AD isAQuest BellmanFord Sol1

public static double solve(int n, int m, int k, int g, Node[] V, Set<Edge> E) {
        double[][] dp = new double[k + 1][n + 1];
        // dp[i][j] = min cost to reach 'j'-th node and having left `i` collobarators
        for (int j = 1; j <= n; ++j) {
          Node node = V[j];
          for (int i = 0; i <= k; ++i)
            dp[i][node.getId()] = Double.MAX_VALUE / 2;
        }

        for (int i = 0; i <= k; ++i)
            dp[i][V[1].getId()] = 0;


        for (int node_idx = 1; node_idx <= n; ++node_idx) {
          Node node = V[node_idx];

          for (Edge edge: E)
            if (edge.from == node) // optimise later
              {
                dp[k][edge.to.getId()] = Math.min(dp[k][edge.to.getId()] , dp[k][edge.from.getId()] + edge.cost);

                for (int j = k - 1; j >= 0; --j)
                    dp[j][edge.to.getId()] = Math.min(dp[j][edge.to.getId()],
                    Math.min(dp[j][edge.from.getId()] + edge.getCost(), dp[j + 1][edge.from.getId()] + edge.getCost() / 2.0));

              }
        }

        // after using all `k` colaborators, min cost of reaching `g`
        double ans = dp[0][V[g].getId()];

        if (ans !=  Double.MAX_VALUE / 2)
          return ans;
        else return -1;
    }