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).  

Reconstruction:  
Keep an auxiliary boolean 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. Provided C++ Solution with Detailed Comments  

#include <bits/stdc++.h>  
using namespace std;

// Maximum sizes: Vases up to 2^10 = 1024, safely covers V ≤ 100  
const int MAXN = 1 << 10;  

int n, m;  
int a[MAXN][MAXN];      // a[i][j] = aesthetic score for flower i in vase j  
int dp[MAXN][MAXN];     // memo table for dp[i][j]  
int best[MAXN][MAXN];   // decision table: 1=take, 0=skip  
int ans[MAXN];          // stores chosen vase index for each flower  

// Read input n (flowers) and m (vases) and the score matrix a  
void read() {  
    cin >> n >> m;  
    for(int i = 0; i < n; i++)  
        for(int j = 0; j < m; j++)  
            cin >> a[i][j];  
}  

// Compute dp[i][j] recursively with memoization  
// returns maximum score for placing flowers i…n-1 into vases j…m-1  
int rec(int i, int j) {  
    // If we've placed all flowers, no more score to add  
    if (i == n)  
        return 0;  
    // If no more vases but still flowers left, impossible => -inf  
    if (j == m)  
        return -1000000000;  

    int &res = dp[i][j];  
    if (res != -1)  
        return res;  // already computed  

    // Option 1: skip this vase  
    res = rec(i, j + 1);  
    best[i][j] = 0;  

    // Option 2: place flower i in vase j  
    int take = a[i][j] + rec(i + 1, j + 1);  
    if (take > res) {  
        res = take;  
        best[i][j] = 1;  
    }  
    return res;  
}  

void solve() {  
    // Initialize dp table to -1 (uncomputed)  
    memset(dp, -1, sizeof(dp));  

    // Compute optimal total score  
    int maxScore = rec(0, 0);  
    cout << maxScore << '\n';  

    // Reconstruct assignment  
    int i = 0, j = 0;  
    while (i < n) {  
        rec(i, j);  // ensure best[i][j] is set  
        if (best[i][j] == 1) {  
            ans[i] = j;  // assign vase j to flower i  
            i++;  
            j++;  
        } else {  
            j++;       // skip vase j  
        }  
    }  

    // Output 1-based vase indices  
    for (int k = 0; k < n; k++)  
        cout << (ans[k] + 1) << (k + 1 < n ? ' ' : '\n');  
}  

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

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

4. Python Solution with Detailed Comments  

```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 memo/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.