POJ - 2031 Building a Space Station

标签：radius   gre   return   iostream   first   uil   intersect   val   ase   

You are a member of the space station engineering team, and are assigned a task in the construction process of the station. You are expected to write a computer program to complete the task.
The space station is made up with a number of units, called cells. All cells are sphere-shaped, but their sizes are not necessarily uniform. Each cell is fixed at its predetermined position shortly after the station is successfully put into its orbit. It is quite strange that two cells may be touching each other, or even may be overlapping. In an extreme case, a cell may be totally enclosing another one. I do not know how such arrangements are possible.

All the cells must be connected, since crew members should be able to walk from any cell to any other cell. They can walk from a cell A to another cell B, if, (1) A and B are touching each other or overlapping, (2) A and B are connected by a `corridor‘, or (3) there is a cell C such that walking from A to C, and also from B to C are both possible. Note that the condition (3) should be interpreted transitively.

You are expected to design a configuration, namely, which pairs of cells are to be connected with corridors. There is some freedom in the corridor configuration. For example, if there are three cells A, B and C, not touching nor overlapping each other, at least three plans are possible in order to connect all three cells. The first is to build corridors A-B and A-C, the second B-C and B-A, the third C-A and C-B. The cost of building a corridor is proportional to its length. Therefore, you should choose a plan with the shortest total length of the corridors.

You can ignore the width of a corridor. A corridor is built between points on two cells‘ surfaces. It can be made arbitrarily long, but of course the shortest one is chosen. Even if two corridors A-B and C-D intersect in space, they are not considered to form a connection path between (for example) A and C. In other words, you may consider that two corridors never intersect.
Input

Output

Sample Input

3
10.000 10.000 50.000 10.000
40.000 10.000 50.000 10.000
40.000 40.000 50.000 10.000
2
30.000 30.000 30.000 20.000
40.000 40.000 40.000 20.000
5
5.729 15.143 3.996 25.837
6.013 14.372 4.818 10.671
80.115 63.292 84.477 15.120
64.095 80.924 70.029 14.881
39.472 85.116 71.369 5.553
0

Sample Output

20.000
0.000
73.834

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 
 7 using namespace std;
 8 #define INF 0x3f3f3f3f
 9 
10 double cost[105][105];
11 double dian[105][4];
12 
13 int main()
14 {
15     int n,m;
16     double x,y,z,r,w;
17     double lowc[1005];
18     while(scanf("%d",&n)!=EOF,n!=0)
19     {
20         memset(cost,0x3f,sizeof(cost));//???
21         for(int i=0;i<n;i++)
22             cin>>dian[i][0]>>dian[i][1]>>dian[i][2]>>dian[i][3];
23         for(int i=0;i<n;i++)
24             for(int j=i+1;j<n;j++)
25             {
26                 w=sqrt((dian[i][0]-dian[j][0])*(dian[i][0]-dian[j][0])+(dian[i][1]-dian[j][1])*(dian[i][1]-dian[j][1])+(dian[i][2]-dian[j][2])*(dian[i][2]-dian[j][2]));
27                 w=w-dian[i][3]-dian[j][3];
28                 if(w<0)
29                     w=0;
30                 cost[i][j]=w;
31                 cost[j][i]=w;
32             }
33         for(int i=1;i<n;i++)
34             lowc[i]=cost[i][0];
35         //vis[0]=true;
36         lowc[0]=-1;
37         double ans=0;
38         int flag=0;
39         for(int i=1;i<n;i++)
40         {
41             double maxx=INF;
42             int p=-1;
43             for(int j=1;j<n;j++)
44                 if(lowc[j]!=-1&&maxx>lowc[j])
45                 {
46                     maxx=lowc[j];
47                     p=j;
48                     //cout<<p<<" "<<maxx<<endl;
49                 }
50             if(maxx==INF)
51             {
52                 flag=1;
53                 break;
54             }    
55             ans+=maxx;
56             //vis[p]=true;
57             lowc[p]=-1;
58             for(int j=1;j<n;j++)
59                 if(lowc[j]>cost[p][j])
60                     lowc[j]=cost[p][j];
61         }
62         if( flag == 1 ){
63             cout << "impossible" << endl;
64         }else{
65             printf("%.3f\n",ans);
66         }
67     }
68     
69     
70     return 0;
71 }

POJ - 2031 Building a Space Station

标签：radius   gre   return   iostream   first   uil   intersect   val   ase   

原文地址：http://www.cnblogs.com/xibeiw/p/7340547.html