10 wizards, C++ (gcc)

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/*30 10 Wizards
There are 10 wizards, 0‑9, you are given a list that 
each entry is a list of wizards known by wizard. 
Define the cost between wizards and wizard as square of different of i and j. 
To find the min cost between 0 and 9.
For example:
wizard[0] list: 1, 4, 5
wizard[4] list: 9
wizard 0 to 9 min distance is (0‑4)^2+(4‑9)^2=41 (wizard[0]‑>wizard[4]‑>wizard[9])
*/
#include <iostream>
#include <vector>
#include <set> 
using namespace std;

// DFS to get the minimum cost with the path vector return .... 
class Wizard{
    int miniCost; 
public: 
    Wizard(){
    }
    int findCost(vector<vector<int>>& costs, int src, int dst){
        miniCost = 1<<30; 
        int cost = 0; 
        set<int> visited;//
        DFS(costs, src, dst, cost, visited);
        if(miniCost == 1<<30) return -1;
        return miniCost;
    }
 private:
    void DFS(vector<vector<int>>& costs, int src, int dst, int cost, set<int>& visited){
        if(cost > miniCost) return; // too expensive 
        if(src == dst){ // find a possible solution
          if(cost < miniCost) 
            miniCost = min(cost, miniCost); 
          return; // always return,  
        }
        
        visited.insert(src); // visit node
        for(auto next: costs[src]){ // visit node's neighbors 
            if(visited.count(next)) continue; 
            DFS(costs, next, dst, cost + (next-src)*(next-src), visited);  
        }
    }   
};

int main(){
    vector<vector<int>> costs(10);
    costs[0] = {1,4,5};
    costs[4] = {9};
    Wizard w; 
    int cost = w.findCost(costs, 0, 4);//
    cout << "cost should equal (41) = " << cost << endl;
    return 0;
}

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Compilation time: 0.63 sec, absolute running time: 0.07 sec, cpu time: 0.01 sec, memory peak: 3 Mb, absolute service time: 0,71 sec