给定半径圆心未定的圆和若干点_pesky mosquitoes csdn-优快云博客

Pesky Mosquitoes

Time Limit: 4000ms, Special Time Limit:10000ms,Memory Limit:65536KB

Total submit users: 9, Accepted users:9

Problem 13239 : No special judgement

Problem description

Mosquitoes are relentless this time of year! They have absolutely ruined your attempt at a picnic and it is time to take your revenge. Unfortunately, you are not well equipped to ward off these pests. All you have got at your disposal is an empty bowl that previously held potato salad. As you glance down at the picnic table, you see a number of mosquitoes waiting idly for you to let your guard down. This is your chance to fight back.

Your task is to determine the maximum number of mosquitoes that can be trapped by quickly bringing down the inverted bowl onto the table. You will be provided with the diameter of the bowl and the exact location of each mosquito on the table. In this exercise you can assume that the mosquitoes are incredibly small and can simply be modeled as a point. A mosquito that lies exactly under the edge of the bowl is considered trapped.

Input

The first number in the input will be an integer 1 ≤ n ≤ 100 that denotes the number of mosquito- trapping scenarios that follow. A blank line comes at the beginning of each scenario. Then follows a line containing an integer 1 ≤ m ≤ 32 (the number of mosquitoes) and a real number 0 < d ≤ 200 (the diameter of the bowl). Each of the following m lines will specify the location of a mosquito in the form of real coordinates −100 ≤ x ≤ 100 and −100 ≤ y ≤ 100.

Output

For each scenario, you are to print the maximum number of mosquitoes that can be caught under the bowl in that scenario. You may assume that the answer would not change if the diameter of the bowl is increased by at most 10−5.

Sample Input

2
4 1.5
1.0 3.75
3.0 1.0
1.0 2.25
1.5 3.0
8 3.0
-1.0 3.0
-1.0 2.0
-2.0 1.0
0.0 1.0
1.0 0.0
1.0 -1.0
2.0 -2.0
3.0 -1.0

Sample Output

3
4

Problem Source

ACM-ICPC North America Qualifier 2014

一个覆盖最多点的圆，必然至少有两个点在圆上。枚举两个点，求过这两个点的单位圆，判断有多少个点在圆中，枚举N^2，判断N

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<map>
#include<string>
#include<vector>
#include<set>
#define long long LL
using namespace std;
const double eps=1e-5;
const double PI=acos(-1);
struct Point
{
double x,y;
Point(double x=0,double y=0):x(x),y(y){}
};
typedef Point Vector;
Vector operator +(Vector A,Vector B)
{
return Vector(A.x+B.x,A.y+B.y);
}
Vector operator -(Vector A,Vector B)
{
return Vector(A.x-B.x,A.y-B.y);
}
Vector operator *(Vector A,double p)
{
return Vector(A.x*p,A.y*p);
}
Vector operator /(Vector A,double p)
{
return Vector(A.x/p,A.y/p);
}
bool operator <(const Point& a,const Point& b)
{
return a.x<b.x||(a.x==b.x&&a.y<b.y);
}
int dcmp(double x)
{
if(fabs(x)<eps)
return 0;
else
return x<0?-1:1;
}
bool operator ==(const Point& a,const Point& b)
{
return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;
}
double Dot(Vector A,Vector B)
{
return A.x*B.x+A.y*B.y;
}
double Length(Vector A)
{
return sqrt(Dot(A,A));
}

Vector Normal(Vector A)
{
double L=Length(A);
return Vector(-A.y/L,A.x/L);
}
double angle(Vector v)
{
return atan2(v.y,v.x);
}
double r;
Point Get_circleCenter(Point p1,Point p2)
{
Vector p=p2-p1;
Point mid=(p1+p2)/2;
double d=sqrt(r*r/4-Length(mid-p1)*Length(mid-p1));
double rad=angle(Normal(p2-p1));
return Point(mid.x+d*cos(rad),mid.y+d*sin(rad));
}
int main()
{
int kase;
cin>>kase;
while(kase--)
{
int m;
cin>>m;
cin>>r;
Point mo[40];
for(int i=0;i<m;i++)
cin>>mo[i].x>>mo[i].y;
int ans=1;
for(int i=0;i<m;i++)
for(int j=i+1;j<m;j++)
{
if(Length(mo[i]-mo[j])>r)
continue;
Point center=Get_circleCenter(mo[i],mo[j]);
int cnt=0;
for(int k=0;k<m;k++)
if(Length(center-mo[k])<r/2+1e-5) cnt++;
ans=max(ans,cnt);
}
cout<<ans<<endl;
}
return 0;
}