Convex Fence

Convex Fence

I have a land consisting of n trees. Since the trees are favorites to cows, I have a big problem saving them. So, I have planned to make a fence around the trees. I want the fence to be convex (curves are allowed) and the minimum distance from any tree to the fence is at least d units. And definitely I want a single big fence that covers all trees.

You are given all the information of the trees, to be specific, the land is shown as a 2D plane and the trees are plotted as 2D points. You have to find the perimeter of the fence that I need to create as described above. And you have to minimize the perimeter.

One tree, a circular fence is needed           Two trees, the fence is shown

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing two integers n (1 ≤ n ≤ 50000), d (1 ≤ d ≤ 1000). Each of the next lines contains two integers x_i y_i (-10⁸ ≤ x_i, y_i ≤ 10⁸) denoting a position of a tree. You can assume that all the positions are distinct.

Output

For each case, print the case number and the minimum possible perimeter of the fence. Errors less than 10^-3 will be ignored.

Sample Input

3

1 2

0 0

2 1

0 -1

0 2

3 5

0 0

5 0

0 5

Sample Output

Case 1: 12.566370614

Case 2: 12.2831853

Case 3: 48.4869943478

Hint

Dataset is huge, use faster i/o methods.

题目就是说,给定几个点,要用围栏围住所有点,且围栏与每个点的距离不小于D.

样例太水,再举几个例子:

多画几个图,你很快就会发现,所求答案就是原图凸包的周长+以D为半径的园的周长,水一水就过了.

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<iostream>
 #define LL long long
 #define PI (acos(-1.0))
 using namespace std;
 int n,top; double R;
 ],ch[];
 point operator - (point u,point v){point ret; ret.x=u.x-v.x,ret.y=u.y-v.y; return ret;}
 double dis(point u,point v){return sqrt((u.x-v.x)*(u.x-v.x)+(u.y-v.y)*(u.y-v.y));}
 LL cross(point u,point v){return u.x*v.y-v.x*u.y;}
 inline int read(){
     ,f=; char ch=getchar();
     '){if (ch=='-') f=-f; ch=getchar();}
     +ch-',ch=getchar();
     return x*f;
 }
 bool cmp(const point &u,const point &v){
     ],v-a[])>||cross(u-a[],v-a[])==&&dis(u,a[])<dis(v,a[]);
 }
 void Out_(){
     ;
     ){
         ; i<top; i++) ans+=dis(ch[i],ch[i+]);
         ans+=dis(ch[top],ch[]);
     }
     ans+=PI*R*;
     printf("%.7lf\n",ans);
 }
 void Graham(){
     ;
     ; i<=n; i++) if (a[i].y<a[now].y||a[i].y==a[now].y&&a[i].x<a[now].x) now=i;
     swap(a[now],a[]);
     sort(a+,a++n,cmp);
     ch[]=a[],ch[]=a[],ch[]=a[],top=;
     ; i<=n; i++){
         ],ch[top]-ch[top-])>) top--;
         ch[++top]=a[i];
     }
 }
 int main(){
     ; Ts<=T; Ts++){
         scanf(,,sizeof ch);
         ; i<=n; i++) a[i].x=read(),a[i].y=read();
         printf("Case %d: ",Ts);
         Graham(),Out_();
     }
     ;
 }