版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/cys460714380/article/details/82534960
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;
}