hdu 3934 Summer holiday （凸包+旋转卡壳）

　　晚上为了演示给师弟看水平序的凸包是多么的好写，于是就随便找了一题凸包，25min居然1y掉了。。

代码如下：

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <iostream>
 4 #include <cstring>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 const int N = 1111111;
10 const double EPS = 1e-8;
11 inline int sgn(double x) { return (x > EPS) - (x < -EPS);}
12 struct Point {
13     double x, y;
14     Point() {}
15     Point(double x, double y) : x(x), y(y) {}
16     Point operator - (Point a) { return Point(x - a.x, y - a.y);}
17     bool operator < (Point a) const { return sgn(x - a.x) < 0 || sgn(x - a.x) == 0 && y < a.y;}
18 } ;
19 
20 inline double cross(Point a, Point b) { return a.x * b.y - a.y * b.x;}
21 
22 int andrew(Point *pt, Point *ch, int n) {
23     sort(pt, pt + n);
24     int m = 0;
25     for (int i = 0; i < n; i++) {
26         while (m > 1 && sgn(cross(pt[i] - ch[m - 2], ch[m - 1] - ch[m - 2])) <= 0) m--;
27         ch[m++] = pt[i];
28     }
29     int k = m;
30     for (int i = n - 2; i >= 0; i--) {
31         while (m > k && sgn(cross(pt[i] - ch[m - 2], ch[m - 1] - ch[m - 2])) <= 0) m--;
32         ch[m++] = pt[i];
33     }
34     if (n > 1) m--;
35     return m;
36 }
37 
38 Point ch[N], pt[N];
39 inline double area(Point a, Point b, Point c) { return fabs(cross(c - a, b - a));}
40 
41 double work(Point *pt, int n) {
42     if (n < 3) return 0;
43     double ans = area(pt[0], pt[1], pt[2]);
44     for (int i = 0, j = 1, k = 2; i < n; i++) {
45         while (true) {
46             bool df = false;
47             if (i == j) j = (j + 1) % n;
48             if (j == k) k = (k + 1) % n;
49             while (true) {
50                 double a1 = area(pt[i], pt[j], pt[k]);
51                 double a2 = area(pt[i], pt[j], pt[(k + 1) % n]);
52                 ans = max(ans, a1);
53                 if (sgn(a1 - a2) > 0) break; 
54                 k = (k + 1) % n;
55                 df = true;
56             }
57             while (true) {
58                 double a1 = area(pt[i], pt[j], pt[k]);
59                 double a2 = area(pt[i], pt[(j + 1) % n], pt[k % n]);
60                 ans = max(ans, a1);
61                 if (sgn(a1 - a2) > 0) break; 
62                 j = (j + 1) % n;
63                 df = true;
64             }
65             if (!df) break;
66         }
67     }
68     return ans / 2;
69 }
70 
71 int main() {
72     int n;
73     while (~scanf("%d", &n)) {
74         for (int i = 0; i < n; i++) scanf("%lf%lf", &pt[i].x, &pt[i].y);
75         n = andrew(pt, ch, n);
76         printf("%.2f\n", work(ch, n));
77     }
78     return 0;
79 }

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——written by Lyon

posted @ 2013-07-30 03:04 LyonLys 阅读(288) 评论(0) 收藏举报

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