题目传送门
1 /*
2 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
3 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，
4 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
5 对点离原点排序竟然submit failed，别人的代码有毒！
6 */
7 #include <cstdio>
8 #include <cstring>
9 #include <algorithm>
10 #include <map>
11 #include <cmath>
12 using namespace std;
13
14 const int MAXN = 1e3 + 10;
15 const int INF = 0x3f3f3f3f;
16 int x[MAXN], y[MAXN];
17 int dir[4][2] = {1, 0, 0, -1, -1, 0, 0, 1};
18 int n;
19 int vis[4][MAXN];
20
21 int work(void)
22 {
23 int nx = 0, ny = 0; int d = 0; int res = 0;
24 memset (vis, 0, sizeof (vis));
25
26 while (true)
27 {
28 int mx = INF, p = -1;
29 for (int i=1; i<=n; ++i)
30 {
31 if (d == 0 || d == 2)
32 {
33 if (ny != y[i]) continue;
34 if ((x[i] - nx) * dir[d][0] < 0) continue;
35 }
36 else
37 {
38 if (nx != x[i]) continue;
39 if ((y[i] - ny) * dir[d][1] < 0) continue;
40 }
41
42 int tmp = abs (x[i] - nx) + abs (y[i] - ny);
43 if (tmp < mx) {mx = tmp; p = i;}
44 }
45
46 if (p == -1) return res;
47 if (vis[d][p]) return -1;
48 vis[d][p] = 1;
49 nx = x[p] - dir[d][0]; ny = y[p] - dir[d][1];
50 d = (d + 1) % 4; res++;
51 }
52
53 }
54
55 int main(void) //SCU 4445 Right turn
56 {
57 // freopen ("E.in", "r", stdin);
58
59 while (scanf ("%d", &n) == 1)
60 {
61 for (int i=1; i<=n; ++i)
62 {
63 scanf ("%d%d", &x[i], &y[i]);
64 }
65 printf ("%d
", work ());
66 }
67
68 return 0;
69 }