Problem
A courier starts at city 0, visits every other city exactly once and returns to city 0. distances[i][j] is the cost from i to j. With at most 15 cities, a bitmask dynamic program (Held–Karp) over visited sets is fast enough.
Implement shortest_tour(distances).
Input
distances: An n × n matrix of travel costs; distances[i][i] = 0.
Output
Return the cost of the cheapest round trip.
Constraints
1 ≤ n ≤ 15; 0 ≤ cost ≤ 1,000,000
Examples
Input:
distances = [
[0, 10, 15, 20],
[10, 0, 35, 25],
[15, 35, 0, 30],
[20, 25, 30, 0]
]
Expected output:
expected = 80Input:
std::vector<std::vector<long long>> distances = {
{0, 10, 15, 20},
{10, 0, 35, 25},
{15, 35, 0, 30},
{20, 25, 30, 0}
};
Expected output:
long long expected = 80;Input:
long[][] distances = new long[][] {
new long[] {0, 10, 15, 20},
new long[] {10, 0, 35, 25},
new long[] {15, 35, 0, 30},
new long[] {20, 25, 30, 0}
};
Expected output:
long expected = 80;Explanation
0→1→3→2→0 costs 10 + 25 + 30 + 15 = 80, the cheapest tour.
Input:
distances = [
[0]
]
Expected output:
expected = 0Input:
std::vector<std::vector<long long>> distances = {
{0}
};
Expected output:
long long expected = 0;Input:
long[][] distances = new long[][] {
new long[] {0}
};
Expected output:
long expected = 0;Explanation
With one city there is nowhere to go: 0.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Since the city count is small, try tracking which cities have been visited instead of enumerating all tours directly.
Hint 2
Use dynamic programming over subsets of cities and remember the best cost to end at each city with that visited set.
Hint 3
Be careful to start and end at city 0, and to handle the cases where some paths are extremely expensive but still valid.
Expected complexity
O(n^2 2^n) time, O(n 2^n) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.