[BZOJ1913][Apio2010]signaling 信号覆盖（计算几何+组合数学）

题目描述

题解

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 1505

const double pi=acos(-1.0);
const double eps=1e-9;
int dcmp(double x)
{
if (x<=eps&&x>=-eps) return 0;
return (x>0)?1:-1;
}
struct Point{double x,y;};
int n;
double ang[N+N],ao,tu,ans;
Point p
;

double C(int n,int m)
{
if (m>n) return 0.0;
double ans=1.0;
for (int i=n-m+1;i<=n;++i)
ans*=(double)i;
for (int i=1;i<=m;++i)
ans/=(double)i;
return ans;
}
double calc(int id)
{
int cnt=0;
for (int i=1;i<=n;++i)
if (id!=i)
ang[++cnt]=atan2(p[i].y-p[id].y,p[i].x-p[id].x);
sort(ang+1,ang+cnt+1);
for (int i=1;i<=cnt;++i) ang[cnt+i]=ang[i]+2*pi;
double ans=0;
int l=1,r=1;
while (l<=cnt)
{
r=max(r,l+1);
while (r<=cnt*2&&ang[r]<ang[l]+pi)
++r;
ans+=C(r-l-1,2);
++l;
}
return C(n-1,3)-ans;
}
int main()
{
scanf("%d",&n);
if (n==3) {puts("3.000000");return 0;}
for (int i=1;i<=n;++i) scanf("%lf%lf",&p[i].x,&p[i].y);
for (int i=1;i<=n;++i)
ao+=calc(i);
tu=C(n,4)-ao;
ans=(ao+2*tu)/C(n,3)+3.0;
printf("%.6lf\n",ans);
}