poj 1981 Circle and Points

题目链接：http://poj.org/problem?id=1981

解题思路：枚举两个点，求过这两点的单位圆，判断有多少个点在圆中。

///////////////////////////////////////////////////////////////////////////
//problem_id: poj 1981
//user_id: SCNU20102200088
///////////////////////////////////////////////////////////////////////////

#include <algorithm>
#include <iostream>
#include <iterator>
#include <iomanip>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
using namespace std;

///////////////////////////////////////////////////////////////////////////
typedef long long LL;
const double EPS=1e-8;
const double PI=acos(-1.0);

const int x4[]={-1,0,1,0};
const int y4[]={0,1,0,-1};
const int x8[]={-1,-1,0,1,1,1,0,-1};
const int y8[]={0,1,1,1,0,-1,-1,-1};

typedef int T;
T max(T a,T b){ return a>b? a:b; }
T min(T a,T b){ return a<b? a:b; }
///////////////////////////////////////////////////////////////////////////

///////////////////////////////////////////////////////////////////////////
//Add Code:
struct Point{
    double x,y;
    Point(double dx=0,double dy=0):x(dx),y(dy){}
}p[305];

double dist(Point a,Point b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

Point circle(Point a,Point b){
    Point mid((a.x+b.x)/2,(a.y+b.y)/2);
    double angle=atan2(a.x-b.x,b.y-a.y);
    double d=sqrt(1-dist(a,mid)*dist(a,mid));
    Point ret(mid.x+d*cos(angle),mid.y+d*sin(angle));
    return ret;
}
///////////////////////////////////////////////////////////////////////////

int main(){
    ///////////////////////////////////////////////////////////////////////
    //Add Code:
    int n,i,j,k;
    while(scanf("%d",&n)!=EOF){
        if(n==0) break;
        for(i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);
        int Max=1;
        for(i=0;i<n;i++){
            for(j=i+1;j<n;j++){
                if(dist(p[i],p[j])>2.0) continue;
                Point center=circle(p[i],p[j]);
                int num=0;
                for(k=0;k<n;k++){
                    if(dist(center,p[k])<1.0+EPS) num++;
                }
                Max=max(Max,num);
            }
        }
        printf("%d
",Max);
    }
    ///////////////////////////////////////////////////////////////////////
    return 0;
}

///////////////////////////////////////////////////////////////////////////
/*
Testcase:
Input:
3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
0
Output:
2
5
5
11
*/
///////////////////////////////////////////////////////////////////////////