[LeetCode] 787. Cheapest Flights Within K Stops K次起色内的最便宜航班

2021年09月15日 阅读数:2
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There are n cities connected by m flights. Each fight starts from city and arrives at v with a price w.html

Now given all the cities and fights, 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.java

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 0 to city 2 with 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:

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

Note:node

  • 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.

解法:Dijkstra's algorithm, python

Java:app

class Solution {
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) {
        Map<Integer, Map<Integer, Integer>> prices = new HashMap<>();
        for (int[] f : flights) {
            if (!prices.containsKey(f[0])) prices.put(f[0], new HashMap<>());
            prices.get(f[0]).put(f[1], f[2]);
        }
        Queue<int[]> pq = new PriorityQueue<>((a, b) -> (Integer.compare(a[0], b[0])));
        pq.add(new int[] {0, src, k + 1});
        while (!pq.isEmpty()) {
            int[] top = pq.remove();
            int price = top[0];
            int city = top[1];
            int stops = top[2];
            if (city == dst) return price;
            if (stops > 0) {
                Map<Integer, Integer> adj = prices.getOrDefault(city, new HashMap<>());
                for (int a : adj.keySet()) {
                    pq.add(new int[] {price + adj.get(a), a, stops - 1});
                }
            }
        }
        return -1;
    }      
} 

Python:post

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
        """
        adj = collections.defaultdict(list)
        for u, v, w in flights:
            adj[u].append((v, w))
        best = collections.defaultdict(lambda: collections.defaultdict(lambda: float("inf")))
        min_heap = [(0, src, K+1)]
        while min_heap:
            result, u, k = heapq.heappop(min_heap)
            if k < 0 or best[u][k] < result:
                continue
            if u == dst:
                return result
            for v, w in adj[u]:
                if result+w < best[v][k-1]:
                    best[v][k-1] = result+w                    
                    heapq.heappush(min_heap, (result+w, v, k-1))
        return -1

Python:this

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
        """
        f = collections.defaultdict(dict)
        for a, b, p in flights:
            f[a][b] = p
        heap = [(0, src, k + 1)]
        while heap:
            p, i, k = heapq.heappop(heap)
            if i == dst:
                return p
            if k > 0:
                for j in f[i]:
                    heapq.heappush(heap, (p + f[i][j], j, k - 1))
        return -1  

C++:url

// Time:  O((|E| + |V|) * log|V|) = O(|E| * log|V|)
// Space: O(|E| + |V|) = O(|E|)

class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
        using P = pair<int, int>;
        unordered_map<int, vector<P>> adj;
        for (const auto& flight : flights) {
            int u, v, w;
            tie(u, v, w) = make_tuple(flight[0], flight[1], flight[2]);
            adj[u].emplace_back(v, w);
        }
        
        unordered_map<int, unordered_map<int, int>> best;
        using T = tuple<int, int, int>;
        priority_queue<T, vector<T>, greater<T>> min_heap;
        min_heap.emplace(0, src, K + 1);
        while (!min_heap.empty()) {
            int result, u, k;
            tie(result, u, k) = min_heap.top(); min_heap.pop();
            if (k < 0 ||
                (best.count(u) && best[u].count(k) &&  best[u][k] < result)) {
                continue;
            }
            if (u == dst) {
                return result;
            }
            for (const auto& kvp : adj[u]) {
                int v, w;
                tie(v, w) = kvp;
                if (!best.count(v) ||
                    !best[v].count(k - 1) ||
                    result + w < best[v][k - 1]) {
                    best[v][k - 1] = result + w;
                    min_heap.emplace(result + w, v, k - 1);
                }
            }
        }
        return -1;
    }
};

    

 

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