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 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:

• 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.

DFS需要建立一個adjacency list來表示graph，注意helper function裡的pruning。

// DFS
void helper(vector<vector<vector<int> > >& graph, vector<bool>& visited, int cur, int dst, int K, int out, int& res) {
    if (cur == dst) {
        res = out;
        return;
    }
    if (K < 0) return;
    for (auto& a : graph[cur]) {
        if (visited[a[0]] || out + a[1] > res) continue; // pruning
        visited[a[0]] = true;
        helper(graph, visited, a[0], dst, K - 1, out + a[1], res);
        visited[a[0]] = false;
    }
}
int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) { // time: O(E + V); space: O(E + V)
    int res = INT_MAX;
    vector<vector<vector<int> > > graph(n); // adjacency list
    vector<bool> visited(n, false);
    visited[src] = true;
    for (auto flight : flights) {
        graph[flight[0]].push_back({flight[1], flight[2]});
    }
    helper(graph, visited, src, dst, K, 0, res);
    return res == INT_MAX ? -1 : res;
}

BFS也需要建立adjacency list來表示graph，需要queue，注意要做pruning如果當前的價格已經超出目前已求得的res。

// BFS
int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) { // time: O(E + V); space: O(E + V)
    int res = INT_MAX, cnt = 0;
    vector<vector<vector<int> > > graph(n); // adjacency list
    // Build graph
    for (auto flight : flights) {
        graph[flight[0]].push_back({flight[1], flight[2]});
    }
    queue<vector<int> > q({{src, 0}}); // queue used for BFS
    while (!q.empty()) {
        for (int i = q.size() - 1; i >= 0; --i) {
            auto t = q.front(); q.pop();
            if (t[0] == dst) res = min(res, t[1]);
            for (auto& a : graph[t[0]]) {
                if (t[1] + a[1] > res) continue;
                q.push({a[0], t[1] + a[1]});
            }
        }
        if (cnt++ > K) break;
    }
    return res == INT_MAX ? -1 : res;
}

Bellman Ford Algorithm利用Dynamic Programming，其中DP state transition就是relaxation。

// Bellman Ford Algorithm + DP
int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) { // time: (K * n^2); space: O(K * n)
    vector<vector<int> > dp(K + 2, vector<int>(n, 1e9));
    dp[0][src] = 0;
    for (int i = 1; i <= K + 1; ++i) {
        dp[i][src] = 0;
        for (auto& flight : flights) {
            dp[i][flight[1]] = min(dp[i][flight[1]], dp[i - 1][flight[0]] + flight[2]); // relaxation
        }
    }
    return (dp[K + 1][dst] >= 1e9) ? -1 : dp[K + 1][dst];
}

優化空間的使用。

// Bellman Ford Algorithm + space-optimized DP
int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) { // time: (K * n^2); space: O(n)
    vector<int> dp(n, 1e9);
    dp[src] = 0;
    for (int i = 0; i < K + 1; ++i) {
        vector<int> t = dp;
        for (auto& flight : flights) {
            t[flight[1]] = min(t[flight[1]], dp[flight[0]] + flight[2]);
        }
        dp = t;
    }
    return dp[dst] >= 1e9 ? -1 : dp[dst];
}