O（n^3）的做法：
枚举任意两点为弦的圆，然后再枚举其它点是否在圆内。
用到了两个函数
atan2反正切函数，据说可以很好的避免一些特殊情况
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 310 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 struct point 18 { 19 double x,y; 20 point(double x = 0,double y =0 ):x(x),y(y){} 21 }p[N]; 22 double dis(point a,point b) 23 { 24 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 25 } 26 point getcircle(point p1,point p2) 27 { 28 point mid = point((p1.x+p2.x)/2,(p2.y+p1.y)/2); 29 double angle = atan2(p2.y-p1.y,p2.x-p1.x); 30 double d = sqrt(1.0-dis(p1,mid)*dis(p1,mid)); 31 return point(mid.x+d*sin(angle),mid.y-d*cos(angle)); 32 } 33 int dcmp(double x) 34 { 35 if(fabs(x)<eps)return 0; 36 else return x<0?-1:1; 37 } 38 int main() 39 { 40 int i,j,n; 41 while(scanf("%d",&n)&&n) 42 { 43 for(i = 1 ;i <= n; i++) 44 scanf("%lf%lf",&p[i].x,&p[i].y); 45 int maxz = 1; 46 for(i = 1; i <= n; i++) 47 for(j = i+1 ; j <= n ;j++) 48 { 49 if(dis(p[i],p[j])>2.0) continue; 50 int tmax = 0; 51 point cir = getcircle(p[i],p[j]); 52 for(int g = 1; g <= n ;g++) 53 { 54 if(dcmp(dis(cir,p[g])-1.0)>0) 55 continue; 56 tmax++; 57 } 58 maxz = max(maxz,tmax); 59 } 60 printf("%d ",maxz); 61 } 62 return 0; 63 }
O（n^2log(n))
这个类似扫描线的做法，以每一个点为圆心化圆，枚举与其相交得圆，保存交点和角度，按角度排序后，扫一遍。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 310 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 struct point 18 { 19 double x,y; 20 point(double x = 0,double y =0 ):x(x),y(y) {} 21 } p[N]; 22 struct node 23 { 24 double ang; 25 int in; 26 } arc[N*N]; 27 double dis(point a,point b) 28 { 29 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 30 } 31 int dcmp(double x) 32 { 33 if(fabs(x)<eps)return 0; 34 else return x<0?-1:1; 35 } 36 bool cmp(node a,node b) 37 { 38 if(dcmp(a.ang-b.ang)==0) 39 return a.in>b.in; 40 return dcmp(a.ang-b.ang)<0; 41 } 42 int main() 43 { 44 int i,j,n; 45 while(scanf("%d",&n)&&n) 46 { 47 for(i = 1 ; i <= n; i++) 48 scanf("%lf%lf",&p[i].x,&p[i].y); 49 int g = 0; 50 int ans = 0,maxz = 1; 51 for(i = 1; i <= n ; i++) 52 { 53 ans = 0; 54 g = 0; 55 for(j = 1; j <= n ; j++) 56 { 57 if(dis(p[i],p[j])>2.0) continue; 58 double ang1 = atan2(p[j].y-p[i].y,p[j].x-p[i].x); 59 double ang2 = acos(dis(p[i],p[j])/2); 60 arc[++g].ang = ang1-ang2;//这里角度的算法很巧妙 61 arc[g].in = 1; 62 arc[++g].ang = ang1+ang2; 63 arc[g].in = -1; 64 } 65 sort(arc+1,arc+g+1,cmp); 66 67 //cout<<g<<endl; 68 for(j = 1 ; j <= g;j++) 69 { 70 ans+=arc[j].in; 71 maxz = max(maxz,ans); 72 } 73 } 74 printf("%d ",maxz); 75 } 76 return 0; 77 }