hdu 4454 三分+几何

最新推荐文章于 2020-07-28 22:57:28 发布

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本文介绍了一个计算几何问题，即求解从指定点出发，经过一个圆上的某点再到矩形边界上某点的距离之和的最小值。通过三分法搜索最优角度来实现，提供了一种有效解决此类问题的方法。

题目大意：给一个点，一个圆和一个矩形，矩形与圆没有重叠部分，点也在圆和矩形外，求从该点出发经过圆上一点再到矩形边上一点的距离和的最小值。

思路：三分角度求最小距离和。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<string>
#include<algorithm>
#include<set>
#include<map>
#include<cstring>
#include<queue>
#include<stack>
#include<list>


using namespace std;

typedef long long ll;
const int maxn=1000+5;
const int inf=1e9;
const ll mod=1e9+7;
const double pi=acos(-1.0);

struct point{
    double x,y;
    point(){}
    point (double  _x,double  _y){
        x=_x;y=_y;
    }
    point operator-(const point &ne)const{
        return point (x-ne.x,y-ne.y);
    }
    point operator+(const point &ne)const{
        return point (x+ne.x,y+ne.y);
    }
    point operator*(const double t)const{
        return point (x*t,y*t);
    }
};


struct line{
    point a,b;
    line (){}
    line(point   _a,point   _b){a=_a;b=_b;}

};

struct circle{
    point O;
    double r;
};

double dmult(point a,point b){
    return a.x*b.x+a.y*b.y;
}

inline double lenth(point a){
    return sqrt(dmult(a,a));
}

inline double dist(point a,point b){
    return lenth(b-a);
}

inline double dist2(point a,point b){
    return dmult(b-a,b-a);
}

double relation(point p,line l){
    line t1(l.a,p);
    return dmult(t1.b-l.a,l.b-l.a)/dist2(l.a,l.b);
}


point perpend(point p,line l){
    double r=relation (p,l);
    return l.a+((l.b-l.a)*r);
}

double distoline(point p,line l){
    double r=relation(p,l);
    if(r<0){
        return dist(p,l.a);
    }
    if(r>1)return dist(p,l.b);
    point np=perpend(p,l);
    return dist(p,np);
}

const double eps=1e-9;
double sx,sy;
double x,y,r;
circle cir;
point start;
point P[4];
line L[4];


double cal(double alpha){
    double x,y;
    x=cir.O.x+cir.r*cos(alpha);
    y=cir.O.y+cir.r*sin(alpha);
    double res=1e18;
    point pp=point(x,y);
    double tmp=dist(pp,start);
    for(int i=0;i<4;i++){
        res=min(res,distoline(pp,L[i]));
    }
    return res+tmp;
}

double  Solve(double leftvale,double rightvale) {   //三分{
     double Left, Right;
     double mid, midmid;
     double mid_value, midmid_value;
     Left = leftvale; Right = rightvale;
     while (Left + eps< Right)
     {
         //mid = (Left + Right) / 2;
         mid=Left+(Right-Left)/3;
         midmid = Right-(Right-Left)/3;
         mid_value = cal(mid);
         midmid_value = cal(midmid);
         // 假设求解最xiao极值.
         if (mid_value < midmid_value) Right = midmid;
         else Left = mid;
     }
     return cal(Left);
}



int main()
{
    while(scanf("%lf%lf",&sx,&sy)!=EOF){
        if(sx==0&&sy==0)break;
        start.x=sx;start.y=sy;
        scanf("%lf%lf%lf",&x,&y,&r);
        cir.O.x=x;cir.O.y=y;cir.r=r;
        double X1,Y1,X2,Y2;
        scanf("%lf%lf%lf%lf",&X1,&Y1,&X2,&Y2);
        P[0].x=X1;P[0].y=Y1;
        P[1].x=X1;P[1].y=Y2;
        P[2].x=X2;P[2].y=Y2;
        P[3].x=X2;P[3].y=Y1;
        L[0]=line(P[0],P[1]);
        L[1]=line(P[1],P[2]);
        L[2]=line(P[2],P[3]);
        L[3]=line(P[3],P[0]);
        double ans=min(Solve(0,pi),Solve(pi,2*pi));
        printf("%.2f\n",ans);

    }
	return 0;
}