2247. Maximum Cost of Trip With K Highways

Description

A series of highways connect n cities numbered from 0 to n - 1. You are given a 2D integer array highways where highways[i] = [city1_i, city2_i, toll_i] indicates that there is a highway that connects city1_i and city2_i, allowing a car to go from city1_i to city2_i and vice versa for a cost of toll_i.

You are also given an integer k. You are going on a trip that crosses exactly k highways. You may start at any city, but you may only visit each city at most once during your trip.

Return the maximum cost of your trip. If there is no trip that meets the requirements, return -1.

Example 1:

Input: n = 5, highways = [[0,1,4],[2,1,3],[1,4,11],[3,2,3],[3,4,2]], k = 3
Output: 17
Explanation:
One possible trip is to go from 0 -> 1 -> 4 -> 3. The cost of this trip is 4 + 11 + 2 = 17.
Another possible trip is to go from 4 -> 1 -> 2 -> 3. The cost of this trip is 11 + 3 + 3 = 17.
It can be proven that 17 is the maximum possible cost of any valid trip.

Note that the trip 4 -> 1 -> 0 -> 1 is not allowed because you visit the city 1 twice.

Example 2:

Input: n = 4, highways = [[0,1,3],[2,3,2]], k = 2
Output: -1
Explanation: There are no valid trips of length 2, so return -1.

Constraints:

2 <= n <= 15
1 <= highways.length <= 50
highways[i].length == 3
0 <= city1_i, city2_i <= n - 1
city1_i != city2_i
0 <= toll_i <= 100
1 <= k <= 50
There are no duplicate highways.

Solutions

class Solution {
    public int maximumCost(int n, int[][] highways, int k) {
        if (k >= n) {
            return -1;
        }
        List<int[]>[] g = new List[n];
        Arrays.setAll(g, h -> new ArrayList<>());
        for (int[] h : highways) {
            int a = h[0], b = h[1], cost = h[2];
            g[a].add(new int[] {b, cost});
            g[b].add(new int[] {a, cost});
        }
        int[][] f = new int[1 << n][n];
        for (int[] e : f) {
            Arrays.fill(e, -(1 << 30));
        }
        for (int i = 0; i < n; ++i) {
            f[1 << i][i] = 0;
        }
        int ans = -1;
        for (int i = 0; i < 1 << n; ++i) {
            for (int j = 0; j < n; ++j) {
                if ((i >> j & 1) == 1) {
                    for (var e : g[j]) {
                        int h = e[0], cost = e[1];
                        if ((i >> h & 1) == 1) {
                            f[i][j] = Math.max(f[i][j], f[i ^ (1 << j)][h] + cost);
                        }
                    }
                }
                if (Integer.bitCount(i) == k + 1) {
                    ans = Math.max(ans, f[i][j]);
                }
            }
        }
        return ans;
    }
}

class Solution {
public:
    int maximumCost(int n, vector<vector<int>>& highways, int k) {
        if (k >= n) {
            return -1;
        }
        vector<pair<int, int>> g[n];
        for (auto& h : highways) {
            int a = h[0], b = h[1], cost = h[2];
            g[a].emplace_back(b, cost);
            g[b].emplace_back(a, cost);
        }
        int f[1 << n][n];
        memset(f, -0x3f, sizeof(f));
        for (int i = 0; i < n; ++i) {
            f[1 << i][i] = 0;
        }
        int ans = -1;
        for (int i = 0; i < 1 << n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i >> j & 1) {
                    for (auto& [h, cost] : g[j]) {
                        if (i >> h & 1) {
                            f[i][j] = max(f[i][j], f[i ^ (1 << j)][h] + cost);
                        }
                    }
                }
                if (__builtin_popcount(i) == k + 1) {
                    ans = max(ans, f[i][j]);
                }
            }
        }
        return ans;
    }
};

class Solution:
    def maximumCost(self, n: int, highways: List[List[int]], k: int) -> int:
        if k >= n:
            return -1
        g = defaultdict(list)
        for a, b, cost in highways:
            g[a].append((b, cost))
            g[b].append((a, cost))
        f = [[-inf] * n for _ in range(1 << n)]
        for i in range(n):
            f[1 << i][i] = 0
        ans = -1
        for i in range(1 << n):
            for j in range(n):
                if i >> j & 1:
                    for h, cost in g[j]:
                        if i >> h & 1:
                            f[i][j] = max(f[i][j], f[i ^ (1 << j)][h] + cost)
                if i.bit_count() == k + 1:
                    ans = max(ans, f[i][j])
        return ans

func maximumCost(n int, highways [][]int, k int) int {
	if k >= n {
		return -1
	}
	g := make([][][2]int, n)
	for _, h := range highways {
		a, b, cost := h[0], h[1], h[2]
		g[a] = append(g[a], [2]int{b, cost})
		g[b] = append(g[b], [2]int{a, cost})
	}
	f := make([][]int, 1<<n)
	for i := range f {
		f[i] = make([]int, n)
		for j := range f[i] {
			f[i][j] = -(1 << 30)
		}
	}
	for i := 0; i < n; i++ {
		f[1<<i][i] = 0
	}
	ans := -1
	for i := 0; i < 1<<n; i++ {
		for j := 0; j < n; j++ {
			if i>>j&1 == 1 {
				for _, e := range g[j] {
					h, cost := e[0], e[1]
					if i>>h&1 == 1 {
						f[i][j] = max(f[i][j], f[i^(1<<j)][h]+cost)
					}
				}
			}
			if bits.OnesCount(uint(i)) == k+1 {
				ans = max(ans, f[i][j])
			}
		}
	}
	return ans
}

function maximumCost(n: number, highways: number[][], k: number): number {
    if (k >= n) {
        return -1;
    }
    const g: [number, number][][] = Array.from({ length: n }, () => []);
    for (const [a, b, cost] of highways) {
        g[a].push([b, cost]);
        g[b].push([a, cost]);
    }
    const f: number[][] = Array(1 << n)
        .fill(0)
        .map(() => Array(n).fill(-(1 << 30)));
    for (let i = 0; i < n; ++i) {
        f[1 << i][i] = 0;
    }
    let ans = -1;
    for (let i = 0; i < 1 << n; ++i) {
        for (let j = 0; j < n; ++j) {
            if ((i >> j) & 1) {
                for (const [h, cost] of g[j]) {
                    if ((i >> h) & 1) {
                        f[i][j] = Math.max(f[i][j], f[i ^ (1 << j)][h] + cost);
                    }
                }
            }
            if (bitCount(i) === k + 1) {
                ans = Math.max(ans, f[i][j]);
            }
        }
    }
    return ans;
}

function bitCount(i: number): number {
    i = i - ((i >>> 1) & 0x55555555);
    i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
    i = (i + (i >>> 4)) & 0x0f0f0f0f;
    i = i + (i >>> 8);
    i = i + (i >>> 16);
    return i & 0x3f;
}

2247 - Maximum Cost of Trip With K Highways

2247. Maximum Cost of Trip With K Highways

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2247. Maximum Cost of Trip With K Highways

Description

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