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\n", work ()); 66 } 67 68 return 0; 69 }
