問題
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.
制約
1 ≤ n ≤ 15; 0 ≤ cost ≤ 1,000,000
例
例 1
入力：
distances = [
[0, 10, 15, 20],
[10, 0, 35, 25],
[15, 35, 0, 30],
[20, 25, 30, 0]
]
期待される出力：
expected = 80入力：
std::vector<std::vector<long long>> distances = {
{0, 10, 15, 20},
{10, 0, 35, 25},
{15, 35, 0, 30},
{20, 25, 30, 0}
};
期待される出力：
long long expected = 80;入力：
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}
};
期待される出力：
long expected = 80;解説
0→1→3→2→0 costs 10 + 25 + 30 + 15 = 80, the cheapest tour.
例 2
入力：
distances = [
[0]
]
期待される出力：
expected = 0入力：
std::vector<std::vector<long long>> distances = {
{0}
};
期待される出力：
long long expected = 0;入力：
long[][] distances = new long[][] {
new long[] {0}
};
期待される出力：
long expected = 0;解説
With one city there is nowhere to go: 0.
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。