Hdu 4454 Stealing a Cake（枚举或三分）

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=4454

思路：枚举角度，确定圆上点的位置，取最小值。

点到矩形最短距离：若点在矩形边所表示的范围内，则到矩形最短距离为x或y坐标到矩形边的距离。否则为点到矩形顶点的距离。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#define debu
using namespace std;
const double pi=acos(-1.0);
struct Point
{
    double x,y,r;
};
Point C,P,st;
double x1,y1,x2,y2;
double miny,maxy,minx,maxx;
double dist(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double Rdist(Point a)
{
    double x=0.0,y=0.0;
    if(a.x<minx) x=minx-a.x;
    else if(a.x>maxx) x=a.x-maxx;
    if(a.y<miny) y=miny-a.y;
    else if(a.y>maxy) y=a.y-maxy;
    return sqrt(x*x+y*y);
}
int main()
{
#ifdef debug
    freopen("in.in","r",stdin);
#endif // debug
    while(scanf("%lf%lf",&st.x,&st.y)!=EOF)
    {
        if(st.x==0&&st.y==0) break;
        double ans=1e10;
        scanf("%lf%lf%lf",&C.x,&C.y,&C.r);
        scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
        minx=min(x1,x2),maxx=max(x1,x2);
        miny=min(y1,y2),maxy=max(y1,y2);
        for(double rad=0; rad<=360; rad+=0.005)
        {
            Point P;
            P.x=C.x+C.r*cos(rad/180*pi);
            P.y=C.y+C.r*sin(rad/180*pi);
            ans=min(ans,dist(P,st)+Rdist(P));
        }
        printf("%.2f\n",ans);
    }
    return 0;
}