1. 程式人生 > >poj 2931 Building a Space Station <克魯斯卡爾>

poj 2931 Building a Space Station <克魯斯卡爾>

accep for each ppi ons cee ont line 求解 0.11

Building a Space Station

Time Limit: 1000MS

Memory Limit: 30000K

Total Submissions: 5869

Accepted: 2910

Description

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

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.

Output

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.

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

Source

field=source&key=Japan+2003+Domestic" target="_blank">Japan 2003 Domestic 題意: 首先輸入一個數n。代表空間站的個數。然後輸入n行數,每行4個數(double型)。分別相應那個空間站的坐標(x,y,z),和它的半徑(由於空間站是圓形的,所以會有半徑),然後題目要求是讓在每一個空間站間安裝一個走廊。能夠從隨意一個空間站到另外一個。也就是構造最小生成樹。(可是須要註意的是這是圓形的你算過圓心之間的距離之後還須要將兩個半徑減去,才是須要修的走廊的長度,而且假設兩個圓心之間的距離有可能小於兩個半徑之和,這個你就須要另外處理,將他處理成0即可了,不能是負數,由於走廊的長度不會是負數)。 思路: 先將隨意兩點的之間的距離(減去兩個半徑)求出來,然後將他們進行排序,用克魯斯卡爾算法求解即可了!(記住距離不能是負數。假設求的是負數,就將其改成0)。 代碼:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int n;
double x[105];
double y[105];
double z[105];
double r[105];
int pre[105];
struct node 
{
	int u,v;
	double w; 
}map[10005];
int cmp(node a,node b)
{
	return a.w<b.w;
}
void init()
{

}
int find(int x)
{
	int r=x;
	while(r!=pre[r])
	{
		r=pre[r];
	}
	int i,j;
	i=x;
	while(i!=r)
	{
		j=pre[i];
		pre[i]=r;
		i=j;
	}
	return r;
}
int join(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		pre[fx]=fy;
		return 1;
	}
	return 0;
}
int main()
{
    
	while(scanf("%d",&n)&&n)
	{
	   	for(int i=1;i<=101;i++)
	    {
		pre[i]=i;
	    }
		
		for(int i=1;i<=n;i++)
	    {
		scanf("%lf%lf%lf%lf",&x[i],&y[i],&z[i],&r[i]);
	    }
		int t=0;
		double d;
		for(int i=1;i<n;i++)
		{
			for(int j=i+1;j<=n;j++)
			{
				t++;
				map[t].u=i;
				map[t].v=j;
				d=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])+(z[i]-z[j])*(z[i]-z[j]));//三維坐標求距離!

if(d<(r[i]+r[j]))//這一點須要特殊粗粒 map[t].w=0; else map[t].w=d-(r[i]+r[j]); } } sort(map+1,map+t+1,cmp); double sum=0; for(int i=1;i<=t;i++) { if(join(map[i].u,map[i].v)) { sum+=map[i].w; } } printf("%.3f\n",sum);//註意輸出的時候,這一道 題有個坑。就是必須用%f輸出! } return 0; }


poj 2931 Building a Space Station &lt;克魯斯卡爾&gt;