Single source shortest path between two cities - GeeksforGeeks

24 Nov, 2021
Given a graph of N Nodes and E edges in form of {U, V, W} such that there exists an edge between U and V with weight W. You are given an integer K and source src and destination dst. The task is to find the cheapest cost path from given source to destination from K stops.Examples:  
Input: N = 3, edges = [[0, 1, 100], [1, 2, 100], [0, 2, 500]], src = 0, dst = 2, k = 1 Output: 200 Explanation: The Cheapest Price is from City 0 to City 2. With just one stop, it just costs 200 which is our Output.Input: N = 3, edges = [[0, 1, 100], [1, 2, 100], [0, 2, 500]], src = 0, dst = 2, k = 0 Output: 500  
 
Method 1: Using Depth First Search 
Explore the current node and keep exploring its children.Use a map to store the visited node in pair with stops and distance as value.Break the recursion tree if the key is present in the map.Find the answer (minimum cost) for each recursion tree and return it.
Explore the current node and keep exploring its children.
Use a map to store the visited node in pair with stops and distance as value.
Break the recursion tree if the key is present in the map.
Find the answer (minimum cost) for each recursion tree and return it.
 
Below is the implementation of our approach. 
C++
Python3
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Structure to implement hashing to// stores pairs in mapstruct pair_hash {    template <class T1, class T2>    std::size_t    operator()(const std::pair<T1, T2>& pair) const    {        return std::hash<T1>()(pair.first)               ^ std::hash<T2>()(pair.second);    }}; // DFS memoizationvector<vector<int> > adjMatrix;typedef std::pair<int, int> pair1;unordered_map<pair1, int, pair_hash> mp; // Function to implement DFS Traversallong DFSUtility(int node, int stops,                int dst, int cities){    // Base Case    if (node == dst)        return 0;     if (stops < 0) {        return INT_MAX;    }     pair<int, int> key(node, stops);     // Find value with key in a map    if (mp.find(key) != mp.end()) {        return mp[key];    }     long ans = INT_MAX;     // Traverse adjacency matrix of    // source node    for (int neighbour = 0;         neighbour < cities;         ++neighbour) {         long weight            = adjMatrix[node][neighbour];         if (weight > 0) {             // Recursive DFS call for            // child node            long minVal                = DFSUtility(neighbour,                             stops - 1,                             dst, cities);             if (minVal + weight > 0)                ans = min(ans,                          minVal                              + weight);        }    }     mp[key] = ans;     // Return ans    return ans;} // Function to find the cheapest price// from given source to destinationint findCheapestPrice(int cities,                      vector<vector<int> >& flights,                      int src, int dst, int stops){     // Resize Adjacency Matrix    adjMatrix.resize(cities + 1,                     vector<int>(cities + 1, 0));     // Traverse flight[][]    for (auto item : flights) {        // Create Adjacency Matrix        adjMatrix[item[0]][item[1]] = item[2];    }     // DFS Call to find shortest path    long ans = DFSUtility(src, stops,                          dst, cities);     // Return the cost    return ans >= INT_MAX ? -1 : (int)ans;    ;} // Driver Codeint main(){    // Input flight : {Source,    // Destination, Cost}    vector<vector<int> > flights        = { { 4, 1, 1 }, { 1, 2, 3 }, { 0, 3, 2 }, { 0, 4, 10 }, { 3, 1, 1 }, { 1, 4, 3 } };     // vec, n, stops, src, dst    int stops = 2;    int totalCities = 5;    int sourceCity = 0;    int destCity = 4;     // Function Call    long ans = findCheapestPrice(        totalCities, flights,        sourceCity, destCity,        stops);     cout << ans;    return 0;}
# Python3 program for the above approach       # DFS memoizationadjMatrix=[]mp=dict() # Function to implement DFS Traversaldef DFSUtility(node, stops, dst, cities):    # Base Case    if (node == dst):        return 0     if (stops < 0) :        return 1e9          key=(node, stops)     # Find value with key in a map    if key in mp:        return mp[key]          ans = 1e9     # Traverse adjacency matrix of    # source node    for neighbour in range(cities):        weight = adjMatrix[node][neighbour]         if (weight > 0) :             # Recursive DFS call for            # child node            minVal = DFSUtility(neighbour, stops - 1, dst, cities)             if (minVal + weight > 0):                ans = min(ans, minVal + weight)                   mp[key] = ans     # Return ans    return ans  # Function to find the cheapest price# from given source to destinationdef findCheapestPrice(cities, flights, src, dst, stops):    global adjMatrix    # Resize Adjacency Matrix    adjMatrix=[[0]*(cities + 1) for _ in range(cities + 1)]     # Traverse flight[][]    for item in flights:        # Create Adjacency Matrix        adjMatrix[item[0]][item[1]] = item[2]          # DFS Call to find shortest path    ans = DFSUtility(src, stops, dst, cities)     # Return the cost    return -1 if ans >= 1e9 else int(ans)       # Driver Codeif __name__ == '__main__':    # Input flight : :Source,    # Destination, Cost    flights = [[ 4, 1, 1 ],[ 1, 2, 3] , [ 0, 3, 2] , [ 0, 4, 10] , [ 3, 1, 1] , [ 1, 4, 3]]      # vec, n, stops, src, dst    stops = 2    totalCities = 5    sourceCity = 0    destCity = 4     # Function Call    ans = findCheapestPrice(        totalCities, flights,        sourceCity, destCity,        stops)     print(ans)
6
 
Time Complexity: O(V + E), where V is the number of nodes and E is the edges.Auxiliary Space: O(V)
Method 2: Using Breadth-First Search 
The idea is to use Queue to perform BFS Traversal.Perform traversal from current node and insert root node in the queue.Traverse the queue and explore the current node and its siblings. Then insert the siblings of the node in the queue.While exploring each sibling and keep pushing the entries in the queue if the cost is less and update the minimum cost.Print the minimum cost after the above traversal.
The idea is to use Queue to perform BFS Traversal.
Perform traversal from current node and insert root node in the queue.
Traverse the queue and explore the current node and its siblings. Then insert the siblings of the node in the queue.
While exploring each sibling and keep pushing the entries in the queue if the cost is less and update the minimum cost.
Print the minimum cost after the above traversal.
Below is the implementation of our approach: 
C++
Python3
// C++ program of the above approach#include <bits/stdc++.h>#include <iostream>#include <queue>#include <tuple>#include <unordered_map> using namespace std;// BSF Memoizationtypedef tuple<int, int, int> tupl; // Function to implement BFSint findCheapestPrice(int cities,                      vector<vector<int> >& flights,                      int src, int dst, int stops){    unordered_map<int,                  vector<pair<int, int> > >        adjList;     // Traverse flight[][]    for (auto flight : flights) {         // Create Adjacency Matrix        adjList[flight[0]].push_back(            { flight[1], flight[2] });    }     // < city, distancefromsource > pair    queue<pair<int, int> >        q;    q.push({ src, 0 });     // Store the Result    int srcDist = INT_MAX;     // Traversing the Matrix    while (!q.empty() && stops-- >= 0) {         int qSize = q.size();         for (int i = 0; i < qSize; i++) {            pair<int, int> curr = q.front();            q.pop();             for (auto next : adjList[curr.first]) {                 // If source distance is already                // least the skip this iteration                if (srcDist < curr.second                                  + next.second)                    continue;                 q.push({ next.first,                         curr.second                             + next.second });                 if (dst == next.first) {                    srcDist = min(                        srcDist, curr.second                                     + next.second);                }            }        }    }     // Returning the Answer    return srcDist == INT_MAX ? -1 : srcDist;} // Driver Codeint main(){    // Input flight : {Source,    // Destination, Cost}    vector<vector<int> > flights        = { { 4, 1, 1 }, { 1, 2, 3 }, { 0, 3, 2 }, { 0, 4, 10 }, { 3, 1, 1 }, { 1, 4, 3 } };     // vec, n, stops, src, dst    int stops = 2;    int totalCities = 5;     // Given source and destination    int sourceCity = 0;    int destCity = 4;     // Function Call    long ans = findCheapestPrice(        totalCities, flights,        sourceCity, destCity,        stops);    cout << ans;    return 0;}
# Python3 program of the above approach from queue import Queue as Q# Function to implement BFSdef findCheapestPrice(cities, flights, src, dst, stops):    adjList=dict()     # Traverse flight[][]    for flight in flights :         # Create Adjacency Matrix        if flight[0] in adjList:            adjList[flight[0]].append(                (flight[1], flight[2]))        else:            adjList[flight[0]]=[(flight[1], flight[2])]         q=Q()    q.put((src, 0))     # Store the Result    srcDist = 1e9    # Traversing the Matrix    while (not q.empty() and stops >= 0) :        stops-=1         for i in range(q.qsize()) :            curr = q.get()             for nxt in adjList[curr[0]]:                 # If source distance is already                # least the skip this iteration                if (srcDist < curr[1] + nxt[1]):                    continue                 q.put((nxt[0],curr[1] + nxt[1]))                 if (dst == nxt[0]):                    srcDist = min( srcDist, curr[1] + nxt[1])                     # Returning the Answer    return -1 if srcDist == 1e9 else srcDist  # Driver Codeif __name__ == '__main__':    # Input flight : :Source,    # Destination, Cost    flights= [[ 4, 1, 1 ],[ 1, 2, 3] , [ 0, 3, 2] , [ 0, 4, 10] , [ 3, 1, 1] , [ 1, 4, 3]]      # vec, n, stops, src, dst    stops = 2    totalCities = 5     # Given source and destination    sourceCity = 0    destCity = 4     # Function Call    ans = findCheapestPrice(        totalCities, flights,        sourceCity, destCity,        stops)    print(ans)
6
 
Time Complexity: O(Stops* N * N) where N is the Product of Cities and Size in QueueAuxiliary Space: O(N)
Method 3: Using Dijkstra Algorithm
Update the distance of all the vertices from the source.The vertices that are not directly connected from the source are marked with infinity and vertices that are directly connected are updated with the correct distance.Start from the source, and extract next min vertices. This will ensure the minimum cost.Do Edge Relaxation at each step: D denotes Distance and w denotes weightsIf D[u] + w(u, z) < D[z] thenD[z] = D[u] + w(u, z)
Update the distance of all the vertices from the source.
The vertices that are not directly connected from the source are marked with infinity and vertices that are directly connected are updated with the correct distance.
Start from the source, and extract next min vertices. This will ensure the minimum cost.
Do Edge Relaxation at each step: D denotes Distance and w denotes weightsIf D[u] + w(u, z) < D[z] thenD[z] = D[u] + w(u, z)
If D[u] + w(u, z) < D[z] thenD[z] = D[u] + w(u, z)
If D[u] + w(u, z) < D[z] then
D[z] = D[u] + w(u, z)
Here is the implementation of our approach:
C++
Python3
// C++ program for the above approach#include <bits/stdc++.h>#include <tuple>using namespace std; typedef tuple<int, int, int> tupl;long findCheapestPrice(int cities,                       vector<vector<int> >& flights,                       int src, int dst, int stops){    // Adjacency Matrix    vector<vector<pair<int, int> > > adjList(cities);     // Traverse flight[][]    for (auto flight : flights) {        // Create Adjacency Matrix        adjList[flight[0]].push_back(            { flight[1], flight[2] });    }     // Implementing Priority Queue    priority_queue<tupl, vector<tupl>,                   greater<tupl> >        pq;     tupl t = make_tuple(0, src, stops);    pq.push(t);     // While PQ is not empty    while (!pq.empty()) {        tupl t = pq.top();        pq.pop();         if (src == dst)            return 0;         int cost = get<0>(t);        int current = get<1>(t);        int stop = get<2>(t);         if (current == dst)             // Return the Answer            return cost;         if (stop >= 0) {            for (auto next : adjList[current]) {                 tupl t = make_tuple((cost                                     + next.second),                                    next.first,                                    stop - 1);                pq.push(t);            }        }    }    return -1;} int main(){    // Input flight : {Source,    // Destination, Cost}    vector<vector<int> > flights        = { { 4, 1, 1 }, { 1, 2, 3 }, { 0, 3, 2 }, { 0, 4, 10 }, { 3, 1, 1 }, { 1, 4, 3 } };     // vec, n, stops, src, dst    int stops = 2;    int totalCities = 5;     // Given source and destination    int sourceCity = 0;    int destCity = 4;     // Function Call    long ans = findCheapestPrice(        totalCities, flights,        sourceCity, destCity, stops);    cout << ans;    return 0;}
# Python3 program for the above approachimport heapq as hq  def findCheapestPrice(cities, flights, src, dst, stops):    # Adjacency Matrix    adjList = [[] for _ in range(cities)]    # Traverse flight[][]    for flight in flights:        # Create Adjacency Matrix        adjList[flight[0]].append((flight[1], flight[2]))     # Implementing Priority Queue    pq = []     t = (0, src, stops)    hq.heappush(pq, t)     # While PQ is not empty    while (pq):        t = hq.heappop(pq)         if (src == dst):            return 0         cost, current, stop = t         if (current == dst):             # Return the Answer            return cost         if (stop >= 0):            for nxt in adjList[current]:                 t = ((cost + nxt[1]), nxt[0], stop - 1)                hq.heappush(pq, t)    return -1  if __name__ == '__main__':    # Input flight : :Source,    # Destination, Cost    flights = [[4, 1, 1], [1, 2, 3], [0, 3, 2],               [0, 4, 10], [3, 1, 1], [1, 4, 3]]     # vec, n, stops, src, dst    stops = 2    totalCities = 5     # Given source and destination    sourceCity = 0    destCity = 4     # Function Call    ans = findCheapestPrice(        totalCities, flights,        sourceCity, destCity, stops)    print(ans)
6
 
Time Complexity: O(E+V log V) where V is the number of nodes and E is the edges.Auxiliary Space: O(V)
Method 4: Using Bellmon Ford
Initialize distances from the source to all vertices as infinite and distance to the source itself as 0.Do Edge Relaxation at each step: D denotes Distance and w denotes weightsIf D[u] + w(u, z) < D[z] then D[z] = D[u] + w(u, z)The algorithm has been modified to solve the given problem instead of relaxing the graph V-1 times, we will do it for the given number of stops.
Initialize distances from the source to all vertices as infinite and distance to the source itself as 0.
Do Edge Relaxation at each step: D denotes Distance and w denotes weightsIf D[u] + w(u, z) < D[z] then D[z] = D[u] + w(u, z)
If D[u] + w(u, z) < D[z] then D[z] = D[u] + w(u, z)
The algorithm has been modified to solve the given problem instead of relaxing the graph V-1 times, we will do it for the given number of stops.
Below is the implementation of the above approach 
C++
Python3
// C++ program of the above Approach#include <bits/stdc++.h>#include <vector>using namespace std; // Function to find the cheapest cost// from source to destination using K stopsint findCheapestPrice(int cities,                      vector<vector<int> >& flights,                      int src, int dst, int stops){    // Distance from source to all other nodes    vector<int> dist(cities, INT_MAX);    dist[src] = 0;     // Do relaxation for V-1 vertices    // here we need for K stops so we    // will do relaxation for k+1 stops    for (int i = 0; i <= stops; i++) {         vector<int> intermediate(dist);         for (auto flight : flights) {             if (dist[flight[0]] != INT_MAX) {                intermediate[flight[1]]                    = min(intermediate[flight[1]],                          dist[flight[0]]                              + flight[2]);            }        }         // Update the distance vector        dist = intermediate;    }     // Return the final cost    return dist[dst] == INT_MAX ? -1 : dist[dst];} // Driver Codeint main(){    // Input flight : {Source, Destination, Cost}    vector<vector<int> > flights        = { { 4, 1, 1 }, { 1, 2, 3 }, { 0, 3, 2 }, { 0, 4, 10 }, { 3, 1, 1 }, { 1, 4, 3 } };     // vec, n, stops, src, dst    int stops = 2;    int totalCities = 5;     // Given source and destination    int sourceCity = 0;    int destCity = 4;     // Function Call    long ans = findCheapestPrice(        totalCities, flights, sourceCity,        destCity, stops);    cout << ans;    return 0;}
# Python3 program of the above Approach # Function to find the cheapest cost# from source to destination using K stopsdef findCheapestPrice(cities, flights, src, dst, stops):    # Distance from source to all other nodes    dist=[1e9]*cities    dist[src] = 0     # Do relaxation for V-1 vertices    # here we need for K stops so we    # will do relaxation for k+1 stops    for i in range(stops):         intermediate=dist         for flight in flights:             if (dist[flight[0]] != 1e9) :                intermediate[flight[1]] = min(intermediate[flight[1]], dist[flight[0]] + flight[2])                               # Update the distance vector        dist = intermediate          # Return the final cost    return -1 if dist[dst] == 1e9 else dist[dst]  # Driver Codeif __name__ == '__main__':    # Input flight : :Source, Destination, Cost    flights = [[4, 1, 1], [1, 2, 3], [0, 3, 2],               [0, 4, 10], [3, 1, 1], [1, 4, 3]]     # vec, n, stops, src, dst    stops = 2    totalCities = 5     # Given source and destination    sourceCity = 0    destCity = 4     # Function Call    ans = findCheapestPrice(        totalCities, flights,        sourceCity, destCity, stops)    print(ans)
6
 
Time Complexity: O(E*V) where V is the number of nodes and E is the edges.Auxiliary Space:O(V)
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