LightOJ 1203 Guarding Bananas (Convex Hull)

Jan's LightOJ :: Problem 1203 - Guarding Bananas

　　简单凸包。处理凸包的时候吧ch写成了pt，wa了两次。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 
 7 using namespace std;
 8 
 9 struct Point {
10     double x, y;
11     Point() {}
12     Point(double x, double y) : x(x), y(y) {}
13 } ;
14 
15 bool operator < (Point x, Point y) { return x.x < y.x || (x.x == y.x && x.y < y.y);}
16 bool operator == (Point x, Point y) { return x.x == y.x && x.y == y.y;}
17 Point operator - (Point x, Point y) { return Point(x.x - y.x, x.y - y.y);}
18 
19 const double EPS = 1e-8;
20 inline int sgn(double x) { return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);}
21 inline double crossDet(Point x, Point y) { return x.x * y.y - x.y * y.x;}
22 inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);}
23 
24 int andrew(Point *pt, int n, Point *ch) {
25     sort(pt, pt + n);
26     n = (int) (unique(pt, pt + n) - pt);
27     int m = 0;
28     for (int i = 0; i < n; i++) {
29         while (m > 1 && sgn(crossDet(ch[m - 2], ch[m - 1], pt[i])) <= 0) m--;
30         ch[m++] = pt[i];
31     }
32     int k = m;
33     for (int i = n - 2; i >= 0; i--) {
34         while (m > k && sgn(crossDet(ch[m - 2], ch[m - 1], pt[i])) <= 0) m--;
35         ch[m++] = pt[i];
36     }
37     if (n > 1) m--;
38     return m;
39 }
40 
41 const int N = 111111;
42 Point pt[N], ch[N];
43 const double PI = acos(-1.0);
44 
45 inline double angle(Point x) { return atan2(x.y, x.x);}
46 
47 int main() {
48 //    freopen("in", "r", stdin);
49     int T, n;
50     cin >> T;
51     for (int cas = 1; cas <= T; cas++) {
52         cin >> n;
53         for (int i = 0; i < n; i++) {
54             scanf("%lf%lf", &pt[i].x, &pt[i].y);
55         }
56         if (n <= 2) {
57             printf("Case %d: 0.00000000\n", cas);
58             continue;
59         }
60         n = andrew(pt, n, ch);
61         for (int i = 0; i < 10; i++) {
62             ch[i + n] = ch[i];
63         }
64         double mini = 1e10;
65         for (int i = 0; i < n; i++) {
66             double ang1 = angle(ch[i] - ch[i + 1]);
67             double ang2 = angle(ch[i + 2] - ch[i + 1]);
68             double da = fabs(ang1 - ang2);
69             if (da > PI) da = 2.0 * PI - da;
70             mini = min(mini, da * 180.0 / PI);
71         }
72         printf("Case %d: %.8f\n", cas, mini);
73     }
74     return 0;
75 }

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——written by Lyon

posted @ 2013-05-29 03:10 LyonLys 阅读(349) 评论(0) 编辑收藏举报

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