POJ 1106 Transmitters

来源：互联网发布：2016淘宝达人怎么赚钱编辑：程序博客网时间：2024/04/28 13:07

给定一些点，和一个圆心坐标，求一个以这个点为圆心的半圆能最多能圈下多少点。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;#define hPI 1.5707963267949#define eps 1e-8struct point{    double x;    double y;};double cross(point a,point b,point c){    a.x-=c.x;a.y-=c.y;    b.x-=c.x;b.y-=c.y;    return a.x*b.y-a.y*b.x;}double mul(point a,point b,point c){    a.x-=c.x;a.y-=c.y;    b.x-=c.x;b.y-=c.y;    return a.x*b.x+a.y*b.y;}double dist(point a,point b){    return sqrt(mul(a,a,b));}double angle(point a,point b,point c){    return cross(a,b,c)/dist(a,c)/dist(b,c);}int main(){    double r,t;    point p[200],o;    int lp,i,j,ans,t1,t2;    while (scanf("%lf%lf%lf",&o.x,&o.y,&r) != EOF)    {        if (r < 0)            break;        ans=0;        scanf("%d",&lp);        for (i=0; i<lp; i++)        {            scanf("%lf%lf",&p[i].x,&p[i].y);        }        for (i=0; i<lp; i++)        {            t1=1;            t2=1;            t=dist(p[i],o);            if (t > r)                continue;            for (j=0; j<lp; j++)            {                if (i == j)                    continue;                t=dist(p[j],o);               // printf("%d %d %lf ",i,j,t);                if (t > r)                    continue;                t=angle(p[i],p[j],o);                //printf(" %lf\n",t);                if (t > 0)                    t1++;                else if (t < 0)                    t2++;                else if (fabs(t-0.0) < eps)                {                    t1++;                    t2++;                }            }           // printf("\nt1=%d t2=%d\n",t1,t2);            t1=t1>t2?t1:t2;            ans=ans>t1?ans:t1;        }        printf("%d\n",ans);    }}