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):
1 //STATUS:C++_AC_4032MS_208KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //using namespace __gnu_cxx; 25 //define 26 #define pii pair<int,int> 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define lson l,mid,rt<<1 29 #define rson mid+1,r,rt<<1|1 30 #define PI acos(-1.0) 31 //typedef 32 typedef long long LL; 33 typedef unsigned long long ULL; 34 //const 35 const int N=310; 36 const int INF=0x3f3f3f3f; 37 const int MOD=100000,STA=8000010; 38 const LL LNF=1LL<<60; 39 const double EPS=1e-8; 40 const double OO=1e15; 41 const int dx[4]={-1,0,1,0}; 42 const int dy[4]={0,1,0,-1}; 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 44 //Daily Use ... 45 inline int sign(double x){return (x>EPS)-(x<-EPS);} 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 49 template<class T> inline T Min(T a,T b){return a<b?a:b;} 50 template<class T> inline T Max(T a,T b){return a>b?a:b;} 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 55 //End 56 57 struct Node{ 58 double x,y; 59 }nod[N],O; 60 int n; 61 62 double dist(Node &a,Node &b) 63 { 64 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 65 } 66 67 void getO(Node &a,Node &b,int dir) 68 { 69 double t=dist(a,b)/2.0; 70 t=dir*sqrt((1.0-t*t)); 71 if(a.y==b.y){ 72 O.x=(a.x+b.x)/2.0; 73 O.y=a.y+t; 74 } 75 else if(a.x==b.x){ 76 O.y=(a.y+b.y)/2.0; 77 O.x=a.x+t; 78 } 79 else { 80 double kt; 81 kt=atan(-(a.x-b.x)/(a.y-b.y)); 82 O.x=(a.x+b.x)/2.0+cos(kt)*t; 83 O.y=(a.y+b.y)/2.0+sin(kt)*t; 84 } 85 } 86 87 int main() 88 { 89 // freopen("in.txt","r",stdin); 90 int i,j,k,ans,tot; 91 while(scanf("%d",&n) && n) 92 { 93 ans=1; 94 for(i=0;i<n;i++){ 95 scanf("%lf%lf",&nod[i].x,&nod[i].y); 96 } 97 for(i=0;i<n;i++){ 98 for(j=i+1;j<n;j++){ 99 if(dist(nod[i],nod[j])<2.0){ 100 getO(nod[i],nod[j],1); 101 for(tot=2,k=0;k<n;k++){ 102 if(k==i || k==j)continue; 103 if(dist(O,nod[k])-1.0<EPS)tot++; 104 } 105 if(tot>ans)ans=tot; 106 } 107 } 108 } 109 110 printf("%d ",ans); 111 } 112 return 0; 113 }
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。
1 //STATUS:C++_AC_750MS_212KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //using namespace __gnu_cxx; 25 //define 26 #define pii pair<int,int> 27 #define mem(a,b) memset(a,b,sizeof(a)) 28 #define lson l,mid,rt<<1 29 #define rson mid+1,r,rt<<1|1 30 #define PI acos(-1.0) 31 //typedef 32 typedef long long LL; 33 typedef unsigned long long ULL; 34 //const 35 const int N=310; 36 const int INF=0x3f3f3f3f; 37 const int MOD=100000,STA=8000010; 38 const LL LNF=1LL<<60; 39 const double EPS=1e-8; 40 const double OO=1e15; 41 const int dx[4]={-1,0,1,0}; 42 const int dy[4]={0,1,0,-1}; 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 44 //Daily Use ... 45 inline int sign(double x){return (x>EPS)-(x<-EPS);} 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 49 template<class T> inline T Min(T a,T b){return a<b?a:b;} 50 template<class T> inline T Max(T a,T b){return a>b?a:b;} 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 55 //End 56 57 struct Node{ 58 double x,y; 59 }nod[N]; 60 struct Point{ 61 double angle; 62 int id; 63 bool operator < (const Point& a)const{ 64 return angle!=a.angle?angle<a.angle:id>a.id; 65 } 66 }p[N*2]; 67 int n; 68 69 double dist(Node &a,Node &b) 70 { 71 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 72 } 73 74 int slove() 75 { 76 int i,j,ans,tot,k,cnt; 77 ans=1; 78 for(i=0;i<n;i++){ 79 for(j=k=0;j<n;j++){ 80 if(j==i || dist(nod[i],nod[j])>2.0)continue; 81 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 82 double phi=acos(dist(nod[i],nod[j])/2); 83 p[k].angle=angle-phi;p[k++].id=1; 84 p[k].angle=angle+phi;p[k++].id=-1; 85 } 86 sort(p,p+k); 87 for(tot=1,j=0;j<k;j++){ 88 tot+=p[j].id; 89 ans=Max(ans,tot); 90 } 91 } 92 return ans; 93 } 94 95 int main() 96 { 97 // freopen("in.txt","r",stdin); 98 int i; 99 while(~scanf("%d",&n) && n) 100 { 101 for(i=0;i<n;i++) 102 scanf("%lf%lf",&nod[i].x,&nod[i].y); 103 104 printf("%d ",slove()); 105 } 106 return 0; 107 }