poj 2031 Building a Space Station（最小生成树，三维，基础）

只是坐标变成三维得了，而且要减去两边的半径而已

题目

//最小生成树，只是变成三维的了
#define  _CRT_SECURE_NO_WARNINGS
#include<stdlib.h>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;

#define inf 999999999
#define M 110
double mat[M][M];
struct tt
{
    double x,y,z,r;
}d[M];

double prim(int n,int sta)
{
    int mark[M],i,j;
    double dis[M],sum=;
    for(i=;i<n;i++)
    {
        dis[i]=mat[sta][i];
        mark[i]=;
    }
    mark[sta]=;
    for(i=;i<n;i++)
    {
        double minn=inf*1.0;
        int flag=-;
        for(j=;j<n;j++)
        {
            if(mark[j]==&&dis[j]<minn)
            {
                minn=dis[j];
                flag=j;
            }
        }
        if(flag!=-)
        {
            mark[flag]=;
            sum=sum+dis[flag];
            for(j=;j<n;j++)
            {
                if(dis[j]>mat[flag][j])
                    dis[j]=mat[flag][j];
            }
        }
    }
    return sum;
}

int main()
{
    int i,j,n;
    double aa;
    while(scanf("%d",&n),n)
    {
        for(i=;i<n;i++)
        {
            scanf("%lf%lf%lf%lf",&d[i].x,&d[i].y,&d[i].z,&d[i].r);
            for(j=;j<i;j++)
            {
                aa=sqrt((d[i].x-d[j].x)*(d[i].x-d[j].x)+(d[i].y-d[j].y)*(d[i].y-d[j].y)+(d[i].z-d[j].z)*(d[i].z-d[j].z))-d[i].r-d[j].r;
                aa=aa>? aa:;
                mat[i][j]=mat[j][i]=aa;
            }
        }
        printf("%.3lf\n",prim(n,));
    }
    return ;
}