ACM-ICPC 2018 沈阳赛区网络预赛 E. The cake is a lie(最小圆覆盖)

题目链接

解析：

最小圆覆盖模板+二分答案

详解博客

知乎讲解

∠1是phi

∠2是theta

按逆时针来算角度的

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
#define Mn 300
const double eps = 1e-9;
const double pi = acos(-1.0);
#define sqr(x) ((x) * (x))
struct point
{
    double x,y;
    void read()
    {
    scanf("%lf%lf",&x,&y);
    }
    void print()
    {
        printf("%lf%lf\n",x,y);
    }
    double friend dis(const point &a,const point &b)
    {
        return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
    }
} p[Mn + 5];
struct alpha
{
    double v;
    bool flag;
    bool friend operator <(const alpha &a,const alpha &b)
    {
        return a.v < b.v;
    }
} alp[Mn * Mn + 5];
int solve(int n,double R)//半径为R的圆最多能覆盖多少个点
{
    int MAX = 0;
    for(int i = 0; i < n; i++)
    {
        int t = 0;
        for(int j = 0; j < n; j++)
        {
            if(i == j)
                continue;
            double theta,phi,D;
            D = dis(p[i],p[j]);
            if(D > 2.0 * R)
                continue;
            theta = atan2(p[j].y - p[i].y,p[j].x - p[i].x);
            if(theta < 0)
                theta += 2 * pi;
            phi = acos(D / (2.0 * R));
            alp[t].v = theta - phi + 2 * pi;
            alp[t].flag = true;
            alp[t + 1].v = theta + phi + 2 * pi;
            alp[t + 1].flag = false;
            t += 2;
        }
        sort(alp,alp + t);
        int sum = 0;
        for(int j = 0; j < t; j++)
        {
            if(alp[j].flag)
                sum ++;
            else
                sum --;
            if(sum > MAX)
                MAX = sum;
        }
    }
    return MAX+1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,s;
        int R;
        scanf("%d%d",&n,&s);
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        scanf("%d",&R);
        double l=0;
        double r=30000;
        double ans=0;
		while(r-l>eps)
        {
            double mid=(l+r)/2;
            if(solve(n,mid)>=s) ans=mid,r=mid;
            else l=mid+0.00001;
        }
		
        if(ans==0) printf("The cake is a lie.\n");
        else printf("%.4lf\n",(double )ans+R);
    }
    return 0;
}