# Shortest path in a graph from a source S to destination D with exactly K edges for multiple Queries

Given a graph with N nodes, a node S and Q queries each consisting of a node D and K, the task is to find the shortest path consisting of exactly K edges from node S to node D for each query. If no such path exists then print -1.

Note: K will always be lesser than 2 * N.

Examples:

Input: N = 3, edges[][] = {{1, 2, 5}, {2, 3, 3}, {3, 1, 4}}, S = 1, Q = {{1, 0}, {2, 1}, {3, 1}, {3, 2}, {3, 5}}
Output: 0 5 -1 8 20
1. The shortest path from 1 to 1 using 0 edge will be 0.
2. The shortest path from 1 to 2 using 1 edge will be 5 i.e 1->2.
3. No path of 1 edge exists between nodes 1 and 3.
4. The shortest path from 1 to 3 using 2 edges will be 8 i.e 1->2->3.
5. The shortest path from 1 to 3 using 5 edges will be 20 i.e 1->2->3->1->2->3.

Input: N = 4, edges[][] = {{1, 2, 8}, {2, 3, 5}, {3, 4, 7}}, S = 1, Q = {{1, 0}, {2, 1}, {3, 1}, {3, 2}, {4, 5}}
Output: 0 8 -1 13 -1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• This problem can be solved with the help of dynamic programming to create a linear solution.
• Initialise a 2-d array, dp[N][2*N] with initial value as ‘inf’ except dp[S][0] as 0.
• Pre-process the graph to find the shortest distance of each and every node from the source for every edge length between {0 to N-1}. The array dp[][] will be used to store the results of the pre processing.
• For the pre-processing, run a loop for J in range [1, 2*N-1] to find the dp[X][J] for each edge in the graph, where dp[X][J] be the shortest path from node ‘S’ to node ‘X’ using exactly ‘J’ edges in total.
• We can find dp[X][J+1] with the help of a recurrence relation:

dp[ edge.second ][ i ] = min(dp[ edge.second ][ i ], dp[ edge.first ][ i-1 ] + weight(edge))

• For every query in Q, if(dp[X][k] == inf) then return -1, else return dp[X][k]

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach    #include using namespace std;    #define inf 100000000    // Function to find the shortest path // between S and D with exactly K edges void ansQueries(int s,                 vector, int> > ed,                 int n, vector > q) {        // To store the dp states     int dp[n + 1][2 * n];        // Initialising the dp[][] array     for (int i = 0; i <= n; i++)         dp[i][0] = inf;     dp[s][0] = 0;        // Pre-Processing     for (int i = 1; i <= 2 * n - 1; i++) {            // Initialising current state         for (int j = 0; j <= n; j++)             dp[j][i] = inf;            // Updating current state         for (auto it : ed) {             dp[it.first.second][i]                 = min(                     dp[it.first.second][i],                     dp[it.first.first][i - 1] + it.second);         }     }        for (int i = 0; i < q.size(); i++) {         if (dp[q[i].first][q[i].second] == inf)             cout << -1 << endl;         else             cout << dp[q[i].first][q[i].second]                  << endl;     } }    // Driver code int main() {     int n = 3;     vector, int> > ed;        // Edges     ed = { { { 1, 2 }, 5 },            { { 2, 3 }, 3 },            { { 3, 1 }, 4 } };        // Source     int s = 1;        // Queries     vector > q = { { 1, 0 },                                   { 2, 1 },                                   { 3, 1 },                                   { 3, 2 },                                   { 3, 5 } };        // Function to answer queries     ansQueries(s, ed, n, q);        return 0; }

## Python3

 # Python3 implementation of the approach  import sys,numpy as np    inf = sys.maxsize;    # Function to find the shortest path  # between S and D with exactly K edges  def ansQueries(s, ed, n, q) :        # To store the dp states      dp = np.zeros((n + 1, 2 * n));         # Initialising the dp[][] array      for i in range(n + 1) :          dp[i][0] = inf;                 dp[s][0] = 0;         # Pre-Processing      for i in range( 1, 2 * n) :            # Initialising current state          for j in range( n + 1) :             dp[j][i] = inf;             # Updating current state          for it in ed :             dp[it[1]][i] = min( dp[it[1]][i],                                  dp[it[0]][i - 1] + ed[it]);             for i in range(len(q)) :         if (dp[q[i][0]][q[i][1]] == inf) :             print(-1);          else :             print(dp[q[i][0]][q[i][1]]);    # Driver code  if __name__ == "__main__" :         n = 3;         # Edges      ed = { ( 1, 2 ) : 5 ,          ( 2, 3 ) : 3 ,          ( 3, 1 ) : 4 };         # Source      s = 1;         # Queries      q = [          ( 1, 0 ),          ( 2, 1 ),          ( 3, 1 ),          ( 3, 2 ),          ( 3, 5 )         ];         # Function to answer queries      ansQueries(s, ed, n, q);         # This code is contributed by AnkitRai01

Output:

0
5
-1
8
20

Time Complexity: O(Q + N*E)
Space Complexity: O(N*N)

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