poj 1106 Transmitters (计算几何，叉积||极角排序)

  1 /*************************************************************************
  2     > File Name: code/poj/1106.cpp
  3     > Author: 111qqz
  4     > Email: rkz2013@126.com 
  5     > Created Time: 2015年11月09日 星期一 09时28分51秒
  6  ************************************************************************/
  7 
  8 #include<iostream>
  9 #include<iomanip>
 10 #include<cstdio>
 11 #include<algorithm>
 12 #include<cmath>
 13 #include<cstring>
 14 #include<string>
 15 #include<map>
 16 #include<set>
 17 #include<queue>
 18 #include<vector>
 19 #include<stack>
 20 #include<cctype>
 21 #define fst first              
 22 #define sec second      
 23 #define lson l,m,rt<<1
 24 #define rson m+1,r,rt<<1|1
 25 #define ms(a,x) memset(a,x,sizeof(a))
 26 using namespace std;
 27 const double eps = 1E-8;
 28 const int dx4[4]={1,0,0,-1};
 29 const int dy4[4]={0,-1,1,0};
 30 typedef long long LL;
 31 const int inf = 0x3f3f3f3f;
 32 const  int N=2E4+7;
 33 
 34 int n;
 35 double r;
 36 int dblcmp( double d)
 37 {
 38     return d<-eps?-1:d>eps;
 39 }
 40 struct point
 41 {
 42     double x,y;
 43     point(){}
 44     point(double _x,double _y):
 45     x(_x),y(_y){};
 46 
 47     void input()
 48     {
 49     scanf("%lf %lf",&x,&y);
 50     }
 51     
 52 
 53     point operator - (const point &p)const{
 54     return point(x-p.x,y-p.y);
 55     }
 56     double operator ^(const point &p)const{
 57     return x*p.y-y*p.x;
 58     }
 59     double dis(point p)
 60     {
 61     return sqrt((x-p.x)*(x-p.x)+(y-p.y)*(y-p.y));
 62     }
 63 }tmp[N],c,p[N];
 64 int main()
 65 {
 66   #ifndef  ONLINE_JUDGE 
 67    freopen("in.txt","r",stdin);
 68   #endif
 69     while (~scanf("%lf %lf %lf",&c.x,&c.y,&r))
 70     {
 71     int cnt = 0 ;
 72     if (dblcmp(r)<=0) break;
 73     scanf("%d",&n);
 74     for ( int i = 0 ;i  < n ; i ++)
 75     {
 76         tmp[i].input();
 77         double d =c.dis(tmp[i]);
 78         if (dblcmp(d-r)<=0)
 79         {
 80         //p[cnt]=tmp[i];
 81         p[cnt].x=tmp[i].x;
 82         p[cnt].y=tmp[i].y;
 83         cnt++;
 84         }
 85     }
 86     int ans = 0;
 87     for ( int i = 0 ; i < cnt ; i++)
 88     {
 89         int sum = 1 ;
 90         for ( int j = 0 ; j < cnt ; j++)
 91         {
 92         double cross = (c-p[i])^(c-p[j]);
 93         if (cross>=0&&i!=j)
 94         {
 95             sum++;
 96         }
 97         }
 98         ans = max(sum,ans);
 99     }
100     printf("%d\n",ans);
101     }
102   
103    
104  #ifndef ONLINE_JUDGE  
105   #endif
106   fclose(stdin);
107     return 0;
108 }