Условие
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.
Ограничения
1 ≤ n ≤ 15; 0 ≤ cost ≤ 1,000,000
Примеры
Пример 1
Ввод:
distances = [
[0, 10, 15, 20],
[10, 0, 35, 25],
[15, 35, 0, 30],
[20, 25, 30, 0]
]
Ожидаемый вывод:
expected = 80Ввод:
std::vector<std::vector<long long>> distances = {
{0, 10, 15, 20},
{10, 0, 35, 25},
{15, 35, 0, 30},
{20, 25, 30, 0}
};
Ожидаемый вывод:
long long expected = 80;Ввод:
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}
};
Ожидаемый вывод:
long expected = 80;Пояснение
0→1→3→2→0 costs 10 + 25 + 30 + 15 = 80, the cheapest tour.
Пример 2
Ввод:
distances = [
[0]
]
Ожидаемый вывод:
expected = 0Ввод:
std::vector<std::vector<long long>> distances = {
{0}
};
Ожидаемый вывод:
long long expected = 0;Ввод:
long[][] distances = new long[][] {
new long[] {0}
};
Ожидаемый вывод:
long expected = 0;Пояснение
With one city there is nowhere to go: 0.
Ссылки
Это задание адаптировано для CodeUtility. Следуйте условию и ограничениям выше.