hdu 1392 Surround the Trees

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1392

题意：给出一些点的坐标，求最小的凸多边形把所有点包围时此多边形的周长。

解法：凸包ConvexHull 模板题

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<cmath>

 5 #include<cstdlib>

 6 #include<algorithm>

 7 #define inf 0x7fffffff

 8 #define PI 3.141592654

 9 #define exp 1e-10

10 using namespace std;

11 struct Point

12 {

13     double x,y;

14     Point (double x=0,double y=0):x(x),y(y){}

15     friend bool operator < (Point A,Point B)

16     {

17         if (A.x != B.x) return A.x < B.x;

18         return A.y<B.y;

19     }

20 }an[111];

21 typedef Point Vector;

22 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }

23 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }

24 

25 int dcmp(double x)

26 {

27     if (fabs(x)<exp) return 0;

28     else return x<0 ? -1 : 1;

29 }

30 double cross(Vector A,Vector B)

31 {

32     return A.x*B.y-B.x*A.y;

33 }

34 int ConvexHull(Point *p,int n,Point *ch)

35 {

36     sort(p,p+n);

37     int m=0;

38     for (int i=0 ;i<n ;i++)

39     {

40         while (m>1 && dcmp(cross(ch[m-1]-ch[m-2] , p[i]-ch[m-2]))<=0) m--;

41         ch[m++]=p[i];

42     }

43     int k=m;

44     for (int i=n-2 ;i>=0 ;i--)

45     {

46         while (m>k && dcmp(cross(ch[m-1]-ch[m-2] , p[i]-ch[m-2]))<=0) m--;

47         ch[m++]=p[i];

48     }

49     ch[m++]=ch[0];

50     if (n>1) m--;

51     return m;

52 }

53 double dis(Point a,Point b)

54 {

55     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));

56 }

57 int main()

58 {

59     int n;

60     while (cin>>n,n)

61     {

62         for (int i=0 ;i<n ;i++)

63         scanf("%lf%lf",&an[i].x,&an[i].y);

64         if (n==1) {printf("0.00\n");continue;}

65         else if (n==2) {printf("%.2f\n",dis(an[0],an[1]));continue;}

66         Point bn[111];

67         int m=ConvexHull(an,n,bn);

68         double sum=0;

69         for (int i=0 ;i<m ;i++)

70         {

71             sum += dis(bn[i],bn[i+1]);

72         }

73         printf("%.2f\n",sum);

74     }

75     return 0;

76 }