HDU 6097 数学几何
Mindis
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1519 Accepted Submission(s): 237
Special Judge
Problem Description
The center coordinate of the circle C is O, the coordinate of O is (0,0) , and the radius is r.
P and Q are two points not outside the circle, and PO = QO.
You need to find a point D on the circle, which makes PD+QD minimum.
Output minimum distance sum.
P and Q are two points not outside the circle, and PO = QO.
You need to find a point D on the circle, which makes PD+QD minimum.
Output minimum distance sum.
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with r : the radius of the circle C.
Next two line each line contains two integers x , y denotes the coordinate of P and Q.
Limits
T≤500000
−100≤x,y≤100
1≤r≤100
Each case begins with one line with r : the radius of the circle C.
Next two line each line contains two integers x , y denotes the coordinate of P and Q.
Limits
T≤500000
−100≤x,y≤100
1≤r≤100
Output
For each case output one line denotes the answer.
The answer will be checked correct if its absolute or relative error doesn't exceed 10−6.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if |a−b|max(1,b)≤10−6.
The answer will be checked correct if its absolute or relative error doesn't exceed 10−6.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if |a−b|max(1,b)≤10−6.
Sample Input
4 4 4 0 0 4 4 0 3 3 0 4 0 2 2 0 4 0 1 1 0
Sample Output
5.6568543 5.6568543 5.8945030 6.7359174
Source
2017 Multi-University Training Contest - Team 6
题意 ： 给一个圆，两个点，这两个点不在圆外，两个点到圆心距离相等，求圆上点到两个点距离和的最小值。
思路 ： 首先要了解反演点
一般指二维反演中的点。
二维上反演以一个特定的反演圆为基础:圆心O为反演中心,圆半径为常数k,把点P反演为点P'就是使得OP×OP'=r^2 （来自百度百科）
首先做出两个点的反演点 P' 和 Q' ，ODP与ODP' 是相似三角型（这个不难证明）
所以极小的 PD + QD 可转化为极小的P'D + Q'D
当P'Q'与圆有交点的时候可以使P'DQ'在同一直线上，这时P'Q'之间的距离就是 ODP' + ODQ'的最小值，那么ODP + ODQ的最小值就等于 P'Q' * （相似的比例）
当P'Q与圆无交点的时候，可以把P' 和 Q'当作一个椭圆的两个交点，我们知道椭圆上的点到椭圆上任意一点的距离之和都相等，所以我们慢慢放大椭圆直到与圆相切的时候，那个点就是最短距离点，你会发现，那个点一定在中垂线上，所以就去求中垂线与圆的交点的距离就可以了
#include
#include
#define eps 1e-8
int main(){
int T;
double r,x1,y1,x2,y2,x3,y3,x4,y4;
double d1,d2,d,k1,k2,dist,ans;
double mx,my,smx,smy;
scanf("%d",&T);
while(T--){
scanf("%lf %lf %lf %lf %lf",&r,&x1,&y1,&x2,&y2);
d1 = sqrt(pow(x1,2) + pow(y1,2)); // 计算出其中一点到圆心距离
k1 = pow(r,2) / pow(d1,2); // 这个 k 由 d(反演点) * d(p点) = r ^ 2 推出
x3 = x1 * k1,y3 = y1 * k1; //用这个比例得出反演点坐标
x4 = x2 * k1,y4 = y2 * k1;
mx = (x3 + x4) / 2; //求出两反演点的中点
my = (y3 + y4) / 2;
d2 = sqrt(pow(mx,2) + pow(my,2)); //中点到圆心的距离可以判断是否与圆相交
if(fabs(d1) <= eps){
ans = 2 * r;
}else if(d2 <= r){
dist = sqrt(pow(x3 - x4,2) + pow(y3 - y4,2));
k2 = d1 / r;
ans = dist * k2;
}else if(d2 > r){
k2 = r / d2;
smx = mx * k2,smy = my * k2;
ans = 2 * (sqrt(pow(smx - x1,2) + pow(smy - y1,2)));
}
printf("%.7f\n",ans);
}
return 0;
}