CodeForces - 908C New Year and Curling

C. New Year and Curling

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Carol is currently curling.

She has n disks each with radius r on
the 2D plane.

Initially she has all these disks above the line y = 10100.

She then will slide the disks towards the line y = 0 one by one in order from 1 to n.

When she slides the i-th disk, she will place its center at the point (xi, 10100).
She will then push it so the disk’s y coordinate continuously decreases, and x coordinate
stays constant. The disk stops once it touches the line y = 0 or it touches any previous disk. Note that once a disk stops moving,
it will not move again, even if hit by another disk.

Compute the y-coordinates of centers of all the disks after all disks have been pushed.

Input

The first line will contain two integers n and r (1 ≤ n, r ≤ 1 000),
the number of disks, and the radius of the disks, respectively.

The next line will contain n integers x1, x2, ..., xn (1 ≤ xi ≤ 1 000) —
the x-coordinates of the disks.

Output

Print a single line with n numbers. The i-th
number denotes the y-coordinate of the center of the i-th
disk. The output will be accepted if it has absolute or relative error at most 10 - 6.

Namely, let's assume that your answer for a particular value of a coordinate is a and the answer of the jury is b.
The checker program will consider your answer correct if 

 for
all coordinates.

Example

input

6 2
5 5 6 8 3 12

output

2 6.0 9.87298334621 13.3370849613 12.5187346573 13.3370849613

Note

The final positions of the disks will look as follows:



In particular, note the position of the last disk.

思路：一开始的思路是用一个set容器存圆心坐标，按照y从大到小排序。每次输入一个x，从头遍历set，直到找到一个点的x与当前x的差值<=2*r，计算这个y值，存进set。若没有找到，则y=r。但这个思路是错的。因为寻找圆心的时候不能只考虑y的大小，还需要考虑圆心与当前圆心的夹角，显然需要取夹角较小的，这样得到的圆心才越高。

这样一来我们就需要比较所有的圆，得到的y值取最大。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int n,r;
double x[1005],y[1005];
int main()
{
while(~scanf("%d%d",&n,&r))
{
for(int i=0;i<n;i++)
{
scanf("%lf",&x[i]);
y[i]=r;
for(int j=0;j<i;j++)
{
if(fabs(x[j]-x[i])<=2*r)
{
y[i]=max(y[i],y[j]+sqrt(4*r*r-(x[j]-x[i])*(x[j]-x[i])));
}
}
printf("%.10f ",y[i]);
}
printf("\n");
}
return 0;
}