poj1106 Transmitters

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <algorithm>
 5 
 6 
 7 using namespace std;
 8 const double PI = acos(-1.0);
 9 const double eps = 1e-10;
10 
11 /****************常用函数***************/
12 //判断ta与tb的大小关系
13 int sgn( double ta, double tb)
14 {
15     if(fabs(ta-tb)<eps)return 0;
16     if(ta<tb)   return -1;
17     return 1;
18 }
19 
20 //点
21 class Point
22 {
23 public:
24 
25     double x, y;
26 
27     Point(){}
28     Point( double tx, double ty){ x = tx, y = ty;}
29 
30     bool operator < (const Point &_se) const
31     {
32         return x<_se.x || (x==_se.x && y<_se.y);
33     }
34     friend Point operator + (const Point &_st,const Point &_se)
35     {
36         return Point(_st.x + _se.x, _st.y + _se.y);
37     }
38     friend Point operator - (const Point &_st,const Point &_se)
39     {
40         return Point(_st.x - _se.x, _st.y - _se.y);
41     }
42     //点位置相同(double类型)
43     bool operator == (const Point &_off)const
44     {
45         return  sgn(x, _off.x) == 0 && sgn(y, _off.y) == 0;
46     }
47 
48 };
49 
50 /****************常用函数***************/
51 //点乘
52 double dot(const Point &po,const Point &ps,const Point &pe)
53 {
54     return (ps.x - po.x) * (pe.x - po.x) + (ps.y - po.y) * (pe.y - po.y);
55 }
56 //叉乘
57 double xmult(const Point &po,const Point &ps,const Point &pe)
58 {
59     return (ps.x - po.x) * (pe.y - po.y) - (pe.x - po.x) * (ps.y - po.y);
60 }
61 //两点间距离的平方
62 double getdis2(const Point &st,const Point &se)
63 {
64     return (st.x - se.x) * (st.x - se.x) + (st.y - se.y) * (st.y - se.y);
65 }
66 //两点间距离
67 double getdis(const Point &st,const Point &se)
68 {
69     return sqrt((st.x - se.x) * (st.x - se.x) + (st.y - se.y) * (st.y - se.y));
70 }
71 
72 Point pt[200],cr;
73 double r;
74 int main(void)
75 {
76     while(~scanf("%lf%lf%lf",&cr.x,&cr.y,&r))
77     {
78         if(sgn(r,0)<0) break;
79         int n,ans=0;
80         scanf("%d",&n);
81         r*=r;
82         for(int i=1;i<=n;i++)
83             scanf("%lf%lf",&pt[i].x,&pt[i].y);
84         for(int i=1;i<=n;i++)
85         if(sgn(getdis2(cr,pt[i]),r)<=0)
86         {
87             int cnt=1;
88             for(int j=1;j<=n;j++)
89             if(i!=j && sgn(getdis2(cr,pt[j]),r)<=0 && sgn(xmult(cr,pt[i],pt[j]),0)>=0)
90                 cnt++;
91             ans=max(ans,cnt);
92         }
93         printf("%d\n",ans);
94     }
95     return 0;
96 }