【POJ 1981 】Circle and Points

当两个点距离小于直径时，由它们为弦确定的一个单位圆（虽然有两个圆，但是想一想知道只算一个就可以）来计算覆盖多少点。

#include <cstdio>
#include <cmath>
#define N 301
#define eps 1e-5
using namespace std;
int n,ans,tol;
double x[N],y[N],dx,dy;
inline double sqr(double x)
{
    return x*x;
}
inline double dis(int a,int b)
{
    return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b]));
}
inline void solve()
{
    tol=0;
    for(int i=1; i<=n; i++)
        if(dis(0,i)<=1+eps)
            tol++;
    if(tol>ans)ans=tol;
}
inline void get(int a,int b){
    double v=atan((x[b]-x[a])/(y[a]-y[b]));//等于0时会返回inf，v=pi/2
    double c=sqrt(1-sqr(dis(a,b)/2.0));
    dx= c*cos(v);
    dy= c*sin(v);
}
int main()
{
    while(scanf("%d",&n)&&n)
    {
        ans=1;
        for (int i=1; i<=n; i++)
            scanf("%lf%lf",&x[i],&y[i]);
        for (int i=1; i<=n; i++)
            for (int j=i+1; j<=n; j++)
            if(dis(i,j)<2){
                double mx=(x[i]+x[j])/2.0,my=(y[i]+y[j])/2.0;
                get(i,j);
                x[0]=mx+dx,y[0]=my+dy;
                solve();
            }
        printf("%d
",ans);
    }
}