1354 - Mobile Computing

There is a mysterious planet called Yaen,whose space is 2-dimensional. There are many beautiful stones on the planet,and the Yaen people love to collect them. They bring the stones back home andmake nice mobile arts of them to decorate their 2-dimensional living rooms.

In their 2-dimensional world, a mobile isdefined recursively as follows:

• a stone hung by a string, or
• a rod of length 1 with two sub-mobiles at both ends; the rod is hung by a string at the center of gravity of sub-mobiles. When the weights of the sub-mobiles are n and m , and their distances from the center of gravity are a and b respectively, the equation n x a =m x b holds.

For example, if you got three stones withweights 1, 1, and 2, here are some possible mobiles and their widths:

Given the weights of stones and the widthof the room, your task is to design the widest possible mobile satisfying bothof the following conditions.

• It uses all the stones.
• Its width is less than the width of the room.

You should ignore the widths of stones. Insome cases two sub-mobiles hung from both ends of a rod might overlap (see thefigure on the right). Such mobiles are acceptable. The width of the exampleis (1/3) + 1 + (1/4) .

Input 

The first line of the input gives thenumber of datasets. Then the specified number of datasets follow. A dataset hasthe following format.

r

s

w₁

w_s

r is a decimal fraction representing the width of the room, whichsatisfies 0 < r < 10 . s isthe number of the stones. You may assume 1s6 . w_i isthe weight of the i -th stone, which is an integer. You mayassume 1w_i1000 .

You can assume that no mobiles whose widthsare between r - 0.00001 and r +0.00001 can be made of given stones.

Output 

For each dataset in the input, one linecontaining a decimal fraction should be output. The decimal fraction shouldgive the width of the widest possible mobile as defined above. An output lineshould not contain extra characters such as spaces.

In case there is no mobile which satisfiesthe requirement, answer `-1' instead.

The answer should not have an error greaterthan 0.00000001. You may output any numb er of digits after the decimal point,provided that the ab ove accuracy condition is satisfied.

SampleInput 

5

1.3

3

1

2

1

1.4

3

1

2

1

2.0

3

1

2

1

1.59

4

2

1

1

3

1.7143

4

1

2

3

5

SampleOutput 

-1

1.3333333333333335

1.6666666666666667

1.5833333333333335

1.7142857142857142

代码:

#include<cstdio>

#include<cstring>

#include<vector>

using namespace std;

struct Tree

{

   double L, R; // distance from the root to the leftmost/rightmost point

   Tree():L(0),R(0) {}

};

const int maxn = 6;

int n, vis[1<<maxn];

double r, w[maxn], sum[1<<maxn];

vector<Tree> tree[1<<maxn];

void dfs(int subset)

{

   if(vis[subset]) return;

   vis[subset] = true;

   bool have_children = false;

   for(int left = (subset-1)&subset; left; left = (left-1)&subset)

    {

       have_children = true;

       int right = subset^left;

       double d1 = sum[right] / sum[subset];

       double d2 = sum[left] / sum[subset];

       dfs(left);

       dfs(right);

       for(int i = 0; i < tree[left].size(); i++)

           for(int j = 0; j < tree[right].size(); j++)

           {

                Tree t;

                t.L = max(tree[left][i].L + d1,tree[right][j].L - d2);

                t.R = max(tree[right][j].R +d2, tree[left][i].R - d1);

                if(t.L + t.R < r)tree[subset].push_back(t);

           }

    }

   if(!have_children) tree[subset].push_back(Tree());

}

int main()

{

   int T;

   scanf("%d", &T);

   while(T--)

    {

       scanf("%lf%d", &r, &n);

       for(int i = 0; i < n; i++) scanf("%lf", &w[i]);

       for(int i = 0; i < (1<<n); i++)

       {

           sum[i] = 0;

           tree[i].clear();

           for(int j = 0; j < n; j++)

                if(i & (1<<j)) sum[i]+= w[j];

       }

       int root = (1<<n)-1;

       memset(vis, 0, sizeof(vis));

       dfs(root);

       double ans = -1;

       for(int i = 0; i < tree[root].size(); i++)

           ans = max(ans, tree[root][i].L + tree[root][i].R);

       printf("%.10lf\n", ans);

    }

   return 0;

}