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):

  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\n",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\n",slove());
105     }
106     return 0;
107 }

posted @ 2013-07-15 21:43 zhsl 阅读(596) 评论(0) 编辑收藏举报

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POJ-1981 Circle and Points 单位圆覆盖

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