787. Cheapest Flights Within K Stops

https://leetcode.com/problems/cheapest-flights-within-k-stops/description/

DFS (slow)

class Solution {
public:
    vector<vector<pair<int,int>>> v;     // city: <connected city, price>
    vector<bool> visited;
    int res = INT_MAX;
    
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
        v = vector<vector<pair<int,int>>>(n);
        visited = vector<bool>(n, false);
        
        for (const auto& flight : flights)
            v[flight[0]].push_back( { flight[1], flight[2] });
        
        visited[src] = true;
        dfs(n, src, dst, K+1, 0);
        return res == INT_MAX ? -1 : res;
    }
    
    void dfs(int n, int cur, int dst, int K, int cost) {
        if (cur == dst) {
            res = min(res, cost);
            return;
        }
        if (cost >= res)    return;     // if cost >= res, no point to search further
        if (K == 0) return;
        
        for (const auto& pCityPrice : v[cur]) {
            if (!visited[pCityPrice.first]) {
                visited[cur] = true;
                dfs(n, pCityPrice.first, dst, K-1, cost + pCityPrice.second);
                visited[cur] = false;
            }
        }
    }
};

BFS (seems not able to use BFS as below, we may have to record each path. Looks like paths in BFS is not independent, we can't use BFS. In this case, if A->B->C and A->C, we are not able to determine cost[C] unless we keep track the path to C. Then we may use DFS directly.)

class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
        vector<vector<pair<int,int>>> v = vector<vector<pair<int,int>>>(n);
        vector<int> cost = vector<int>(n, -1);
        
        for (const auto& flight : flights)
            v[flight[0]].push_back( { flight[1], flight[2] });
        
        queue<int> q;
        q.push(src);
        cost[src] = 0;
        
        int stop = 0;
        while (!q.empty()) {
            if (stop++ > K)
                break;
            
            int qsz = q.size();
            while (qsz-- > 0) {
                int cur = q.front(); q.pop();
                for (const auto& pCityPrice : v[cur]) {
                    int curDest = pCityPrice.first;
                    if (cost[curDest] == -1 || cost[curDest] > cost[cur] + pCityPrice.second) {
                        q.push(curDest);
                        cost[curDest] = cost[cur] + pCityPrice.second;
                    }
                }
            }
        }
        return cost[dst];
    }
};

Expand stops from src.

class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
        vector<vector<pair<int,int>>> v = vector<vector<pair<int,int>>>(n);
        
        for (const auto& flight : flights)
            v[flight[0]].push_back( { flight[1], flight[2] });
        
        vector<vector<int>> dp(n, vector<int>(K+1, INT_MAX));
        
        for (const auto& pCityPrice : v[src]) {
            int curDest = pCityPrice.first;
            int curPrice = pCityPrice.second;
            dp[curDest][0] = min(dp[curDest][0], curPrice);
        }
        for (int k = 1; k <= K; k++) {
            for (int i = 0; i < n; i++) {
                if (dp[i][k-1] == INT_MAX)   continue;
                for (const auto& pCityPrice : v[i]) {
                    int curDest = pCityPrice.first;
                    int curPrice = pCityPrice.second;
                    dp[curDest][k] = min(dp[curDest][k], dp[i][k-1] + curPrice);
                }
            }
        }
        int res = INT_MAX;
        for (int k = 0; k <= K; k++)
            res = min(res, dp[dst][k]);
        return res == INT_MAX ? -1 : res;
    }
};

Think backward, expand stops from dst.

class Solution {
public:
    int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int K) {
        vector<int> bestfrom(n, INT_MAX);     // Current best from n to dst
        
        for (const auto& flight : flights) {        // non-stop from each city to dst
            if (flight[1] == dst)
                bestfrom[flight[0]] = min(bestfrom[flight[0]], flight[2]);
        }
        // from each city to dst with one more stop.
        // say we know bestfrom[A] is reachable, now if we find price[B->A] + bestfrom[A] < bestfrom[B],
        // we find a better route from B to dst.
        for (int k = 1; k <= K; k++) {              
            for (const auto& flight : flights) {
                if (bestfrom[flight[1]] != INT_MAX)
                    bestfrom[flight[0]] = min(bestfrom[flight[0]], bestfrom[flight[1]] + flight[2]);
            }
        }
        return bestfrom[src] == INT_MAX ? -1 : bestfrom[src];
    }
};