模拟+贪心 SCU 4445 Right turn

 

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 }

 

posted @ 2015-07-14 10:08  Running_Time  阅读(256)  评论(0)  编辑  收藏  举报