# Stealing a Cake

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 80    Accepted Submission(s): 15

Problem Description
There is a big round cake on the ground. A small ant plans to steal a small piece of cake. He starts from a certain point, reaches the cake, and then carry the piece back home. He does not want to be detected, so he is going to design a shortest path to achieve his goal.

The big cake can be considered as a circle on a 2D plane. The ant’s home can be considered as a rectangle. The ant can walk through the cake. Please find out the shortest path for the poor ant.

Input
The input consists of several test cases.
The first line of each test case contains x,y, representing the coordinate of the starting point. The second line contains x, y, r. The center of the cake is point (x,y) and the radius of the cake is r. The third line contains x1,y1,x2,y2, representing the coordinates of two opposite vertices of the rectangle --- the ant's home.
All numbers in the input are real numbers range from -10000 to 10000. It is guaranteed that the cake and the ant's home don't overlap or contact, and the ant's starting point also is not inside the cake or his home, and doesn't contact with the cake or his home.
If the ant touches any part of home, then he is at home.
Input ends with a line of 0 0. There may be a blank line between two test cases.

Output
For each test case, print the shortest distance to achieve his goal. Please round the result to 2 digits after decimal point.

Sample Input
1 1 -1 1 1 0 -1 1 0 0 2 -1 1 1 0 -1 1 0 0 0

Sample Output
1.75 2.00

Source

//============================================================================
// Name        : HDU4454.cpp
// Author      : kuangbin
// Version     :
// Description : 三分法
//============================================================================

#include <iostream>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;
const double eps=1e-6;
const double PI=acos(-1.0);
struct Point
{
double x,y;
Point(double xx=0,double yy=0):x(xx),y(yy){}
Point operator -(const Point p1)
{
return Point(x-p1.x,y-p1.y);
}
double operator ^(const Point p1)
{
return x*p1.x+y*p1.y;
}
};
inline int sign(double d)
{
if(d>eps)return 1;
else if(d<-eps)return -1;
else return 0;
}
double dis(Point a,Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double cross(Point a,Point b)
{
return a.x*b.y-a.y*b.x;
}

double ptoline(Point tp,Point st,Point ed)//求点到线段的距离
{
double t1=dis(tp,st);
double t2=dis(tp,ed);
double ans=min(t1,t2);
if(sign((ed-st)^(tp-st))>=0 && sign((st-ed)^(tp-ed))>=0)//锐角
{
double h=fabs(cross(st-tp,ed-tp))/dis(st,ed);
ans=min(ans,h);
}
return ans;
}
double xx1,yy1,xx2,yy2;

double ptoRec(Point tp)
{
Point p1(xx1,yy1);
Point p2(xx1,yy2);
Point p3(xx2,yy1);
Point p4(xx2,yy2);
double ans=ptoline(tp,p1,p2);
ans=min(ans,ptoline(tp,p2,p4));
ans=min(ans,ptoline(tp,p4,p3));
ans=min(ans,ptoline(tp,p3,p1));
return ans;
}
double x,y,R;
double xx0,yy0;
Point get_point(double A)
{
return Point(x+R*cos(A),y+R*sin(A));
}
double solve()
{
double l,r,mid,midmid;
Point p0(xx0,yy0);
Point p;
l=0;r=PI;
while(r-l>=1e-8)
{
mid=(r+l)/2;
midmid=(mid+r)/2;
Point p1=get_point(mid);
Point p2=get_point(midmid);
double t1=ptoRec(p1)+dis(p1,p0);
double t2=ptoRec(p2)+dis(p2,p0);
if(t1>t2)l=mid;
else r=midmid;
}
p=get_point(l);
double ans1=dis(p,p0)+ptoRec(p);
l=PI;r=2*PI;
while(r-l>=1e-8)
{
mid=(r+l)/2;
midmid=(mid+r)/2;
Point p1=get_point(mid);
Point p2=get_point(midmid);
double t1=ptoRec(p1)+dis(p1,p0);
double t2=ptoRec(p2)+dis(p2,p0);
if(t1>t2)l=mid;
else r=midmid;
}
p=get_point(l);
double ans2=dis(p,p0)+ptoRec(p);
return min(ans1,ans2);
}

int main() {
while(scanf("%lf%lf",&xx0,&yy0)==2)
{
if(xx0==0 && yy0==0)break;
scanf("%lf%lf%lf",&x,&y,&R);
scanf("%lf%lf%lf%lf",&xx1,&yy1,&xx2,&yy2);
printf("%.2lf\n",solve());
}
return 0;
}

posted on 2012-11-08 22:24  kuangbin  阅读(1823)  评论(3编辑  收藏  举报

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