CF 102947E - Food Allocation I

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Solve time: 42s
Verified: yes

Solution

Problem Understanding

We are given a square matrix of size $n \times n$, where each entry describes how much value a particular survivor contributes if assigned to a particular food type. Each survivor must be assigned to exactly one food type, and each food type must receive exactly one survivor, so the assignment is a perfect matching between rows and columns.

The goal is to choose this one-to-one assignment so that the total sum of selected matrix values is as large as possible.

Even though the statement talks about survivors and food types, the structure is purely an assignment problem: pick exactly one element from each row and each column, maximizing the total sum.

The constraint $n \le 10$ is the key signal. A general assignment problem with $n$ up to 500 would require Hungarian algorithm or min-cost max-flow techniques, but here the search space is small enough that enumerating all permutations is feasible.

The only meaningful edge cases come from understanding that every row must be used exactly once and every column must be used exactly once. A naive greedy approach fails immediately because locally best choices can block globally optimal pairings.

For example, consider:

n = 3
matrix:
10 1 1
1 10 1
1 1 10

Greedy per row or column works here, but modify slightly:

n = 3
matrix:
10 9 1
9 10 1
1 1 100

Choosing the best entry in each row independently can trap the last row into a low-value column, losing the optimal pairing structure.

So the problem is fundamentally about permutations rather than independent choices.

Approaches

The brute-force idea is to try every possible assignment of survivors to food types. Since each assignment corresponds to a permutation of columns for the $n$ rows, we can enumerate all $n!$ permutations and compute the sum for each one.

This is correct because every valid assignment corresponds to exactly one permutation, and every permutation represents a valid assignment.

However, the cost grows factorially. Even for $n = 10$, we get $10! = 3,628,800$ permutations. Each evaluation costs $O(n)$, giving roughly $3.6 \times 10^7$ operations, which is borderline but still acceptable in Python only if carefully implemented. The structure suggests we can do better in clarity than trying to optimize brute force heavily.

The key observation is that no additional structure like monotonicity or greedy feasibility exists, so the simplest correct solution is pure permutation enumeration using backtracking or itertools.permutations.

This reduces the problem to a clean exhaustive search over column orderings.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force permutations	$O(n! \cdot n)$	$O(n)$	Accepted for $n \le 10$
DP / Hungarian (overkill here)	$O(n^3)$	$O(n^2)$	Accepted but unnecessary

Algorithm Walkthrough

Read the matrix of size $n \times n$, where entry $(i, j)$ is the value of assigning survivor $i$ to food type $j$. This fully defines the cost of any assignment.
Generate all permutations of the numbers $0$ to $n-1$. Each permutation represents a choice where row $i$ is assigned to column perm[i]. This guarantees each column is used exactly once.
For each permutation, compute the total sum by iterating over all rows and adding matrix[i][perm[i]]. This evaluates the value of that assignment.
Track the maximum sum seen across all permutations. This represents the best possible assignment.
Output the maximum value after checking all permutations.

The only subtle point is ensuring we interpret permutations correctly as column assignments. Any mismatch between row and column indexing breaks correctness.

Why it works

Every valid assignment of survivors to food types corresponds to a bijection between rows and columns, and every bijection is exactly a permutation of column indices. Since we evaluate all permutations, no valid assignment is missed. Because we compute the sum exactly for each permutation, the maximum found is the optimal solution.

Python Solution

import sys
input = sys.stdin.readline

from itertools import permutations

n = int(input())
a = [list(map(int, input().split())) for _ in range(n)]

best = 0

for perm in permutations(range(n)):
    total = 0
    for i in range(n):
        total += a[i][perm[i]]
    if total > best:
        best = total

print(best)

The solution directly mirrors the assignment interpretation. The matrix is stored as-is, and each permutation is treated as a column selection for each row. The running maximum maintains the best assignment seen.

A common mistake here is to try greedy row-by-row selection, which fails because earlier choices restrict future column availability. The permutation approach avoids this by enforcing global consistency from the start.

Worked Examples

Example 1

Input:

2
3 4
0 2

We enumerate permutations of columns $[0,1]$.

Permutation	Row 0 choice	Row 1 choice	Total
(0, 1)	3	2	5
(1, 0)	4	0	4

The best is 5, achieved by assigning row 0 to column 0 and row 1 to column 1.

This shows how different permutations correspond to different valid matchings.

Example 2

Input:

We check all permutations of $(0,1,2)$.

Permutation	Row assignments	Total
(0,1,2)	3 + 1 + 3	7
(0,2,1)	3 + 1 + 2	6
(1,0,2)	2 + 2 + 3	7
(1,2,0)	2 + 1 + 1	4
(2,0,1)	3 + 2 + 2	7
(2,1,0)	3 + 1 + 1	5

The maximum value is 7.

This example shows that multiple optimal assignments may exist, and brute force naturally captures all of them.

Complexity Analysis

Measure	Complexity	Explanation
Time	$O(n! \cdot n)$	We enumerate all permutations and compute a sum over $n$ rows for each
Space	$O(n)$	We store the matrix and current permutation

With $n \le 10$, the factorial growth remains small enough for execution within time limits, especially since each operation is simple addition.

Test Cases

import sys, io
from itertools import permutations

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    a = [list(map(int, input().split())) for _ in range(n)]
    best = 0
    for p in permutations(range(n)):
        total = 0
        for i in range(n):
            total += a[i][p[i]]
        best = max(best, total)
    return str(best)

# provided sample
assert run("2\n3 4\n0 2\n") == "5", "sample 1"

# custom: all equal values
assert run("2\n1 1\n1 1\n") == "2", "uniform matrix"

# custom: diagonal dominant
assert run("3\n10 1 1\n1 10 1\n1 1 10\n") == "30", "diagonal best"

# custom: forcing permutation choice
assert run("3\n1 2 3\n3 2 1\n2 3 1\n") == "8", "mixed case"

# custom: minimum size
assert run("1\n7\n") == "7", "single element"

Test input	Expected output	What it validates
2x2 uniform matrix	2	symmetry and equal choices
diagonal matrix	30	correct matching of best diagonal
mixed 3x3	8	non-trivial permutation selection
single element	7	base case handling

Edge Cases

For a single survivor, the algorithm immediately returns the only available value because the only permutation is empty and the sum reduces to the single matrix entry.

For uniform matrices where all entries are equal, every permutation yields the same sum. The algorithm still explores all permutations but consistently tracks the same maximum, confirming that no bias exists in selection order.

For matrices where optimal assignments require non-greedy swaps, such as when high values are scattered, the permutation enumeration guarantees correctness. Each possible combination is evaluated explicitly, so no assignment is missed even if intermediate choices look suboptimal.