787 Cheapest Flights Within K Stops¶

here 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 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.

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 <code>0</code> to city <code>2</code> with at most 1 stop costs 200, as marked red in the picture.

alt

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 <code>0</code> to city <code>2</code> with at most 0 stop costs 500, as marked blue in the picture.

alt

Solution 1: DFS¶

w/o prunning TLE w/ prunning Accepted

// Running time: 36 ms
class Solution {
public:
  int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
    g_.clear();
    for (const auto& e : flights)
      g_[e[0]].emplace_back(e[1], e[2]);
    vector<int> visited(n, 0);
    visited[src] = 1;
    int ans = INT_MAX;
    dfs(src, dst, K + 1, 0, visited, ans);
    return ans == INT_MAX ? - 1 : ans;
  }
private:
  unordered_map<int, vector<pair<int,int>>> g_;

  void dfs(int src, int dst, int k, int cost, vector<int>& visited, int& ans) {
    if (src == dst) {
      ans = cost;
      return;
    }

    if (k == 0) return;

    for (const auto& p : g_[src]) {
      if (visited[p.first]) continue; // do not visit the same city twice.
      if (cost + p.second > ans) continue; // IMPORTANT!!! prunning
      visited[p.first] = 1;
      dfs(p.first, dst, k - 1, cost + p.second, visited, ans);
      visited[p.first] = 0;
    }
  }
};

Solution 2: BFS¶

// Running time: 20 ms
class Solution {
public:
  int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
    unordered_map<int, vector<pair<int,int>>> g;
    for (const auto& e : flights)
      g[e[0]].emplace_back(e[1], e[2]);

    int ans = INT_MAX;
    queue<pair<int,int>> q;
    q.push({src, 0});
    int steps = 0;

    while (!q.empty()) {
      int size = q.size();
      while (size--) {
        int curr = q.front().first;
        int cost = q.front().second;
        q.pop();
        if (curr == dst)
          ans = min(ans, cost);
        for (const auto& p : g[curr]) {
          if (cost + p.second > ans) continue; // Important: prunning
          q.push({p.first, cost + p.second});
        }
      }
      if (steps++ > K) break;
    }

    return ans == INT_MAX ? - 1 : ans;
  }
};

Solution 3: Bellman-Ford algorithm¶

dp[k][i]: min cost from src to i taken up to k flights (k-1 stops)

init: dp[0:k+2][src] = 0

transition: dp[k][i] = min(dp[k-1][j] + price[j][i])

ans: dp[K+1][dst]

Time complexity: O(k * |flights|) / O(k*n^2)

Space complexity: O(k*n) -> O(n)

w/o space compression O(k*n)

// Running time: 11 ms
class Solution {
  public int findCheapestPrice(int n, int[][] flights, int src, int dst, int K) {
    final int kInfCost = 1<<30;
    int[] cost = new int[n];
    Arrays.fill(cost, kInfCost);
    cost[src] = 0;

    for (int i = 0; i <= K; ++i) {
      int[] tmp = cost.clone();
      for(int[] p: flights)
        tmp[p[1]] = Math.min(tmp[p[1]], cost[p[0]] + p[2]);
      cost = tmp;
    }

    return cost[dst] >= kInfCost ? -1 : cost[dst];
  }
}