Hdu4987a simple probability problem. (convex hull)

Link

In the last session of multiple schools, I didn't understand the meaning of the question at the time. I thought it was a probability problem and I didn't have to look into it.

Official question

For the first one he said, considering that the probability of a line segment with a length of L is 2L/(pI * D), we can understand that we do not know what we are talking about ..

According to my initial idea, I want to enumerate the angle and calculate the probability based on the height difference of the convex hull. However, there is a simpler way to find the perimeter of the convex hull in the question. In this way, I understand that, the probability of a line in a convex hull passing through a straight line is as 2L/(pI * d) As mentioned above. However, because he is a polygon, there must be a forward line, so passing through a line segment is equivalent to passing through two lines. The whole is/2... I don't know if this method is correct.

 1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set>10 using namespace std;11 #define N 11112 #define LL long long13 #define INF 0xfffffff14 const double eps = 1e-8;15 const double pi = acos(-1.0);16 const double inf = ~0u>>2;17 struct point18 {19     double x,y;20     point(double x=0,double y =0 ):x(x),y(y){}21 }p[N],ch[N];22 typedef point pointt;23 point operator -(point a,point b)24 {25     return point(a.x-b.x,a.y-b.y);26 }27 double cross(point a,point b)28 {29     return a.x*b.y-a.y*b.x;30 }31 int dcmp(double x)32 {33     if(fabs(x)<eps) return 0;34      else return x<0?-1:1;35 }36 double mul(point p0,point p1,point p2)37 {38     return cross(p1-p0,p2-p0);39 }40 double dis(point a)41 {42     return sqrt(a.x*a.x+a.y*a.y);43 }44 bool cmp(point a,point b)45 {46     if(dcmp(mul(p[0],a,b))==0)47         return dis(a-p[0])<dis(b-p[0]);48     else49         return dcmp(mul(p[0],a,b))>0;50 }51 int Graham(int n)52 {53     int i,k = 0,top;54     point tmp;55     for(i = 0 ; i < n; i++)56     {57         if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x))58             k = i;59     }60     if(k!=0)61     {62         tmp = p[0];63         p[0] = p[k];64         p[k] = tmp;65     }66     sort(p+1,p+n,cmp);67     ch[0] = p[0];68     ch[1] = p[1];69     top = 1;70     for(i = 2; i < n ; i++)71     {72         while(top>0&&dcmp(mul(ch[top-1],ch[top],p[i]))<0)73             top--;74         top++;75         ch[top] = p[i];76     }77     return top;78 }79 80 int main()81 {82     int t,i,n,d;83     int kk =0 ;84     cin>>t;85     while(t--)86     {87         scanf("%d%d",&n,&d);88         for(i =0 ; i < n;  i++)89         scanf("%lf%lf",&p[i].x,&p[i].y);90         int m = Graham(n);91         ch[m+1] = ch[0];92         double ans =0 ;93         for(i = 0 ; i <= m ; i++)94         ans += dis(ch[i]-ch[i+1]);95         printf("Case #%d: %.4f\n",++kk,ans/(pi*d));96     }97     return 0;98 }

View code

 

Hdu4987a simple probability problem. (convex hull)