模拟+贪心 SCU 4445 Right turn

题目传送门

 /*
     题意；从原点出发，四个方向，碰到一个点向右转，问多少次才能走出，若不能输出-1
     模拟：碰到的点横坐标相等或纵坐标相等，然而要先满足碰到点最近，
         当没有转向或走到之前走过的点结束循环。dir数组使得代码精简巧妙
     对点离原点排序竟然submit failed，别人的代码有毒！
 */
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <map>
 #include <cmath>
 using namespace std;

 const int MAXN = 1e3 + ;
 const int INF = 0x3f3f3f3f;
 int x[MAXN], y[MAXN];
 int dir[][] = {, , , -, -, , , };
 int n;
 int vis[][MAXN];

 int work(void)
 {
     int nx = , ny = ;    int d = ;    int res = ;
     memset (vis, , sizeof (vis));

     while (true)
     {
         int mx = INF, p = -;
         for (int i=; i<=n; ++i)
         {
             if (d ==  || d == )
             {
                 if (ny != y[i])    continue;
                 if ((x[i] - nx) * dir[d][] < )    continue;
             }
             else
             {
                 if (nx != x[i])    continue;
                 if ((y[i] - ny) * dir[d][] < )    continue;
             }

             int tmp = abs (x[i] - nx) + abs (y[i] - ny);
             if (tmp < mx)    {mx = tmp; p = i;}
         }

         if (p == -)    return res;
         if (vis[d][p])    return -;
         vis[d][p] = ;
         nx = x[p] - dir[d][];    ny = y[p] - dir[d][];
         d = (d + ) % ;    res++;
     }

 }

 int main(void)        //SCU 4445 Right turn
 {
 //    freopen ("E.in", "r", stdin);

     while (scanf ("%d", &n) == )
     {
         for (int i=; i<=n; ++i)
         {
             scanf ("%d%d", &x[i], &y[i]);
         }
         printf ("%d\n", work ());
     }

     return ;
 }