Solution Notes: This is a fairly standard shortest path problem, where we wish to find the shortest path from the initial location of the tractor to the outer boundary of the square with coordinates in the range 1..1000. Squares containing haybales cost 1 unit to cross, and all other squares cost 0 units. We could solve this problem using Dijkstra's algorithm, although the fact that costs are either zero or one allows us to use a slightly different (perhaps slightly simpler) approach using a pair of queues, one containing all the squares we intend to visit that are at distance zero away from our current location, and the other containing a list of all squares we intend to visit that are at distance one away from our current location. We always visit squares from the "zero away" quque, and when this empties out, we refill it with the contents of the "one away" queue. The total running time is therefore linear in the area of the square we are searching.

#include <cstdio>
#include <queue>

using namespace std;

struct Point {
  int x, y;
  Point (int _x, int _y) { x = _x; y = _y; }
  Point (void) { x=y=0; }
};

queue<Point> zero_away, one_away;
int A[1002][1002], D[1002][1002];

int relax(int curx, int cury, int x, int y)
{
  if (x>=0 && x<=1001 && y>=0 && y<=1001 && (D[x][y]==0 || D[curx][cury]+A[x][y]<D[x][y])) {
    D[x][y] = D[curx][cury]+A[x][y];
    if (A[x][y]==0) zero_away.push(Point(x,y));
    else            one_away.push(Point(x,y));
  }
}

int main(void)
{
  Point p;
  int i, n;

  freopen ("tractor.in", "r", stdin);
  freopen ("tractor.out", "w", stdout);
  
  scanf ("%d %d %d", &n, &p.x, &p.y);
  D[p.x][p.y] = 1;
  zero_away.push(p);
  for (i=0; i<n; i++) {
    scanf ("%d %d", &p.x, &p.y);
    A[p.x][p.y] = 1;
  }

  while (!zero_away.empty() || !one_away.empty()) {
    if (zero_away.empty()) 
      while (!one_away.empty()) {
	zero_away.push(one_away.front()); one_away.pop();
      }
    p = zero_away.front(); zero_away.pop();
    relax(p.x,p.y,p.x-1,p.y);
    relax(p.x,p.y,p.x+1,p.y);
    relax(p.x,p.y,p.x,p.y-1);
    relax(p.x,p.y,p.x,p.y+1);
  }

  printf ("%d\n", D[0][0]-1);

  return 0;
}