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2013
11-10

POJ 2031 Building a Space Station [解题报告] Java

Building a Space Station

问题描述 :

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.

输入:

The input consists of multiple data sets. Each data set is given in the following format.

n

x1 y1 z1 r1

x2 y2 z2 r2



xn yn zn rn

The first line of a data set contains an integer n, which is the number of cells. n is positive, and does not exceed 100.

The following n lines are descriptions of cells. Four values in a line are x-, y- and z-coordinates of the center, and radius (called r in the rest of the problem) of the sphere, in this order. Each value is given by a decimal fraction, with 3 digits after the decimal point. Values are separated by a space character.

Each of x, y, z and r is positive and is less than 100.0.

The end of the input is indicated by a line containing a zero.

输出:

For each data set, the shortest total length of the corridors should be printed, each in a separate line. The printed values should have 3 digits after the decimal point. They may not have an error greater than 0.001.

Note that if no corridors are necessary, that is, if all the cells are connected without corridors, the shortest total length of the corridors is 0.000.

样例输入:

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

样例输出:

20.000
0.000
73.834

解题代码:

//* @author: [email protected]
import java.io.*;
import java.util.Arrays;
public class Main
{
 static int[] p;
 public static void main(String[] args) throws NumberFormatException, IOException
 {
  InputStreamReader is=new InputStreamReader(System.in);
  BufferedReader in=new BufferedReader(is);
  while(true)
  {
	int a=Integer.parseInt(in.readLine());
	if(a==0)break;
	double[][] dd=new double[a][4];
	for(int i=0;i< a;i++)
	{
	  String[] ss=in.readLine().split(" ");
	  for(int j=0;j< 4;j++)
	    dd[i][j]=Double.parseDouble(ss[j]);
	}
	p=new int[a];
	for(int i=0;i< a;i++)
		p[i]=i;
	rir[] yy=new rir[7000];
	int len=0;
	for(int i=0;i< a;i++)
	 for(int j=i+1;j< a;j++)
	 {
	  if(i==j) continue;
	    double dis=(dd[i][0]-dd[j][0])*(dd[i][0]-dd[j][0])+
		(dd[i][1]-dd[j][1])*(dd[i][1]-dd[j][1])+
		(dd[i][2]-dd[j][2])*(dd[i][2]-dd[j][2]);
	   dis=Math.sqrt(dis);
	   dis-=dd[i][3];
	   dis-=dd[j][3];
	   if(dis< 0) dis=0;
	     yy[len]=new rir(i,j,dis);
	   len++;
	}
	Arrays.sort(yy,0,a*(a-1)/2);
	//for(int i=0;i< a*(a-1)/2;i++)
	//System.out.println(yy[i].juli+" "+yy[i].x+" "+yy[i].y);
	double sum=0;
	for(int i=0;i< len;i++)
	{
	  if(un(yy[i].x,yy[i].y))
	  {
	   sum+=yy[i].juli;
	  }
	}
	System.out.printf("%.3f\n",sum);
			
  }
 }

  static int root(int x)
  {
   int b=x;
   while(p[b]!=b)
    b=p[b];
   return p[x]=b;
  }

  static boolean un(int a,int b)
  {
   int aa=root(a);
   int bb=root(b);
   if(aa==bb)return false;
   if(aa>bb)
	p[aa]=bb;
	else p[bb]=aa;
	return true;
  }
}

class rir implements Comparable< rir>
{
 int x;
 int y;
 double juli;
 public rir(int xx,int yy,double j)
 {
	x=xx;y=yy;juli=j;
 }
 @Override
 public int compareTo(rir o) {
	if(juli-o.juli< 0)return -1;
	if(juli==o.juli)return 0;
	else return 1;
 }
}

  1. 算法是程序的灵魂,算法分简单和复杂,如果不搞大数据类,程序员了解一下简单点的算法也是可以的,但是会算法的一定要会编程才行,程序员不一定要会算法,利于自己项目需要的可以简单了解。

  2. 5.1处,反了;“上一个操作符的优先级比操作符ch的优先级大,或栈是空的就入栈。”如代码所述,应为“上一个操作符的优先级比操作符ch的优先级小,或栈是空的就入栈。”

  3. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮