1. Abridged Problem Statement
Given F flowers (numbered 1…F) and V vases in a row (numbered 1…V, V ≥ F), each vase j yields an aesthetic score A[i][j] if flower i is placed there (or 0 if left empty). You must place each flower in exactly one vase so that the flower indices remain in increasing vase positions (i.e., flower 1’s vase < flower 2’s vase < … < flower F’s vase). Empty vases are allowed. Maximize the total aesthetic score and output the maximum sum and one valid assignment.

2. Detailed Editorial

Let n = F and m = V, and let A[i][j] be the given scores (0-based indexed here). We need to choose an increasing sequence of vase indices 0 ≤ c0 < c1 < … < c_{n-1} < m, assigning flower i to vase c_i, to maximize
   sum_{i=0 to n−1} A[i][c_i].

This is a classic dynamic-programming on two indices:

Define DP state
   dp[i][j] = maximum attainable score if we are to place flowers i, i+1, …, n−1 using only vases j, j+1, …, m−1.

Transitions:
1. Skip vase j: dp[i][j] ≥ dp[i][j+1]
2. Use vase j for flower i: dp[i][j] ≥ A[i][j] + dp[i+1][j+1]

Base cases:
- If i == n (all flowers placed), dp[n][j] = 0 for any j.
- If j == m but i < n, dp[i][m] = −∞ (we cannot place remaining flowers).

We memoize dp[i][j] in an n×m table (n,m ≤ 100 ⇒ 10⁴ states). Each state takes O(1) to compute, so overall O(n·m). The reference code fills this table with a top-down recursion.

Reconstruction:
Keep an auxiliary table best[i][j] that marks whether the optimal choice at state (i,j) was to place flower i in vase j (best[i][j]=1) or to skip j (best[i][j]=0). Starting from (i=0,j=0), walk forward—when best[i][j]=1, record c_i=j and increment both i and j; else j++.

Finally, output dp[0][0] and the 1-based vase indices for each flower.

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, m;
vector<vector<int>> a;

void read() {
    cin >> n >> m;
    a.assign(n, vector<int>(m));
    cin >> a;
}

vector<vector<int>> dp;
vector<vector<int>> best;

int rec(int i, int j) {
    if(i == n) {
        return 0;
    }
    if(j == m) {
        return -1e9;
    }

    int& memo = dp[i][j];
    if(memo != -1) {
        return memo;
    }

    best[i][j] = 0;
    memo = rec(i, j + 1);

    int take = a[i][j] + rec(i + 1, j + 1);
    if(take > memo) {
        memo = take;
        best[i][j] = 1;
    }

    return memo;
}

void solve() {
    // - dp[i][j] = best total aesthetic value placing bunches i..F-1 into vases
    //   j..V-1, preserving the order (bunch i must sit left of bunch i+1).
    //
    // - At each state we either skip vase j (move to dp[i][j+1]) or place bunch
    //   i in vase j and take a[i][j] + dp[i+1][j+1]; best[i][j] records which
    //   choice was optimal so the placement can be reconstructed.
    //
    // - Walk the best[] choices from (0, 0) to read off the chosen vase for
    //   each bunch.

    dp.assign(n, vector<int>(m, -1));
    best.assign(n, vector<int>(m, 0));

    vector<int> ans(n);
    cout << rec(0, 0) << '\n';

    int i = 0, j = 0;
    while(i != n) {
        rec(i, j);
        if(best[i][j] == 0) {
            j++;
        } else {
            ans[i] = j;
            i++;
            j++;
        }
    }

    for(int k = 0; k < n; k++) {
        cout << ans[k] + 1 << " ";
    }
    cout << '\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
```python
import sys
sys.setrecursionlimit(10000)

def main():
    # Read number of flowers n and vases m
    n, m = map(int, sys.stdin.readline().split())
    # Read the aesthetic scores: a[i][j]
    a = [list(map(int, sys.stdin.readline().split())) for _ in range(n)]

    # dp[i][j] = best score placing flowers i..n-1 into vases j..m-1
    dp = [[None] * (m + 1) for _ in range(n + 1)]
    # best_dec[i][j] = True if optimal at (i,j) is to take, False to skip
    best_dec = [[False] * (m + 1) for _ in range(n + 1)]

    NEG_INF = -10**9

    def rec(i, j):
        # Base: all flowers placed
        if i == n:
            return 0
        # No vases left but still flowers => impossible
        if j == m:
            return NEG_INF
        if dp[i][j] is not None:
            return dp[i][j]

        # Option 1: skip vase j
        skip = rec(i, j + 1)
        take = a[i][j] + rec(i + 1, j + 1)

        # Choose better
        if take > skip:
            dp[i][j] = take
            best_dec[i][j] = True
        else:
            dp[i][j] = skip
            best_dec[i][j] = False
        return dp[i][j]

    # Compute answer
    max_score = rec(0, 0)
    print(max_score)

    # Reconstruct chosen vases
    res = []
    i = j = 0
    while i < n and j < m:
        if best_dec[i][j]:
            res.append(j + 1)  # store 1-based index
            i += 1
            j += 1
        else:
            j += 1

    print(" ".join(map(str, res)))

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

5. Compressed Editorial

Define dp[i][j] = max total for assigning flowers i…F−1 into vases j…V−1. Transitions:
- skip j: dp[i][j] = dp[i][j+1]
- take j: dp[i][j] = A[i][j] + dp[i+1][j+1]
Base: dp[F][*]=0, dp[*][V]=−∞. Fill with memoized recursion in O(F·V). Record choices in best[i][j] to reconstruct one optimal placement.
Print dp[0][0] and the chosen sequence of vase indices.
