poj1981Circle and Points(单位圆覆盖最多的点）

链接

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

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 310
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct point
 {
     double x,y;
     point(double x = ,double y = ):x(x),y(y){}
 }p[N];
 double dis(point a,point b)
 {
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }
 point getcircle(point p1,point p2)
 {
     point mid = point((p1.x+p2.x)/,(p2.y+p1.y)/);
     double angle = atan2(p2.y-p1.y,p2.x-p1.x);
     double d = sqrt(1.0-dis(p1,mid)*dis(p1,mid));
     return point(mid.x+d*sin(angle),mid.y-d*cos(angle));
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps)return ;
     else return x<?-:;
 }
 int main()
 {
     int i,j,n;
     while(scanf("%d",&n)&&n)
     {
         for(i =  ;i <= n; i++)
         scanf("%lf%lf",&p[i].x,&p[i].y);
         int maxz = ;
         for(i = ; i <= n; i++)
             for(j = i+ ; j <= n ;j++)
             {
                 if(dis(p[i],p[j])>2.0) continue;
                 int tmax = ;
                 point cir = getcircle(p[i],p[j]);
                 for(int g =  ; g <= n ;g++)
                 {
                     if(dcmp(dis(cir,p[g])-1.0)>)
                     continue;
                     tmax++;
                 }
                 maxz = max(maxz,tmax);
             }
         printf("%d\n",maxz);
     }
     return ;
 }

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。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 310
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct point
 {
     double x,y;
     point(double x = ,double y = ):x(x),y(y) {}
 } p[N];
 struct node
 {
     double ang;
     int in;
 } arc[N*N];
 double dis(point a,point b)
 {
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps)return ;
     else return x<?-:;
 }
 bool cmp(node a,node b)
 {
     if(dcmp(a.ang-b.ang)==)
         return a.in>b.in;
     return dcmp(a.ang-b.ang)<;
 }
 int main()
 {
     int i,j,n;
     while(scanf("%d",&n)&&n)
     {
         for(i =  ; i <= n; i++)
             scanf("%lf%lf",&p[i].x,&p[i].y);
         int g = ;
         int ans = ,maxz = ;
         for(i = ; i <= n ; i++)
         {
             ans = ;
             g = ;
             for(j = ; j <= n ; j++)
             {
                 if(dis(p[i],p[j])>2.0) continue;
                 double ang1 = atan2(p[j].y-p[i].y,p[j].x-p[i].x);
                 double ang2 = acos(dis(p[i],p[j])/);
                 arc[++g].ang = ang1-ang2;//这里角度的算法很巧妙
                 arc[g].in = ;
                 arc[++g].ang = ang1+ang2;
                 arc[g].in = -;
             }
             sort(arc+,arc+g+,cmp);

             //cout<<g<<endl;
             for(j =  ; j <= g;j++)
             {
                 ans+=arc[j].in;
                 maxz = max(maxz,ans);
             }
         }
         printf("%d\n",maxz);
     }
     return ;
 }