hdu 1077 意思懂，没思路，借用大牛的思想

大牛的这么认为，给定两个点，如里它们的距离大于直径，不必考虑，否则，认为这两个点在某一圆上，计算出圆心坐标，依次遍历顶点，记录在这个圆里面的点，这个值就是我们要进行判定的值。。。

#include <iostream>#include <cmath>#include <cstdio>using namespace std;const static double  eps = 1e-6;struct Point{double x,y;};double distancess(Point a,Point b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}Point look_center(Point a,Point b){Point aa,bb,mid;aa.x = b.x-a.x;aa.y = b.y-a.y;mid.x = (a.x+b.x)/2.0;mid.y = (a.y+b.y)/2.0;double dist = distancess(a,mid);double c = sqrt(1.0-dist);if(fabs(aa.y)<eps){bb.x = mid.x;bb.y = mid.y+c;}else{double ang = atan(-aa.x/aa.y);bb.x = mid.x + c*cos(ang);bb.y = mid.y + c*sin(ang);}return bb;}int main(){int test;Point p[305],a,b,c;int n;scanf("%d",&test);while(test--){scanf("%d",&n);for(int i=0; i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);int ans = 1;int temp = 0;for(int i=0; i<n; i++)for(int j=i+1;j<n;j++){   if(distancess(p[i],p[j])>4) continue;a = look_center(p[i],p[j]);temp = 0;for(int k=0;k<n;k++){if(distancess(a,p[k])<=1.000001) temp++;}if(ans<temp) ans = temp;}printf("%d\n",ans);}return 0;}