题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4454
题意:对于给定的点(x,y),求这个点到所给圆再到所给矩形的最短距离
解题思路:可将题目理解为圆上一点到给定点以及矩形的最短距离为多少。这样的话,就可将(0,2π)进行三分,根据弧度求出圆上一点坐标,进而求出到点和矩形的距离和。
ac代码:
#include<bits/stdc++.h>
using namespace std;
const double pi=acos(-1.0);
double sx,sy;
double rx,ry,r;
double x1,x2,z1,z2;
double mmax(double a,double b)
{
return a>b?a:b;
}
double eend(double x,double y)//求到矩形的距离
{
double ll=0;
double lx=min(x1,x2);
double rx=mmax(x1,x2);
double hy=mmax(z1,z2);
double ly=min(z1,z2);
if(x>=lx&&x<=rx&&y>=hy) ll=y-hy;
else if(x>=lx&&x<=rx&&y<=ly) ll=ly-y;
else if(y<=hy&&y>=ly&&x<=lx) ll=lx-x;
else if(y<=hy&&y>=ly&&x>=rx) ll=x-rx;
else if(x<=lx&&y>=hy) ll=sqrt((x-lx)*(x-lx)+(y-hy)*(y-hy));
else if(x>=rx&&y>=hy) ll=sqrt((x-rx)*(x-rx)+(y-hy)*(y-hy));
else if(x<=lx&&y<=ly) ll=sqrt((x-lx)*(x-lx)+(y-ly)*(y-ly));
else if(x>=rx&&y<=ly) ll=sqrt((x-rx)*(x-rx)+(y-ly)*(y-ly));
return ll;
}
double road(double a)//求该弧度下的距离和
{
double length=0;
double tx,ty;
tx=rx+r*cos(a);
ty=ry+r*sin(a);
/*cout<<"**********"<<endl;
cout<<tx<<"&&&"<<ty<<endl;
cout<<"**********"<<endl;*/
length=sqrt((sx-tx)*(sx-tx)+(sy-ty)*(sy-ty));
length+=(eend(tx,ty));
return length;
}
double three()//三分
{
double res;
double p1,p2;
double l=0;
double r=pi*2;
int i=0;
while(fabs(r-l)>0.0001)
{
p1=fabs(r-l)/3.0+l;
p2=r-fabs(r-l)/3.0;
double len1=road(p1);
double len2=road(p2);
res=min(len1,len2);
if(len1>len2) l=p1;
else r=p2;
/*cout<<p1<<"&&&"<<len1<<endl;
cout<<p2<<"&&&"<<len2<<endl;
cout<<l<<"&&&"<<r<<endl;
cout<<"**********"<<endl;*/
}
return res;
}
int main()
{
ios::sync_with_stdio(false);
while(cin>>sx>>sy)
{
if(sx==0&&sy==0) break;
cin>>rx>>ry>>r;
cin>>x1>>z1>>x2>>z2;
double ans=three();
cout<<fixed<<setprecision(2)<<ans<<endl;
}
return 0;
}