“Miyane!” This day Hana asks Miyako for help again. Hana plays the part of angel on the stage show of the cultural festival, and she is going to look for her human friend, Hinata. So she must find the shortest path to Hinata’s house.
The area where angels live is a circle, and Hana lives at the bottom of this circle. That means if the coordinates of circle’s center is (rx, ry)(rx,ry) and its radius is rr, Hana will live at (rx, ry - r)(rx,ry−r).
Apparently, there are many difficulties in this journey. The area which is located both outside the circle and below the line y = ryy=ry is the sea, so Hana cannot pass this area. And the area inside the circle is the holy land of angels, Hana cannot pass this area neither.
However, Miyako cannot calculate the length of the shortest path either. For the loveliest Hana in the world, please help her solve this problem!
Input
Each test file contains several test cases. In each test file:
The first line contains a single integer T(1 le T le 500)T(1≤T≤500) which is the number of test cases.
Each test case contains only one line with five integers: the coordinates of center rxrx 、 ryry, the radius rr, thecoordinates of Hinata’s house xx 、yy. The test data guarantees that y > ryy>ry and (x, y)(x,y) is out of the circle. (-10^2 le rx,ry,x,y le 10^2,0 < r le 10^2)(−102≤rx,ry,x,y≤102,0<r≤102).
Output
For each test case, you should print one line with your answer (please keep 44 decimal places).
样例输入
2 1 1 1 2 2 1 1 1 1 3
样例输出
2.5708 3.8264
题意就是给一个圆，求从圆的最底下，到圆中线上面的圆外一个位置的最短距离。
比赛时，板子套歪了，发现直接手算就可以。
代码:
1 //C-简单的计算几何 2 #include<bits/stdc++.h> 3 using namespace std; 4 typedef long long ll; 5 const int maxn=1e5+10; 6 const double PI=acos(-1.0); 7 8 int main() 9 { 10 int t; 11 scanf("%d",&t); 12 while(t--){ 13 double rx,ry,r,x,y; 14 scanf("%lf%lf%lf%lf%lf",&rx,&ry,&r,&x,&y); 15 double length=0.0; 16 if((x<=rx-r)||(x>=rx+r)){ 17 if(x<=rx-r){ 18 length+=sqrt((x-(rx-r))*(x-(rx-r))+(y-ry)*(y-ry)); 19 length+=0.5*PI*r; 20 } 21 else{ 22 length+=sqrt((x-(rx+r))*(x-(rx+r))+(y-ry)*(y-ry)); 23 length+=0.5*PI*r; 24 } 25 } 26 else{ 27 double d=sqrt((x-rx)*(x-rx)+(y-ry)*(y-ry)); 28 double jiao; 29 if(x!=rx){ 30 double cosr=abs(x-rx)/d; 31 jiao=acos(cosr)-acos(r/d); 32 } 33 else{ 34 jiao=0.5*PI-acos(r/d); 35 } 36 jiao+=0.5*PI; 37 length+=jiao*r; 38 length+=sqrt(d*d-r*r); 39 } 40 printf("%.4f ",length); 41 } 42 }