UVa 1354 Mobile Computing［暴力枚举］

 **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 and make nice mobile arts of them to decorate their 2-dimensional living rooms. 
In their 2-dimensional world, a mobile is defined 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 × a = m × b holds. 
For example, if you got three stones with weights 1, 1, and 2, here are some possible mobiles and their widths: 
Given the weights of stones and the width of the room, your task is to design the widest possible mobile satisfying both of 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. 
In some cases two sub-mobiles hung from both ends of a rod might overlap (see the figure on the right). Such mobiles are acceptable. The width of the example is (1/3) + 1 + (1/4). 
Input 
The first line of the input gives the number of datasets. Then the specified number of datasets follow. A dataset has the following format. 
r s w1 . 
ws 
r is a decimal fraction representing the width of the room, which satisfies 0 < r < 10. s is the number of the stones. You may assume 1 ≤ s ≤ 6. wi is the weight of the i-th stone, which is an integer. You may assume 1 ≤ wi ≤ 1000. 
Input 
The first line of the input gives the number of datasets. Then the specified number of datasets follow. A dataset has the following format. 
r s w1 . 
ws 
r is a decimal fraction representing the width of the room, which satisfies 0 < r < 10. s is the number of the stones. You may assume 1 ≤ s ≤ 6. wi is the weight of the i-th stone, which is an integer. You may assume 1 ≤ wi ≤ 1000. 
You can assume that no mobiles whose widths are between r − 0.00001 and r + 0.00001 can be made of given stones. 
Output 
For each dataset in the input, one line containing a decimal fraction should be output. The decimal fraction should give the width of the widest possible mobile as defined above. An output line should not contain extra characters such as spaces. 
In case there is no mobile which satisfies the requirement, answer ‘-1’ instead. 
The answer should not have an error greater than 0.00000001. You may output any numb er of digits after the decimal point, provided that the ab ove accuracy condition is satisfied. 
Sample Input 
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 
Sample Output 
-1 
1.3333333333333335 
1.6666666666666667 
1.5833333333333335 
1.7142857142857142 

解题思路： 
1.采用自底向上的方法枚举树——每次随机选取两棵子树合并成一棵树，每个结点依次编号。 
2.对于一棵确定的树，其长度必然可以确定。以根结点为坐标轴原点，dfs计算每个结点相对根结点的距离即可求出该树宽度。 
注意：输入只有一块石头时，输出0；

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 const int maxn=16;
 7 int lchild[maxn];//左孩子编号
 8 int rchild[maxn];//右孩子编号
 9 int wight[maxn];//编号对应的质量
10 int vis[maxn];//-1表示编号不存在 0表示编号不在树中 1表示在树中
11 double dis[maxn];
12 
13 double r,ans;
14 int s;
15 void init(){
16     ans=0;
17     memset(lchild, -1, sizeof lchild);
18     memset(rchild, -1, sizeof rchild);
19     memset(wight,0,sizeof wight);
20     memset(vis, -1, sizeof vis);
21 }
22 
23 void calculate(int id){//计算每个编号相对根结点的距离
24     if(lchild[id]!=-1){
25         dis[lchild[id]]=dis[id]-double(wight[rchild[id]])/double(wight[lchild[id]]+wight[rchild[id]]);
26         dis[rchild[id]]=dis[id]+double(wight[lchild[id]])/double(wight[lchild[id]]+wight[rchild[id]]);
27         calculate(lchild[id]);
28         calculate(rchild[id]);
29     }
30 }
31 
32 void search(int cnt,int m){//m为此阶段石头最大编号
33     if(cnt==1){
34         memset(dis, 0, sizeof dis);
35         calculate(0);
36         double a=0,b=0;
37         for(int i=0;i<maxn;i++){
38             if(dis[i]<a) a=dis[i];
39             if(dis[i]>b) b=dis[i];
40         }
41         double c=b-a;
42      //   cout<<"  "<<c<<endl;
43         if(c<r&&c>ans) ans=c;
44         return ;
45     }
46     for(int i=1;i<maxn;i++){
47         if(vis[i]==0){
48             vis[i]=1;
49             for(int j=1;j<maxn;j++){
50                 if(vis[j]==0){
51                     vis[j]=1;
52                     if(cnt==2){
53 
54                         lchild[0]=i;rchild[0]=j;
55                         wight[0]=wight[i]+wight[j];
56                         search(cnt-1,m);
57                     }
58                     else{
59 
60                         vis[m+1]=0;
61                         lchild[m+1]=i;rchild[m+1]=j;
62                         wight[m+1]=wight[i]+wight[j];
63                         search(cnt-1,m+1);
64                         vis[m+1]=-1;
65                     }
66                     vis[j]=0;
67                 }
68             }
69             vis[i]=0;
70         }
71     }
72 }
73 int main() {
74     //freopen("input.txt", "rb", stdin);
75     //freopen("output.txt","wb",stdout);
76     int N;
77     scanf("%d",&N);
78     while(N--){
79         init();
80         scanf("%lf%d",&r,&s);
81 
82         for(int i=1;i<=s;i++){
83            scanf("%d",&wight[i]);
84             vis[i]=0;
85         }
86         if(s==1) {printf("%.16f\n",ans);continue;}
87         search(s,s);
88         if(ans==0) cout<<"-1"<<endl;
89         else printf("%.16f\n",ans);
90     }
91     return 0;
92 }