Goat in the Garden （ural 1348）
http://acm.hust.edu.cn:8080/judge/contest/view.action?cid=11732#problem/E
E - Goat in the Garden 2
Time Limit:1000MS Memory Limit:16384KB 64bit IO Format:%I64d & %I64uSubmit Status Practice URAL 1348
Description
A goat is tied to a peg (in a point C) in a garden with a strong rope of the length L (i.e. a goat may eat a grass that is not farther than Lmeters from the peg). There is a bed of pineapples that he loves very much. The bed is a line segment with the ends A and B.
Humph… We wonder, how much the goat is to stretch the roap in order to reach at least one pine apple? And all the pineapples?
Input
There are points’ A, B and C coordinates and a length of the rope L in the input. All the numbers are integer, L ≥ 0, all the coordinates don’t exceed 10000 by the absolute value. The numbers are separated with spaces or line feeds.
Output
The first line should contain the minimal length that the goat is to elongate the rope in order to reach the pineapples bed. The second line should contain the minimal length that the goat is to elongate the rope in order to eat all the pineapples from the bed. All the numbers are to be outputted within two digits after a decimal point.
Sample Input
| input |
output |
| 8 -6 8 6 0 0 7 |
3.00 |
思路：求点到线段的最短和最长距离；：
方法两种，一直接有几何公式代入
有几个公式的应用：直线方程：Ax+By+C=0;点到直线的距离：d=|Ax+By+C|/(sqrt(A*A+B*B));
二·三分法查找
这里给出我自己写的几何法，及三分法的核心代码
#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
const double eps=1e-8;
int is_online(double x1,double y1,double x2,double y2,double x,double y)//判断（x,y）是否在线段上
{
if(((x-x1)*(x-x2)+(y-y1)*(y-y2))<=eps)//（注意精度问题）
return 1;
else
return 0;
}
int main()
{
int i,j;
double A,B,C1,C2,d1,d2,l,d;
double a1,a2,b1,b2,c1,c2,x,y;
while(scanf("%lf%lf%lf%lf",&a1,&a2,&b1,&b2)!=EOF)
{scanf("%lf%lf%lf",&c1,&c2,&l);
A=b2-a2;
B=a1-b1;
C1=a1*(a2-b2)-a2*(a1-b1);//Ax+By=C1表示直线L1方程
C2=A*c2-B*c1;//Bx-Ay+C2=0表示过圆心且垂直于直线L1的方程
//printf("%.2lf %.2lf %.2lf\n",A,B,C);
if(A==0&&B==0)
d1=d2=sqrt((a1-c1)*(a1-c1)+(a2-c2)*(a2-c2));//两点重合的情况
else
{ d=sqrt((a1-c1)*(a1-c1)+(a2-c2)*(a2-c2));
d2=sqrt((b1-c1)*(b1-c1)+(b2-c2)*(b2-c2));
y=(A*C2-B*C1)/(A*A+B*B);//（x,y）表示L1于L2的交点
x=(A*C1+B*C2)/(A*A+B*B)*(-1.0);
if(is_online(a1,a2,b1,b2,x,y))
d1=fabs(A*c1+B*c2+C1)/(sqrt(A*A+B*B));//点到直线的共识
else
d1=(d<d2? d:d2);
d2=(d2>d? d2:d);
}
//printf("%.2lf %.2lf",x,y);
if(d2<=l)
d2=0;
else
d2=d2-l;
if(d1<=l)
d1=0;
else
d1=d1-l;
printf("%.2lf\n%.2lf\n",d1,d2);
}
return 0;
}
#include<stdio.h>
#include<math.h>
#include<iostream>
#define eps 1e-8
using namespace std;
int is_online(double x1,double y1,double x2,double y2,double x,double y)//判断（x,y）是否在线段上
{
double temp1=(x2-x)*(y1-y)-(y2-y)*(x1-x);//点在直线上
double temp2=(x-x1)*(x-x2);//点在线段中
double temp3=(y-y1)*(y-y2);
if(temp1<=eps&&(temp2<=eps&&temp3<=eps))
return 1;
return 0;
}
int main()
{
int i,j;
double A,B,C1,C2,d1,d2,l,d;
double a1,a2,b1,b2,c1,c2,x,y;
while(scanf("%lf%lf%lf%lf",&a1,&a2,&b1,&b2)!=EOF)
{
scanf("%lf%lf%lf",&c1,&c2,&l);
A=b2-a2;
B=a1-b1;
C1=a1*(a2-b2)-a2*(a1-b1);//A B C1为线段方程的系数
C2=A*c2-B*c1;
if(A==0&&B==0)
d1=d2=sqrt((a1-c1)*(a1-c1)+(a2-c2)*(a2-c2));
else
{
d=sqrt((a1-c1)*(a1-c1)+(a2-c2)*(a2-c2));
d2=sqrt((b1-c1)*(b1-c1)+(b2-c2)*(b2-c2));
y=(A*C2-B*C1)/(A*A+B*B);
x=(A*C1+B*C2)/(A*A+B*B)*(-1.0);
if(is_online(a1,a2,b1,b2,x,y))
d1=fabs(A*c1+B*c2+C1)/(sqrt(A*A+B*B));
else
d1=(d<d2? d:d2);
d2=(d2>d? d2:d);
}
if(d2<=l)
d2=0;
else
d2=d2-l;
if(d1<=l)
d1=0;
else
d1=d1-l;
printf("%.2lf\n%.2lf\n",d1+eps,d2+eps);
}
return 0;
}
三分核心代码
double three()
{
double d1,d2;
Node l,r,mid,midmid;
l.x=l1.x,l.y=l1.y;
r.x=l2.x,r.y=l2.y;
d1=1,d2=0;
while(fabs(d1-d2)>esp){
mid.x=(l.x+r.x)/2;
mid.y=(l.y+r.y)/2;
midmid.x=(mid.x+r.x)/2;
midmid.y=(mid.y+r.y)/2;
d1=dis(mid,o);
d2=dis(midmid,o);
if(d1<d2)
r=midmid;
else l=mid;
}
return d1;
}