Описать структуру с именем airport. Реализовать сортировку записей

#include <iostream>
#include <stdlib.h>
using namespace std;
struct Airport
{
int numberOfRase;
char destination [100];
char typeOfPlane [100]; };
 
void main()
{
Airport a;
cout<<"enter destination:"<< endl;
cin>> a.destination;
Airport routes [10];
int N;
cin>>N;
for (int i; i=0; i<N;i++)
{
cout<< "enter type of plane:"<<endl;
cin>> routes[i].typeOfPlane;
cout<< "enter number of rase:"<<endl;
cin << routes[i].numberOfRase;}
 
}

#include <stdio.h>
#include <conio.h>
 
#define _SIZE 256
#define __countof(ptr) sizeof(ptr)/sizeof(ptr[0]) \
 
typedef struct Airport
{
    int numberOfFlight;
    char destination[_SIZE];
    char typeOfPlane[_SIZE]; 
} AIRPORT;
 
static AIRPORT airport[] = { { 1, "Riga", "1" },
                             { 2, "New York", "2" },
                             { 3, "Moscow", "3" },
                             { 4, "Athens", "4" },
                             { 5, "New Orlean", "5" },
                             { 6, "Milan", "6" } }; 
 
bool iseqkey(char* str1, char* str2, int len)
{
    int index = 0; 
    for (int q = len-1; q >= 0; q--)
        if (str1[q] != str2[q]) return false;
 
    return true;
}
 
void swap(AIRPORT& obj1, AIRPORT& obj2)
 { AIRPORT _tobj = obj1; obj1 = obj2; obj2 = _tobj; }
 
int main(int argc, char* argv[])
{
    for (int q = 0; q < _SIZE; q++)
        for (int i = 0; i != __countof(airport); i++)
        {
            int min = i;
            for (int r = i+1; r != __countof(airport); r++)
                min = (airport[r].destination[q] < airport[min].destination[q]) ? r : min;
 
            if (iseqkey(airport[i].destination, airport[min].destination, q))
                swap(airport[i], airport[min]);
        }
 
    for (int i = 0; i != __countof(airport); i++)
        printf("NumberOfFlight = %d Destination = %s TypeOfPlane = %s\n", 
            airport[i].numberOfFlight, airport[i].destination, airport[i].typeOfPlane);
 
     _getch();
 
    return 0;
}