hoj2798 Globulous Gumdrops

Globulous Gumdrops

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	Source : 2008 Stanford Programming Contest
	Time limit : 1 sec		Memory limit : 64 M

Submitted : 174, Accepted : 94

Description

Gwen just bought a bag of gumdrops! However, she does not like carrying gumdrops in plastic bags; instead, she wants to pack her gumdrops in a cylindrical tube of diameter d. Given that each of her gumdrops are perfect spheres of radii r1, r2, . . . , rn, find the shortest length tube Gwen can use to store her gumdrops.

You should assume that the gumdrop radii are sufficiently large that no three gumdrops can be simultaneously in contact with each other while fitting in the tube. Given this restriction, it may be helpful to realize that the gumdrops will always be packed in such a way that their centers lie on a single two-dimensional plane containing the axis of rotation of the tube.

Input

The input file will contain multiple test cases. Each test case will consist of two lines. The first line of each test case contains an integer n (1 <= n <= 15) indicating the number of gumdrops Gloria has, and a floating point value d (2.0 <= d <= 1000.0) indicating the diameter of the cylindrical tube, separated by a space. The second line of each test case contains a sequence of n space-separated floating point numbers, r1 r2 . . . rn (1.0 <= ri <= d/2) are the radii of the gum drops in Gloria's bag. A blank line separates input test cases. A single line with the numbers "0 0" marks the end of input; do not process this case.

Output

For each input test case, print the length of the shortest tube, rounded to the nearest integer.

Sample Input

2 98.1789
42.8602 28.7622
3 747.702
339.687 191.953 330.811
0 0

Sample Output

138
1628

/*
    可以用dp[state][i]表示所用球状态为state,且最上面一个是第i个球，要放入这些球所需要的最大高度，那么每次加入一个球就可以转移了，状态转移方程为dp[state1][j]=min(dp[state1 ][j],jisuan(dp[state][i],i,j) );state1=state|( 1<<(j-1) );
*/
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<cstdio>
#include<cstring>
using namespace std;
double r[16],d,dp[1<<15][16];
int n;
double count(double pre,int now,int top){
        double x=r[top]+r[now];
        double y=d-r[top]-r[now];
        return pre+r[top]-r[now]+sqrt(x*x-y*y);
}
int main(){
        while(scanf("%d%lf",&n,&d)!=EOF){
                if(n==0&&d==0)return 0;
                for(int i=1;i<=n;i++)scanf("%lf",&r[i]);
                //memset(dp,127/3,sizeof(dp));
                for(int i=1;i<(1<<n);i++){
                        for(int j=1;j<=n;j++){
                                dp[i][j]=0x7fffffff;
                        }
                }
                for(int i=1;i<=n;i++)dp[1<<(i-1)][i]=r[i]*2;
                for(int state=1;state<(1<<n);state++){
                        for(int i=1;i<=n;i++){
                                for(int j=1;j<=n;j++){
                                        if(state&(1<<(j-1)))continue;
                                        int state1=state|(1<<(j-1));
                                        //cout<<count(dp[state][i],i,j)<<endl;
                                        dp[state1][j]=min(dp[state1][j],count(dp[state][i],i,j));
                                }
                        }
                }
                double ans=0x7fffffff;
                for(int i=1;i<=n;i++){
                        ans=min(ans,dp[(1<<n)-1][i]);
                }
                int Ans=(int)(ans+0.5);
                printf("%d\n",Ans);
        }
}