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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 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.

 

Constraints:

• 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.

Related Topics:
Dynamic Programming, Heap, Breadth-first Search

Similar Questions:

• Maximum Vacation Days (Hard)

Solution 1. Bellman ford

With k stops, the longest path has k + 1 edges. So, using Bellman-ford algorithm, we should relax k + 1 times.

The relaxation happens from layer to layer. This is to prevent using distance in layer i when relaxing from layer i-1 to layer i.

// OJ: https://leetcode.com/problems/cheapest-flights-within-k-stops/
// Author: github.com/lzl124631x
// 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];
    }
};

Solution 2. Dijkstra

// OJ: https://leetcode.com/problems/cheapest-flights-within-k-stops/
// Author: github.com/lzl124631x
// Time: O(KElogE)
// Space: O(KN + E)
class Solution {
    typedef array<int, 3> item; // distance, cityId, # stops left
public:
    int findCheapestPrice(int n, vector<vector<int>>& E, int src, int dst, int k) {
        vector<vector<pair<int, int>>> G(n);
        for (auto &e : E) G[e[0]].emplace_back(e[1], e[2]);
        priority_queue<item, vector<item>, greater<>> pq;
        ++k;
        pq.push({0, src, k});
        vector<vector<int>> dist(k + 1, vector<int>(n, INT_MAX));
        for (int i = 0; i <= k; ++i) dist[i][src] = 0;
        while (pq.size()) {
            auto [cost, u, stop] = pq.top();
            pq.pop();
            if (cost > dist[stop][u]) continue;
            if (u == dst) return cost;
            if (stop == 0) continue;
            for (auto &[v, w] : G[u]) {
                if (dist[stop - 1][v] > cost + w) {
                    dist[stop - 1][v] = cost + w;
                    pq.push({dist[stop - 1][v], v, stop - 1});
                }
            }
        }
        return -1;
    }
};

Solution 3. BFS

A plain BFS also works, but it doesn't prioritize low cost paths first so it might be less efficient.

// OJ: https://leetcode.com/problems/cheapest-flights-within-k-stops/
// Author: github.com/lzl124631x
// Time: O(KE)
// Space: O(N + E)
class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& E, int src, int dst, int k) {
        vector<vector<pair<int, int>>> G(n);
        for (auto &e : E) G[e[0]].emplace_back(e[1], e[2]);
        queue<pair<int, int>> q; // cityId, cost
        q.emplace(src, 0);
        vector<int> dist(n, INT_MAX);
        dist[src] = 0;
        int step = 0;
        while (step <= k) {
            int cnt = q.size();
            while (cnt--) {
                auto [u, cost] = q.front();
                q.pop();
                for (auto &[v, w] : G[u]) {
                    if (dist[v] > cost + w) { // Note that we shouldn't use `dist[u]` here. Because `dist[u]` might get updated using two jumps but the `cost` here corresponds to one jump.
                        dist[v] = cost + w;
                        q.emplace(v, dist[v]);
                    }
                }
            }
            ++step;
        }
        return dist[dst] == INT_MAX ? -1 : dist[dst];
    }
};