ACM-ICPC 2018 沈阳赛区网络预赛 E The cake is a lie 几何 单位圆覆盖

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先转化问题 对每个蛋糕只考虑圆心 求圈住s个圆心的最小半径
那么这个最小半径加R一定可以包含s个圆

#include <iostream>
#include <algorithm>
#include <sstream>
#include <string>
#include <queue>
#include <cstdio>
#include <map>
#include <set>
#include <utility>
#include <stack>
#include <cstring>
#include <cmath>
#include <vector>
#include <ctime>
#include <bitset>
#include <assert.h>
using namespace std;
#define pb push_back
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define ss(str) scanf("%s",str)
#define ansn() printf("%d\n",ans)
#define lansn() printf("%lld\n",ans)
#define r0(i,n) for(int i=0;i<(n);++i)
#define r1(i,e) for(int i=1;i<=e;++i)
#define rn(i,e) for(int i=e;i>=1;--i)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define lowbit(a) (a&(-a))
#define all(a) a.begin(),a.end()
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp(aa,bb) make_pair(aa,bb)
#define lrt rt<<1
#define rrt rt<<1|1
#define X first
#define Y second
#define PI (acos(-1.0))
double pi = acos(-1.0);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const ll mod = 1000000007;
const double eps=1e-9;
const int inf=0x3f3f3f3f;
//const ll infl = 100000000000000000;//1e17
const int maxn=  2e6+20;
const int maxm = 5e3+20;
//muv[i]=(p-(p/i))*muv[p%i]%p;
inline int in(int &ret) {
    char c;
    int sgn ;
    if(c=getchar(),c==EOF)return -1;
    while(c!='-'&&(c<'0'||c>'9'))c=getchar();
    sgn = (c=='-')?-1:1;
    ret = (c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');
    ret *=sgn;
    return 1;
}
struct po {
    double x,y;
    void in() {
        sff(x,y);
    }
} p[maxn];
struct node {
    double ang;
    int in;
    bool operator < (const node &o)const {
        return ang<o.ang;
    }
} no[maxn];
double dis(po a,po b) {
    return sqrt( (a.x-b.x) * (a.x-b.x) + (a.y-b.y) * (a.y-b.y) );
}
int main() {
#ifdef LOCAL
    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
#endif // LOCAL

    int t;
    sd(t);
    for(; t--;) {
        int n,s;
        sdd(n,s);
        r1(i,n)p[i].in();
        int R;
        sd(R);
        if(n<s) {
            puts("The cake is a lie.");
            continue;
        }
        double l = 0,r = 20000;
        r0(_,40) {
            double mid = (l+r)/2;
            int mx = 0;
            for(int i=1; i<=n; ++i) {
                int cnt = 0;
                int ans = 0;
                for(int j=1; j<=n; ++j) {
                    if(i==j)continue;
                    double dist = dis(p[i],p[j]);
                    if( dist > 2.0*mid)continue;
                    double theta = atan2( p[j].y - p[i].y , p[j].x - p[i].x);
                    if(theta<0)theta += 2*PI;
                    double phi = acos( dist / (2.0*mid) );
                    no[++cnt].ang = theta - phi + 2*PI;
                    no[cnt].in = 1;
                    no[++cnt].ang = theta + phi + 2*PI;
                    no[cnt].in = -1;
                }
                sort(no+1,no+1+cnt);
                for(int j=1; j<=cnt; ++j) {
                    ans += no[j].in;
                    mx = max(mx,ans);
                }
            }
            if(mx+1>=s)r = mid;
            else l = mid;
        }
        double ans = l;
        ans += R;
        printf("%.4f\n",ans);
    }
    return 0;
}