POJ1106(计算几何 判断折线方向)

题目链接: http://poj.org/problem?id=1106

 

题意: 给出一个半圆形的坐标和半径,另给出一些平面内的点,求此半圆最多覆盖的点的数目.

 

解题思路:很简单的计算几何题,枚举半圆直径所在的边,水之~

8491066dooder_daodao1106Accepted164K0MSC++962B2011-04-15 09:09:40

 

 

 

#include<stdio.h>#include<string.h>#include<math.h>const double eps = 1e-8;const double f = 10e8;#define M 158typedef struct {double x,y;}Point;Point p[M];Point o;double r;int n;double Dis(Point p,Point q){return (p.x-q.x)*(p.x-q.x)+(p.y-q.y)*(p.y-q.y);}double Multi(Point p,Point q,Point o){return (p.x-o.x)*(q.y-o.y)-(q.x-o.x)*(p.y-o.y);}int main(){int i,j,tmpl,tmpr,ans;double d;while(scanf("%lf%lf%lf",&o.x,&o.y,&r),r>=0){r=r*r;scanf("%d",&n);for(i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);ans=0;for(i=0;i<n;i++){tmpl=tmpr=0;for(j=0;j<n;j++){d=Dis(p[j],o);if(d<r||fabs(d-r)<eps){d=Multi(p[j],p[i],o);if(fabs(d)<eps)tmpl++,tmpr++;else if(d>0.0)tmpl++;else tmpr++;}}if(tmpl>ans) ans=tmpl;if(tmpr>ans) ans=tmpr;}printf("%d/n",ans);}return 0;}