For one of my intro to C++ assignments, I must write code where a group of athletes performances is judged by a variable number of judges. The performance score of the athletes is obtained by rejecting both the lowest and the highest score, and taking the average of the remaining scores. I have most of it figured out I think, however the minimum score is a bit over my head; the max score can be reset to 0, for following athletes to be judged, but the minimum cannot, because no score will be less than 0 to be picked up as a low score. My code:

Code:
#include <iostream>
using namespace std;
int main ()
{
    int athleteid, numjudges, startjudges, scorecard;
    double score, total, average, min, min2, max=0;
   cout.setf(ios::fixed,ios::floatfield);
   cout.precision (1);
   cout << "Welcome to the athlete performance score program" << endl;
   cout << "To start, please enter the athlete ID> ";
   cin >> athleteid;
       while (athleteid > 0)
       {
             scorecard=1;
             cout << "How many judges gave entered a score?> ";
             cin >> numjudges;
             for (startjudges=1; startjudges <= numjudges; startjudges++)
             {
             cout << "Please enter score " << scorecard << ">";
             cin >> score;
             if (score > max)
              {
                 max = score;
                 }  
             if (score < min)
             {
                min = score;
                }
             total = total+score;
             scorecard++;
             }
             average= (total-(max+min))/(numjudges-2);
             cout << "The average is " << average << endl;
             total=0;
             max=0;
             min=(I get stuck here);
             cout << "Finished. Enter next athlete ID, 0 to finish> ";
             cin >> athleteid;
       }
           system("pause");
}
Everything works out well for the most part, but I have no clue on how to reset the min variable, to pick up the lowest number again while the loop goes back around.

Any help would be appreciated.