POJ-1981 Circle and Points 单位圆覆盖

　　题目链接：http://poj.org/problem?id=1981

　　容易想到直接枚举两个点，然后确定一个圆来枚举，算法复杂度O(n^3).

　　这题还有O(n^2*lg n)的算法。将每个点扩展为单位圆，依次枚举每个单位圆，枚举剩下的单位圆，如果有交点，每个圆产生两个交点，然后对产生的2n个交点极角排序，判断被覆盖最多的弧，被覆盖相当于这个弧上的点为圆心的圆可以覆盖到覆盖它的那些点，所以被覆盖最多的弧就是答案了。

O(n^3):

 //STATUS:C++_AC_4032MS_208KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 struct Node{
     double x,y;
 }nod[N],O;
 int n;

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

 void getO(Node &a,Node &b,int dir)
 {
     double t=dist(a,b)/2.0;
     t=dir*sqrt((1.0-t*t));
     if(a.y==b.y){
         O.x=(a.x+b.x)/2.0;
         O.y=a.y+t;
     }
     else if(a.x==b.x){
         O.y=(a.y+b.y)/2.0;
         O.x=a.x+t;
     }
     else {
         double kt;
         kt=atan(-(a.x-b.x)/(a.y-b.y));
         O.x=(a.x+b.x)/2.0+cos(kt)*t;
         O.y=(a.y+b.y)/2.0+sin(kt)*t;
     }
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i,j,k,ans,tot;
     while(scanf("%d",&n) && n)
     {
         ans=;
         for(i=;i<n;i++){
             scanf("%lf%lf",&nod[i].x,&nod[i].y);
         }
         for(i=;i<n;i++){
             for(j=i+;j<n;j++){
                 if(dist(nod[i],nod[j])<2.0){
                     getO(nod[i],nod[j],);
                     for(tot=,k=;k<n;k++){
                         if(k==i || k==j)continue;
                         if(dist(O,nod[k])-1.0<EPS)tot++;
                     }
                     if(tot>ans)ans=tot;
                 }
             }
         }

         printf("%d\n",ans);
     }
     return ;
 }    

O(n^2*lg n)： 建立极角的时候，不是以枚举的圆心 i->j 方向的向量，而是 j->i 方向的向量，因为 i->j 方向不能完全判断圆的方向，在极角排序的时候会出错。

 //STATUS:C++_AC_750MS_212KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 struct Node{
     double x,y;
 }nod[N];
 struct Point{
     double angle;
     int id;
     bool operator < (const Point& a)const{
         return angle!=a.angle?angle<a.angle:id>a.id;
     }
 }p[N*];
 int n;

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

 int slove()
 {
     int i,j,ans,tot,k,cnt;
     ans=;
     for(i=;i<n;i++){
         for(j=k=;j<n;j++){
             if(j==i || dist(nod[i],nod[j])>2.0)continue;
             double angle=atan2(nod[i].y-nod[j].y,nod[i].x-nod[j].x);  //注意为i-j的向量方向
             double phi=acos(dist(nod[i],nod[j])/);
             p[k].angle=angle-phi;p[k++].id=;
             p[k].angle=angle+phi;p[k++].id=-;
         }
         sort(p,p+k);
         for(tot=,j=;j<k;j++){
             tot+=p[j].id;
             ans=Max(ans,tot);
         }
     }
     return ans;
 }

 int main()
 {
  //   freopen("in.txt","r",stdin);
     int i;
     while(~scanf("%d",&n) && n)
     {
         for(i=;i<n;i++)
             scanf("%lf%lf",&nod[i].x,&nod[i].y);

         printf("%d\n",slove());
     }
     return ;
 }