1. Abridged Problem Statement  
Given N levels in an underwater station. Each level i has:  
• Wᵢ = initial water weight  
• Lᵢ = maximum water it can hold before leaking  
• Pᵢ = cost to manually depressurize (leak) this level  
When a level leaks—either automatically because accumulated water > Lᵢ or manually by paying Pᵢ—its water (plus any water coming from above) falls to the next level. You must cause the N-th level to leak (depressurize). Find a set of levels to manually depressurize so that, after all cascades, level N leaks, and the total paid cost is minimized. Output the 1-based indices of levels you manually depressurize, in increasing order.

2. Detailed Editorial  
We need to establish a cascade of leaks from some starting level s up through level N, so that water keeps falling until level N leaks. Model the process as follows: choose a first level s to trigger (we will manually depressurize it if it does not overflow automatically). Let accumulated water w = 0. Iterate i from s to N:  
  – Add Wᵢ to w (water that falls to level i).  
  – If w > Lᵢ, the level i leaks automatically—no cost. Otherwise, it holds the water unless we pay Pᵢ to depressurize it; since we need continuous leakage to reach N, we must pay Pᵢ in that case.  
Thus for a chosen s, the total cost is the sum of Pᵢ over all i ∈ [s..N] for which w (just after adding Wᵢ) ≤ Lᵢ. We want to pick s in [1..N] to minimize this cost. After finding the best s, we repeat the above sweep and record exactly those i where w ≤ Lᵢ, which are the indices we manually depressurize.

Complexity: a naive double loop over s and i would be O(N²). However, as soon as w exceeds 15 000 (the maximum Lᵢ), further levels will always overflow automatically, so we can break early. In practice this runs fast for N up to 15 000.

3. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Global variables: number of levels and array of triples (W, L, P)
int n;
vector<tuple<int,int,int>> a;

// Read input
void read() {
    cin >> n;
    a.resize(n);
    // For each level i, read Wᵢ, Lᵢ, Pᵢ
    for(auto &t : a) {
        int W, L, P;
        cin >> W >> L >> P;
        t = make_tuple(W, L, P);
    }
}

void solve() {
    // Initialize answer cost to the (worst) cost of just starting at the last level
    int bestCost = get<2>(a[n-1]);
    int bestStart = n-1;  // zero-based index for the best starting level

    // Try every possible starting level s from 0 to n-1
    for(int s = 0; s < n; s++) {
        int w = 0;        // accumulated falling water
        int cost = 0;     // cost if we start at s

        // Simulate cascade from level s to n-1
        for(int i = s; i < n; i++) {
            auto [W, L, P] = a[i];
            w += W;       // water arrives at level i

            if(w <= L) {
                // It would hold unless we pay to depressurize
                cost += P;
            }
            // If w > L, it leaks automatically (no added cost)

            // Optimization: if w exceeds maximum possible L, break early
            if(w > 15000) break;
        }

        // Update the best starting point if this cost is lower
        if(cost < bestCost) {
            bestCost = cost;
            bestStart = s;
        }
    }

    // Re-simulate from bestStart to collect which levels we paid for
    int w = 0;
    vector<int> answer;
    for(int i = bestStart; i < n; i++) {
        auto [W, L, P] = a[i];
        w += W;
        if(w <= L) {
            // We had to pay here
            answer.push_back(i + 1);  // convert to 1-based index
        }
    }

    // Output the chosen levels, one per line
    for(int idx : answer) {
        cout << idx << "\n";
    }
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
data = sys.stdin.read().split()
# Parse input
n = int(data[0])
# Store levels as tuples (W, L, P)
levels = []
ptr = 1
for _ in range(n):
    W = int(data[ptr]); L = int(data[ptr+1]); P = int(data[ptr+2])
    ptr += 3
    levels.append((W, L, P))

best_cost = float('inf')
best_start = 0

# Try each starting level s
for s in range(n):
    w = 0       # accumulated water
    cost = 0    # total paid cost starting at s

    # Cascade simulation from s to n-1
    for i in range(s, n):
        W, L, P = levels[i]
        w += W
        if w <= L:
            # if it wouldn't leak automatically, pay cost
            cost += P
        # if w > L, it leaks automatically, no cost
        if w > 15000:
            # further L's are ≤ 15000, so all will leak automatically
            break

    # Keep the best starting point
    if cost < best_cost:
        best_cost = cost
        best_start = s

# Re-simulate from best_start to list the paid levels
w = 0
result = []
for i in range(best_start, n):
    W, L, P = levels[i]
    w += W
    if w <= L:
        # we paid at this level
        result.append(str(i+1))  # convert to 1-based index

# Print results, one index per line
sys.stdout.write("\n".join(result))

```

5. Compressed Editorial  
Try every possible first level s to trigger the leak. Maintain a running sum w of falling water. For each level i ≥ s, add Wᵢ to w; if w > Lᵢ it leaks automatically, otherwise you must pay Pᵢ to force a leak. Sum these payments to get the cost for start s. Track the minimum over all s. Finally, rerun for the best s to output exactly those levels where w ≤ Lᵢ (the ones you paid for). Early break when w > max(Lᵢ) yields acceptable performance for N up to 15 000.