# 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] 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] = inf; ` `    ``dp[s] = 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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