CodeForces - 908C New Year and Curling（圆与圆之间的关系的问题）

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 xcoordinate 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 aand 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.

题意： 已知一些圆在(xi,10100)处，然后依次落下，下落时只要碰到某个圆或
达到y=0处，就会自动停止，就算再被其他球碰住也不会动，求这些圆的最后的y值，也即是水平高度

说实话考试时，这道题我没做出来，wrong anwer  7；

做了这道题给我的收获是：1：以后再判断圆与圆的位置关系时，就用两个圆心与半径之间的关系来判断；2：因为这个是依次放球，他可能会和当前坐标左边的碰撞也也能和右边的碰撞，怎么判断它与那个球碰撞？

我开始的思路就是找它前面的球能与它相撞，且圆心位置最高的球；但是 wrong anwer  7；说实话，我现在还不确定这样判断为啥错，希望看了我博客的同学们，找到我的错误，给我留言，告诉我那错了，谢谢！qq邮箱：755672829@qq.com

收获2：我们可以换一种思路，就是这个球与其他某个球相撞了，这个球现在位置一定是当前能达到最高的高度，因为这个球是从高处y  = 10^100下滑的，碰撞住东西就停住；

错误代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define Max 1010
#include<math.h>

struct node
{
double t;
double l,r;
double h;
}stu[Max];
int n;
double r;

void fff(int j,int i)
{
double kk = sqrt((2*r*2*r)-(stu[i].t-stu[j].t)*(stu[i].t-stu[j].t));
stu[i].h = stu[j].h + kk;
}
int main()
{
int i,j;
while(~scanf("%d%lf",&n,&r))
{
for(i= 0 ;i < n;i++)
{
scanf("%lf",&stu[i].t);
stu[i].l = stu[i].t - r;
stu[i].r = stu[i].t + r;
double hh = -1;
int flag = -1;
for(j = 0;j < i;j ++)
{
if(stu[i].l>stu[j].r||stu[j].l>stu[i].r)
continue;
if(stu[j].h>hh)
{
flag = j;
hh = stu[j].h;
}
}
if(flag == -1)
stu[i].h = r;
else
fff(flag,i);
}
for(i=0;i<n;i++)
{
if(i==n-1) printf("%.10f\n",stu[i].h);
else printf("%.10f ",stu[i].h);
}
}
return 0;
}正确的代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define Max 1010
#include<math.h>

struct node
{
double t;
double h;
}stu[Max];
int n;
double r;
int main()
{
int i,j;
while(~scanf("%d%lf",&n,&r))
{
for(i= 0 ;i < n;i++)
{
scanf("%lf",&stu[i].t);
stu[i].h = -1;
for(j = 0;j < i;j ++)
{
// 能落到的一定是最大高度；
if(fabs(stu[i].t-stu[j].t)<=2*r)
stu[i].h = max(stu[i].h,stu[j].h+(sqrt((2*r*2*r)-(stu[i].t-stu[j].t)*(stu[i].t-stu[j].t))));
}
if(stu[i].h==-1)
stu[i].h = r;
}
for(i=0;i<n;i++)
{
if(i==n-1) printf("%.10f\n",stu[i].h);
else printf("%.10f ",stu[i].h);
}
}
return 0;
}