문제
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.
제약 조건
1 ≤ n ≤ 15; 0 ≤ cost ≤ 1,000,000
예시
입력:
distances = [
[0, 10, 15, 20],
[10, 0, 35, 25],
[15, 35, 0, 30],
[20, 25, 30, 0]
]
예상 출력:
expected = 80입력:
std::vector<std::vector<long long>> distances = {
{0, 10, 15, 20},
{10, 0, 35, 25},
{15, 35, 0, 30},
{20, 25, 30, 0}
};
예상 출력:
long long expected = 80;입력:
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}
};
예상 출력:
long expected = 80;설명
0→1→3→2→0 costs 10 + 25 + 30 + 15 = 80, the cheapest tour.
입력:
distances = [
[0]
]
예상 출력:
expected = 0입력:
std::vector<std::vector<long long>> distances = {
{0}
};
예상 출력:
long long expected = 0;입력:
long[][] distances = new long[][] {
new long[] {0}
};
예상 출력:
long expected = 0;설명
With one city there is nowhere to go: 0.
접근 방법
먼저 스스로 풀어 본 뒤 힌트를 하나씩 열어 보세요.
힌트 1
Since the city count is small, try tracking which cities have been visited instead of enumerating all tours directly.
힌트 2
Use dynamic programming over subsets of cities and remember the best cost to end at each city with that visited set.
힌트 3
Be careful to start and end at city 0, and to handle the cases where some paths are extremely expensive but still valid.
예상 복잡도
O(n^2 2^n) time, O(n 2^n) extra space
참고 자료
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