[leetcode] problem 787 – cheapest flights within k stops

There are n cities connected by m flights. Each fight starts from city u and arrives at v with a price w.

Now given all the cities and flights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1.

Example

No.1

Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 1

Output: 200

Explanation:
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 1 stop costs 200, as marked red in the picture.

No.2

Input:
n = 3, edges = [[0,1,100],[1,2,100],[0,2,500]]
src = 0, dst = 2, k = 0

Output: 500

Explanation:
The graph looks like this:

The cheapest price from city 0 to city 2 with at most 0 stop costs 500, as marked blue in the picture.

Note

The number of nodes n will be in range [1, 100], with nodes labeled from 0 to n - 1.
The size of flights will be in range [0, n * (n - 1) / 2].
The format of each flight will be (src, dst, price).
The price of each flight will be in the range [1, 10000].
k is in the range of [0, n - 1].
There will not be any duplicated flights or self cycles.

Code

public int (int n, int[][] flights, int src, int dst, int K) {
    List<int[]>[] graph = new List[n];
    int[] distance = new int[n];
    Queue<int[]> queue = new LinkedList<>();

    Arrays.fill(distance, Integer.MAX_VALUE);
    distance[src] = 0;
    queue.offer(new int[] {src, 0});

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

    for (int[] flight : flights)
        graph[flight[0]].add(new int[] {flight[1], flight[2]});

    while (!queue.isEmpty() && K >= 0) {
        int size = queue.size();

        for (int i = 0; i < size; i++) {
            int[] from = queue.poll();
            int u = from[0];

            for (int[] to : graph[u]) {
                int v = to[0];

                if (distance[v] > from[1] + to[1]) {
                    distance[v] = from[1] + to[1];
                    queue.offer(new int[] {v, distance[v]});
                }
            }
        }

        K--;
    }

    return distance[dst] == Integer.MAX_VALUE ? -1 : distance[dst];
}

[leetcode] problem 787 – cheapest flights within k stops

Example

No.1

No.2

Note

Code

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