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Formatted question description: https://leetcode.ca/all/787.html
787. Cheapest Flights Within K Stops (Medium)
There are n cities connected by m flights. Each flight 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 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
0to city2with at most 1 stop costs 200, as marked red in the picture.
Example 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:0to city2with at most 0 stop costs 500, as marked blue in the picture.
Constraints:
- The number of nodes
nwill be in range[1, 100], with nodes labeled from0ton- 1. - The size of
flightswill 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]. kis in the range of[0, n - 1].- There will not be any duplicated flights or self cycles.
Related Topics:
Dynamic Programming, Heap, Breadth-first Search
Similar Questions:
Solution 1. Bellman ford
// OJ: https://leetcode.com/problems/cheapest-flights-within-k-stops/
// Time: O(K * (N+ E))
// Space: O(N)
class Solution {
public:
int findCheapestPrice(int n, vector<vector<int>>& A, int src, int dst, int K) {
vector<int> dist(n, 1e8);
dist[src] = 0;
for (int i = 0; i <= K; ++i) {
vector<int> tmp = dist;
for (auto &e : A) {
int u = e[0], v = e[1], w = e[2];
dist[v] = min(dist[v], tmp[u] + w);
}
}
return dist[dst] == 1e8 ? -1 : dist[dst];
}
};
-
class Solution { public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) { int[] prices = new int[n]; for (int i = 0; i < n; i++) prices[i] = Integer.MAX_VALUE; prices[src] = 0; Map<Integer, List<int[]>> flightsMap = new HashMap<Integer, List<int[]>>(); for (int[] flight : flights) { int flightSrc = flight[0], flightDst = flight[1], flightPrice = flight[2]; List<int[]> flightsList = flightsMap.getOrDefault(flightSrc, new ArrayList<int[]>()); flightsList.add(new int[]{flightDst, flightPrice}); flightsMap.put(flightSrc, flightsList); } Queue<Integer> queue = new LinkedList<Integer>(); queue.offer(src); int remain = K; while (remain >= 0 && !queue.isEmpty()) { int[] tempPrices = new int[n]; for (int i = 0; i < n; i++) tempPrices[i] = prices[i]; int size = queue.size(); for (int i = 0; i < size; i++) { int curSrc = queue.poll(); int curPrice = prices[curSrc]; List<int[]> curFlightsList = flightsMap.getOrDefault(curSrc, new ArrayList<int[]>()); for (int[] flight : curFlightsList) { int flightDst = flight[0], flightPrice = flight[1]; int newPrice = curPrice + flightPrice; if (newPrice < tempPrices[flightDst]) { tempPrices[flightDst] = newPrice; queue.offer(flightDst); } } } for (int i = 0; i < n; i++) prices[i] = tempPrices[i]; remain--; } int totalPrice = prices[dst]; return totalPrice == Integer.MAX_VALUE ? -1 : totalPrice; } } ############ class Solution { private static final int INF = 0x3f3f3f3f; public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) { int[] dist = new int[n]; int[] backup = new int[n]; Arrays.fill(dist, INF); dist[src] = 0; for (int i = 0; i < k + 1; ++i) { System.arraycopy(dist, 0, backup, 0, n); for (int[] e : flights) { int f = e[0], t = e[1], p = e[2]; dist[t] = Math.min(dist[t], backup[f] + p); } } return dist[dst] == INF ? -1 : dist[dst]; } } -
// OJ: https://leetcode.com/problems/cheapest-flights-within-k-stops/ // Time: O(K * (N+ E)) // Space: O(N) class Solution { public: int findCheapestPrice(int n, vector<vector<int>>& A, int src, int dst, int K) { vector<int> dist(n, 1e8); dist[src] = 0; for (int i = 0; i <= K; ++i) { auto tmp = dist; for (auto &e : A) { int u = e[0], v = e[1], w = e[2]; dist[v] = min(dist[v], tmp[u] + w); } } return dist[dst] == 1e8 ? -1 : dist[dst]; } }; -
class Solution: def findCheapestPrice( self, n: int, flights: List[List[int]], src: int, dst: int, k: int ) -> int: INF = 0x3F3F3F3F dist = [INF] * n dist[src] = 0 for _ in range(k + 1): backup = dist.copy() for f, t, p in flights: dist[t] = min(dist[t], backup[f] + p) return -1 if dist[dst] == INF else dist[dst] ############ class Solution(object): def findCheapestPrice(self, n, flights, src, dst, K): """ :type n: int :type flights: List[List[int]] :type src: int :type dst: int :type K: int :rtype: int """ graph = collections.defaultdict(dict) for u, v, e in flights: graph[u][v] = e visited = [0] * n ans = [float('inf')] self.dfs(graph, src, dst, K + 1, 0, visited, ans) return -1 if ans[0] == float('inf') else ans[0] def dfs(self, graph, src, dst, k, cost, visited, ans): if src == dst: ans[0] = cost return if k == 0: return for v, e in graph[src].items(): if visited[v]: continue if cost + e > ans[0]: continue visited[v] = 1 self.dfs(graph, v, dst, k - 1, cost + e, visited, ans) visited[v] = 0 -
func findCheapestPrice(n int, flights [][]int, src int, dst int, k int) int { const inf = 0x3f3f3f3f dist := make([]int, n) backup := make([]int, n) for i := range dist { dist[i] = inf } dist[src] = 0 for i := 0; i < k+1; i++ { copy(backup, dist) for _, e := range flights { f, t, p := e[0], e[1], e[2] dist[t] = min(dist[t], backup[f]+p) } } if dist[dst] == inf { return -1 } return dist[dst] } func min(a, b int) int { if a < b { return a } return b }