Skip to content

2247. Maximum Cost of Trip With K Highways 👍

  • Time: $O(n^2 \cdot 2^n + |\texttt{highways}|)$
  • Space: $O(n \cdot 2^n + |\texttt{highways}|)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
class Solution {
 public:
  int maximumCost(int n, vector<vector<int>>& highways, int k) {
    if (k + 1 > n)
      return -1;

    int ans = -1;
    // dp[u][mask] := max cost of trip starting from u given mask representing
    // visited cities
    dp.resize(n, vector<int>(1 << n, -1));
    vector<vector<pair<int, int>>> graph(n);

    for (const vector<int>& h : highways) {
      const int u = h[0];
      const int v = h[1];
      const int w = h[2];
      graph[u].emplace_back(v, w);
      graph[v].emplace_back(u, w);
    }

    for (int i = 0; i < n; ++i)
      ans = max(ans, maximumCost(graph, i, 1 << i, k));

    return ans;
  }

 private:
  vector<vector<int>> dp;

  int maximumCost(const vector<vector<pair<int, int>>>& graph, int u, int mask,
                  int k) {
    if (__builtin_popcount(mask) == k + 1)
      return 0;
    if (dp[u][mask] != -1)
      return dp[u][mask];

    int ans = -1;
    for (const auto& [v, w] : graph[u]) {
      if (mask >> v & 1)
        continue;
      const int res = maximumCost(graph, v, mask | 1 << v, k);
      if (res != -1)
        ans = max(ans, w + res);
    }
    return dp[u][mask] = ans;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
class Solution {
  public int maximumCost(int n, int[][] highways, int k) {
    if (k + 1 > n)
      return -1;

    int ans = -1;
    // dp[u][mask] := max cost of trip starting from u given mask representing
    // visited cities
    dp = new Integer[n][1 << n];
    List<Pair<Integer, Integer>>[] graph = new List[n];

    for (int i = 0; i < n; ++i)
      graph[i] = new ArrayList<>();

    for (int[] h : highways) {
      final int u = h[0];
      final int v = h[1];
      final int w = h[2];
      graph[u].add(new Pair<>(v, w));
      graph[v].add(new Pair<>(u, w));
    }

    for (int i = 0; i < n; ++i)
      ans = Math.max(ans, maximumCost(graph, i, 1 << i, k));

    return ans;
  }

  private Integer[][] dp;

  private int maximumCost(List<Pair<Integer, Integer>>[] graph, int u, int mask, int k) {
    if (Integer.bitCount(mask) == k + 1)
      return 0;
    if (dp[u][mask] != null)
      return dp[u][mask];

    int ans = -1;
    for (Pair<Integer, Integer> pair : graph[u]) {
      final int v = pair.getKey();
      final int w = pair.getValue();
      if ((mask >> v & 1) == 1)
        continue;
      final int res = maximumCost(graph, v, mask | 1 << v, k);
      if (res != -1)
        ans = Math.max(ans, w + res);
    }
    return dp[u][mask] = ans;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
class Solution:
  def maximumCost(self, n: int, highways: List[List[int]], k: int) -> int:
    if k + 1 > n:
      return -1

    graph = [[] for _ in range(n)]

    for u, v, w in highways:
      graph[u].append((v, w))
      graph[v].append((u, w))

    # dp(u, mask) := max cost of trip starting from u given mask representing
    # visited cities
    @functools.lru_cache(None)
    def dp(u: int, mask: int) -> int:
      if mask.bit_count() == k + 1:
        return 0

      ans = -1
      for v, w in graph[u]:
        if mask >> v & 1:
          continue
        res = dp(v, mask | 1 << v)
        if res != -1:
          ans = max(ans, w + res)
      return ans

    return max(dp(i, 1 << i) for i in range(n))