题意：
给你几个球的球心以及半径，让你在球之间建立路径。使得每个球之间都可以直接或者间接的到达。
做法：
把每个球看成一个点，求出来每两个点之间的距离，然后最小生成树算法。、
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#define INF 99999999
using namespace std;
struct list
{double x;double y;double z;double r;
}point[100001];
double juli(int i,int j)
{double len;len=sqrt((point[i].x-point[j].x)*(point[i].x-point[j].x)+(point[i].y-point[j].y)*(point[i].y-point[j].y)+(point[i].z-point[j].z)*(point[i].z-point[j].z));if(len-point[i].r-point[j].r<0)return 0;return len-point[i].r-point[j].r;
}
int main()
{int i,j,n;double map[101][101];while(scanf("%d",&n)&&n){memset(map,0,sizeof(map));for(i=0;i<n;i++){cin>>point[i].x>>point[i].y>>point[i].z>>point[i].r;}for(i=0;i<n;i++){for(j=0;j<n;j++){if(i!=j){map[i][j]=juli(i,j);}}}double low[101];int visit[101];memset(visit,0,sizeof(visit));visit[0]=1;for(i=0;i<n;i++){low[i]=map[0][i];}double sum;sum=0;for(i=0;i<n;i++){double min;int ipos;min=INF;for(j=0;j<n;j++){if(!visit[j]&&min>low[j]){min=low[j];ipos=j;}}if(min==INF)break;sum+=min;visit[ipos]=1;for(j=0;j<n;j++){if(low[j]>map[ipos][j]){low[j]=map[ipos][j];}}}printf("%.3f\n",sum);}return 0;
}
