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. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<tuple<int, int, int>> a;

void read() {
    cin >> n;
    a.resize(n);
    for(auto& [x, y, z]: a) {
        cin >> x >> y >> z;
    }
}

void solve() {
    // We pick a single top level to start the chain reaction by paying to
    // depressurize it: its water W pours onto the next level, and from there
    // on the falling water w_fall accumulates downward. At each level, if the
    // accumulated water already exceeds the level's capacity L it breaks for
    // free and we keep falling; otherwise we must pay P to break it (and add
    // its own W to the cascade). The total cost of a given start is the sum of
    // the P of the levels we had to pay for; once w_fall exceeds the maximum
    // possible capacity (15000) the rest collapses for free, so we stop early.
    // We try every start, keep the cheapest, and replay it to list which
    // levels we ended up paying to depressurize.

    int ans = get<2>(a[n - 1]);
    int best_pos = n - 1;

    for(int start = 0; start < n; start++) {
        int w_fall = 0, candidate = 0;
        for(int i = start; i < n; i++) {
            auto [x, y, z] = a[i];
            w_fall += x;
            if(w_fall <= y) {
                candidate += z;
            } else if(w_fall > 15000) {
                break;
            }
        }

        if(candidate < ans) {
            ans = candidate;
            best_pos = start;
        }
    }

    int w_fall = 0;
    vector<int> best;
    for(int i = best_pos; i < n; i++) {
        auto [x, y, z] = a[i];
        w_fall += x;
        if(w_fall <= y) {
            best.push_back(i + 1);
        }
    }

    for(auto x: best) {
        cout << x << '\n';
    }
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        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.