【bzoj1913】 Apio2010—signaling 信号覆盖

题意

Solution

细节

代码

// bzoj1913
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#define LL long long
#define inf 2147483640
#define eps 1e-6
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;

const int maxn=2000;
struct point {int x,y;}p[maxn];
LL C[maxn][maxn];
double a[maxn<<1];
int n;

int main() {
scanf("%d",&n);
for (int i=1;i<=n;i++) scanf("%d%d",&p[i].x,&p[i].y);
for (int i=0;i<=n;i++) C[i][0]=1;
for (int i=1;i<=n;i++)
for (int j=1;j<=i;j++) C[i][j]=C[i-1][j]+C[i-1][j-1];
LL t2,t1=0;
for (int i=1;i<=n;i++) {
int cnt=0;
for (int j=1;j<=n;j++) if (i!=j) {
a[++cnt]=atan2(p[j].y-p[i].y,p[j].x-p[i].x);
if (a[cnt]<0) a[cnt]+=2*Pi;
}
sort(a+1,a+cnt+1);
for (int j=1;j<=cnt;j++) a[j+cnt]=a[j]+2*Pi;
for (int j=1,k=1;j<=cnt;j++) {
while (a[k+1]-a[j]<Pi && k<2*cnt) k++;
t1-=C[k-j][2];   //凸多边形个数
}
t1+=C[n-1][3];
}
t2=C
[4]-t1;
printf("%.6lf",(double)(t1+t2*2)/C
[3]+3);
return 0;
}