos = n, search over states DP[n][last_h][per] such that per + last_h ≥ T (we add last_h for the rightmost side). Pick the state with maximum stolen sum.

Reconstruction  
Store back-pointers during the DP (or recompute by checking which transition led to the optimal). Walk backwards from pos = n, the chosen (last_h, per) to pos = 0, recording which boards were stolen.

Complexities  
n ≤ 50, heights ≤ 100, original P ≤ ~5200 ⇒ DP dimension ~ 51×101×5300 ≈ 27 million states. Each has two transitions ⇒ ~54M updates. Fit into 0.25 s in optimized C++ using int16 arrays.

3. C++ Solution with Line-by-Line Comments  

```cpp
#include <algorithm>
#include <cmath>
#include <iostream>
#include <tuple>
#include <vector>
using namespace std;

// Maximum board height + 1 and an upper bound on possible perimeter
const int16_t MAX_HEIGHT = 101;
const int16_t MAX_PERIMETER = 5300;

// Compute the original fence perimeter from all heights
int16_t calculate_initial_perimeter(const vector<int16_t>& h) {
    int16_t per = 2;  // start with top+bottom edge of the first board
    int n = h.size();
    // for each adjacent pair, add horizontal top+bottom (2) and vertical difference
    for(int16_t i = 0; i < n - 1; ++i)
        per += 2 + abs(h[i] - h[i+1]);
    // add the left side (h[0]) and right side (h[n-1])
    per += h[0] + h[n-1];
    return per;
}

// Trace back through dp to find which boards were stolen
vector<int16_t> reconstruct_stolen_boards(
    const vector<vector<vector<int16_t>>>& dp,
    const vector<int16_t>& h,
    int16_t cur_h, int16_t cur_per) 
{
    vector<int16_t> stolen;
    int n = h.size();
    // walk backward from pos = n down to 1
    for(int16_t pos = n; pos > 0; --pos) {
        int16_t val = dp[pos][cur_h][cur_per];
        // if we arrived here by stealing board pos-1:
        if(cur_h >= 0 && dp[pos-1][cur_h][cur_per] + h[pos-1] == val) {
            stolen.push_back(pos);  // record its 1-based index
            // state remains (cur_h, cur_per)
        } else {
            // otherwise we must have kept board pos-1,
            // so find which previous height led here
            for(int16_t prev_h = 0; prev_h < MAX_HEIGHT; ++prev_h) {
                int16_t prev_per = cur_per - (2 + abs(prev_h - cur_h));
                if(prev_per >= 0 &&
                   dp[pos-1][prev_h][prev_per] == val) {
                    cur_h = prev_h;
                    cur_per = prev_per;
                    break;
                }
            }
        }
    }
    sort(stolen.begin(), stolen.end());
    return stolen;
}

// Main solver: returns (max_stolen_sum, count, list_of_indices)
tuple<int16_t,int16_t,vector<int16_t>>
solve_board_theft(int16_t n, const vector<int16_t>& h) {
    // 3D DP: dp[pos][last_h][per] = max stolen sum, or -1 if impossible
    vector dp(n+1,
        vector<vector<int16_t>>(MAX_HEIGHT,
            vector<int16_t>(MAX_PERIMETER, -1)));
    dp[0][0][0] = 0;  // no boards taken, last height = 0, per = 0

    int16_t initP = calculate_initial_perimeter(h);
    // target half perimeter (rounded up)
    int16_t target = (initP + 1) / 2;

    // Build up the DP
    for(int16_t pos = 0; pos < n; ++pos) {
        int16_t height = h[pos];
        for(int16_t last_h = 0; last_h < MAX_HEIGHT; ++last_h) {
            for(int16_t per = 0; per < MAX_PERIMETER; ++per) {
                int16_t cur = dp[pos][last_h][per];
                if(cur < 0) continue;  // impossible state

                // Option 1: steal this board
                dp[pos+1][last_h][per] = max<int16_t>(
                    dp[pos+1][last_h][per],
                    cur + height
                );

                // Option 2: keep this board
                int16_t new_per = per + 2 + abs(last_h - height);
                if(new_per < MAX_PERIMETER) {
                    dp[pos+1][height][new_per] = max<int16_t>(
                        dp[pos+1][height][new_per],
                        cur
                    );
                }
            }
        }
    }

    // Find best end state satisfying kept_per + last_h ≥ target
    int16_t best_sum = -1, best_last_h = 0, best_per = 0;
    for(int16_t last_h = 0; last_h < MAX_HEIGHT; ++last_h)
        for(int16_t per = 0; per < MAX_PERIMETER; ++per) {
            int16_t val = dp[n][last_h][per];
            if(val >= 0 && per + last_h >= target && val > best_sum) {
                best_sum = val;
                best_last_h = last_h;
                best_per = per;
            }
        }

    // Reconstruct stolen boards
    auto stolen = reconstruct_stolen_boards(
        dp, h, best_last_h, best_per
    );
    return {best_sum,
            static_cast<int16_t>(stolen.size()),
            stolen};
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int16_t n;
    cin >> n;
    vector<int16_t> heights(n);
    for(int16_t i = 0; i < n; ++i)
        cin >> heights[i];

    auto [sumH, k, stolen] = solve_board_theft(n, heights);

    cout << sumH << "\n"
         << k << "\n";
    for(auto idx : stolen)
        cout << idx << " ";
    cout << "\n";
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
# Impudent Thief in Python: DP with backpointers via dictionaries.
# Note: This is a direct translation of the DP idea. For n=50 and P~5300,
# it may be slow in CPython but illustrates the approach.

import sys
from math import ceil

def main():
    input_data = sys.stdin.read().split()
    n = int(input_data[0])
    heights = list(map(int, input_data[1:]))

    # Calculate initial perimeter
    P = 2
    for i in range(n-1):
        P += 2 + abs(heights[i] - heights[i+1])
    P += heights[0] + heights[-1]
    target = (P + 1) // 2

    MAX_PER = P + 1  # we'll never exceed P
    # dp maps (last_h, per) -> max stolen sum
    dp = {(0,0): 0}
    # back pointers: (pos, last_h, per) -> (prev_last_h, prev_per, stole?)
    back = {}

    for pos, h in enumerate(heights, start=1):
        dp_next = {}
        for (last_h, per), stolen_sum in dp.items():
            # Option 1: steal this board
            key1 = (last_h, per)
            new_sum1 = stolen_sum + h
            if dp_next.get(key1, -1) < new_sum1:
                dp_next[key1] = new_sum1
                back[(pos, key1)] = (last_h, per, True)

            # Option 2: keep this board
            new_per = per + 2 + abs(last_h - h)
            if new_per <= MAX_PER:
                key2 = (h, new_per)
                if dp_next.get(key2, -1) < stolen_sum:
                    dp_next[key2] = stolen_sum
                    back[(pos, key2)] = (last_h, per, False)

        dp = dp_next

    # Find best end state meeting the perimeter constraint
    best = (-1, None)  # (stolen_sum, (last_h, per))
    for (last_h, per), stolen_sum in dp.items():
        if per + last_h >= target and stolen_sum > best[0]:
            best = (stolen_sum, (last_h, per))

    best_sum, state = best
    last_h, per = state

    # Reconstruct stolen boards
    stolen_indices = []
    pos = n
    while pos > 0:
        prev_h, prev_per, did_steal = back[(pos, (last_h, per))]
        if did_steal:
            stolen_indices.append(pos)  # 1-based
        # move to previous
        last_h, per = prev_h, prev_per
        pos -= 1

    stolen_indices.reverse()

    # Output
    print(best_sum)
    print(len(stolen_indices))
    print(*stolen_indices)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  

- Compute the original fence perimeter P.  
- Let T = ceil(P/2).  
- DP over positions: state (last_kept_height, current_perimeter) → max stolen-sum.  
- Two transitions per board: “steal” (add height to sum, perimeter unchanged) or “keep” (increase perimeter by 2+|Δh|, sum unchanged).  
- Final valid states satisfy current_perimeter + last_kept_height ≥ T.  
- Pick the one with maximum stolen-sum; backtrack to find stolen indices.