## 1) Abridged problem statement

There are **n (≤ 10)** participants and **m (≤ 1000)** consecutive river segments (riffles) between points \(p_0 \to p_1 \to \dots \to p_m\).

For riffle \(i\):
- If the total weight of people **on the raft** exceeds critical weight \(c_i\), the raft **capsizes** and takes \(D_i\) minutes to go from \(p_{i-1}\) to \(p_i\).
- Otherwise it takes \(d_i\) minutes.

For participant \(j\):
- If they **walk** from \(p_{i-1}\) to \(p_i\), it takes \(t_j\) minutes.
- Each time they **change** between bank and raft (either get on or get off), it costs \(s_j\) minutes.

Before each riffle, you choose who rides and who walks. Everyone must arrive at \(p_i\), then you may perform swaps (paying sum of \(s_j\) for all who change), then proceed. The segment time is the time until **both** raft and all walkers arrive, i.e. the maximum of raft travel time and the slowest walker.

Constraints/requirements:
- Start at \(p_0\) with **everyone on the bank** (raft present).
- Finish at \(p_m\) with **everyone on the bank** (raft also there).
- The raft must never travel empty (so on each riffle at least one person rides).

Compute the **minimum total time**.

---

## 2) Detailed editorial (solution explanation)

### Key observation: the only "state" is who is on the raft
At each point \(p_i\), the situation is fully described by the set of people currently **on the raft** (the rest are on the bank). Since \(n \le 10\), we can represent this set as a bitmask of size \(2^n\).

Let:
- `mask` = bitmask of people on the raft at the current point.
- `0` means everyone on the bank.
- We are not allowed to traverse a riffle with `mask = 0` (raft cannot move empty).

We need the shortest time from:
- start state: at \(p_0\), `mask = 0`
to:
- end state: at \(p_m\), `mask = 0`

### Segment cost when traversing riffle i with a fixed mask
Suppose at point \(p_{i-1}\) we have chosen `mask` (people on raft) and then traverse riffle \(i\).

1) Raft travel time:
- compute total raft weight \(W(mask)\)
- if \(W(mask) > c_i\), raft time = \(D_i\) else \(d_i\)

2) Walkers' travel time:
- walkers are those not in mask
- all walkers move in parallel, so the group is ready only when the slowest walker arrives:
  \[
  T_{\text{walk}}(mask) = \max_{j \notin mask} t_j
  \]
  (If everyone is on the raft, this max is 0.)

3) Because everyone must wait for everyone, the segment duration is:
\[
T_{\text{seg}}(i, mask)=\max(\text{raft time}, T_{\text{walk}}(mask))
\]

So, for each `mask`, we can compute segment time independently.

### Cost for changing composition (swapping people) at a point
At point \(p_i\), you may change who is on the raft before starting the next riffle. If you change from `maskA` to `maskB`, anyone whose bit differs must move between bank and raft, costing their \(s_j\). That is exactly:
\[
\text{swapCost}(maskA, maskB)=\sum_{j \in (maskA \oplus maskB)} s_j
\]
This is like a graph on masks where edges flip one bit with cost \(s_j\).

### Why Dijkstra?
You might try dynamic programming per riffle, but at a fixed point \(p_i\), you can perform swaps in any order, possibly multiple changes; this forms cycles among masks. The clean approach: view mask transitions as a weighted graph and run shortest paths.

However, we also have progression over riffles \(i = 1..m\). The solution does:

For each riffle \(i\):
1) We start with current best times `prev_dp[mask]` meaning: minimum time to be at point \(p_{i-1}\) with raft set `mask`.
2) Before traversing riffle \(i\), we can do any swaps at \(p_{i-1}\). This is exactly shortest paths over the mask graph (bit flips), starting from the initial distances `prev_dp`. Compute `dist[mask]` = best time after arbitrary swaps, still at \(p_{i-1}\) but ready to depart with raft set `mask`.
   - This is done via Dijkstra over `2^n` nodes, each node has `n` neighbors (flip each bit).
3) Now traverse riffle \(i\) with some non-empty `mask`:
   \[
   next\_dp[mask] = dist[mask] + T_{\text{seg}}(i, mask), \quad mask \neq 0
   \]
   (Because raft cannot be empty during traversal.)
4) Set `prev_dp = next_dp` and proceed.

After processing all m riffles, we are at point \(p_m\) but may still have people on the raft. We must end with everyone on the bank (`mask=0`), so we run one final Dijkstra over mask flips at \(p_m\) to allow everyone to get off:
- answer = shortest distance to `mask=0`.

### Precomputation for speed
Since \(2^n \le 1024\), we can precompute for every mask:
- `sum_w[mask] = W(mask)`
- `max_out[mask] = max t_j among j not in mask` (0 if mask includes all)

Then each riffle transition per mask is O(1).

### Complexity
Let \(N = 2^n \le 1024\).
Each Dijkstra run:
- nodes \(N\), edges \(N \cdot n\)
- complexity \(O(N n \log N)\)

We run it:
- once per riffle (m times) + one final time
So total:
\[
O((m+1)\, N\, n \log N)
\]
With \(m \le 1000\), \(N \le 1024\), \(n \le 10\), this is easily fast enough.

---

## 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;
};

const int64_t inf = (int64_t)1e18 + 42;

int n, m;
vector<int> w, t, s;
vector<int> c, D, d;

void read() {
    cin >> n >> m;
    w.resize(n);
    t.resize(n);
    s.resize(n);
    for(int j = 0; j < n; j++) {
        cin >> w[j] >> t[j] >> s[j];
    }
    c.resize(m);
    D.resize(m);
    d.resize(m);
    for(int i = 0; i < m; i++) {
        cin >> c[i] >> D[i] >> d[i];
    }
}

void solve() {
    // We first notice that n <= 10, meaning that something like 2^n is feasible
    // as one of the terms in the complexity. That matches the unique sets that
    // can be in the boat at any time, meaning we can try to encode the full
    // state. Particularly, the set is defined by the people currently in the
    // boat, and the current raft we are at. At every moment we can either
    // remove or add some people to the mask, or proceed to the next raft. This
    // is O(n) transitions which is sufficiently small, but the issue is that we
    // can't do DP as there are cycles (due to the mask transitions). However,
    // we can think of this as a graph problem instead, with states being the
    // same. Then running Dijkstra would be sufficiently fast for the given
    // constraints. There are just some details in terms of the transitions.
    // One small optimization is to do M independent runs of Dijkstra over the
    // 2^n dimension, as that's the only place where we could have cycles.

    int N = 1 << n;
    vector<int> sum_w(N), max_out(N);
    for(int mask = 0; mask < N; mask++) {
        int sw = 0, mo = 0;
        for(int j = 0; j < n; j++) {
            if(mask & (1 << j)) {
                sw += w[j];
            } else {
                mo = max(mo, t[j]);
            }
        }
        sum_w[mask] = sw;
        max_out[mask] = mo;
    }

    vector<int64_t> prev_dp(N, inf);
    prev_dp[0] = 0;

    auto run_dijkstra = [&](vector<int64_t>& dist) {
        priority_queue<
            pair<int64_t, int>, vector<pair<int64_t, int>>, greater<>>
            pq;
        for(int mask = 0; mask < N; mask++) {
            if(dist[mask] < inf) {
                pq.push({dist[mask], mask});
            }
        }
        while(!pq.empty()) {
            int64_t time = pq.top().first;
            int umask = pq.top().second;
            pq.pop();
            if(time > dist[umask]) {
                continue;
            }
            for(int j = 0; j < n; j++) {
                int vmask = umask ^ (1 << j);
                int64_t new_time = time + s[j];
                if(new_time < dist[vmask]) {
                    dist[vmask] = new_time;
                    pq.push({new_time, vmask});
                }
            }
        }
    };

    for(int riff = 0; riff < m; riff++) {
        vector<int64_t> dist(N, inf);
        for(int mask = 0; mask < N; mask++) {
            dist[mask] = prev_dp[mask];
        }
        run_dijkstra(dist);

        vector<int64_t> next_dp(N, inf);
        for(int mask = 1; mask < N; mask++) {
            if(dist[mask] < inf) {
                bool capsize = sum_w[mask] > c[riff];
                int raft_t = capsize ? D[riff] : d[riff];
                int seg_t = max(raft_t, max_out[mask]);
                next_dp[mask] = dist[mask] + seg_t;
            }
        }
        prev_dp = next_dp;
    }

    vector<int64_t> dist(N, inf);
    for(int mask = 0; mask < N; mask++) {
        dist[mask] = prev_dp[mask];
    }
    run_dijkstra(dist);

    cout << dist[0] << endl;
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach, with detailed comments)

```python
import sys
import heapq

INF = 10**30

def solve() -> None:
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    n = int(next(it))
    m = int(next(it))

    # Participant parameters
    w = [0] * n  # weight
    t = [0] * n  # walking time per segment
    s = [0] * n  # swap time to get on/off raft
    for j in range(n):
        w[j] = int(next(it))
        t[j] = int(next(it))
        s[j] = int(next(it))

    # Riffle parameters
    c = [0] * m  # critical weight
    D = [0] * m  # time if capsizes
    d = [0] * m  # time if not
    for i in range(m):
        c[i] = int(next(it))
        D[i] = int(next(it))
        d[i] = int(next(it))

    N = 1 << n  # number of masks

    # Precompute:
    # sum_w[mask]  = total weight of people on raft
    # max_out[mask]= max walking time among walkers (people NOT in mask)
    sum_w = [0] * N
    max_out = [0] * N
    for mask in range(N):
        sw = 0
        mo = 0
        for j in range(n):
            if mask & (1 << j):
                sw += w[j]
            else:
                if t[j] > mo:
                    mo = t[j]
        sum_w[mask] = sw
        max_out[mask] = mo

    def run_dijkstra(dist):
        """
        Dijkstra on the mask graph:
        node = mask
        edge: mask -> mask^(1<<j) with cost s[j]
        Starts with initial distances in 'dist' and relaxes all masks.
        """
        pq = []
        for mask in range(N):
            if dist[mask] < INF:
                heapq.heappush(pq, (dist[mask], mask))

        while pq:
            cur, umask = heapq.heappop(pq)
            if cur != dist[umask]:
                continue

            # Try toggling each bit j
            for j in range(n):
                vmask = umask ^ (1 << j)
                nd = cur + s[j]
                if nd < dist[vmask]:
                    dist[vmask] = nd
                    heapq.heappush(pq, (nd, vmask))

    # DP over points: prev_dp[mask] = best time at current point with raft set mask
    prev_dp = [INF] * N
    prev_dp[0] = 0  # start with everyone on bank

    for i in range(m):
        # Allow any swapping at point p_{i} before traversing riffle i+1
        dist = prev_dp[:]        # copy
        run_dijkstra(dist)

        # Traverse the riffle with a non-empty raft
        next_dp = [INF] * N
        for mask in range(1, N):
            if dist[mask] >= INF:
                continue

            capsize = sum_w[mask] > c[i]
            raft_t = D[i] if capsize else d[i]

            # segment finishes when raft and all walkers arrive
            seg_t = raft_t if raft_t > max_out[mask] else max_out[mask]

            next_dp[mask] = dist[mask] + seg_t

        prev_dp = next_dp

    # At finish point pm, allow swaps to get everyone off (reach mask=0)
    dist = prev_dp[:]
    run_dijkstra(dist)

    print(dist[0])

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

---

## 5) Compressed editorial

Model the configuration at each point by a bitmask `mask` of people on the raft (\(n \le 10 \Rightarrow 2^n \le 1024\)). Swapping at a point from one mask to another costs \(\sum s_j\) over differing bits, i.e. edges `mask -> mask^(1<<j)` with cost `s[j]`. Before each riffle, compute the best achievable times for all masks via Dijkstra on this mask graph starting from current DP values.

Traversing riffle \(i\) with non-empty `mask` takes:
- raft time = \(D_i\) if total raft weight \(> c_i\) else \(d_i\)
- walkers time = \(\max t_j\) over walkers (j not in mask)
- segment time = `max(raft_time, walkers_time)`
Then `next_dp[mask] = dist[mask] + segment_time`. Repeat for all \(m\) riffles.

Finally, at \(p_m\) run Dijkstra once more to reach `mask=0` (everyone on bank). Output the distance to `0`.
