poj2031 Building a Space Station

題目鏈接：http://poj.org/problem?id=2031
題意：給你n個球，讓你用最小的花費，把這n個球聯通，如果兩個球相交，就默認聯通，否者連接這兩個球的花費就是，球面之間的最短距離,輸出最小花費
解析：最小生成樹裸題

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn = 1e5+100;
struct node
{
    int u,v;
    double c;
    node() {}
    node(int _u,int _v,double _c)
    {
        u = _u;
        v = _v;
        c = _c;
    }
    bool operator < (const node &b)const
    {
        return c<b.c;
    }
}res[maxn];
struct point
{
    double x,y,z,r;
}a[105];
int fa[105];
double dis(point p1,point p2)
{
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+
                (p1.y-p2.y)*(p1.y-p2.y)+
                (p1.z-p2.z)*(p1.z-p2.z));
}
int getfa(int x)
{
    if(fa[x]==x)
        return fa[x];
    return fa[x] = getfa(fa[x]);
}
int main(void)
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lf %lf %lf %lf",&a[i].x,&a[i].y,&a[i].z,&a[i].r);
            fa[i] = i;
        }
        int cnt = 0;
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                double d = dis(a[i],a[j])-a[i].r-a[j].r;
                res[cnt++] = node(i,j,max(d,0.0));
            }
        }
        double ans = 0;
        sort(res,res+cnt);
        for(int i=0;i<cnt;i++)
        {
            int t1 = getfa(res[i].u);
            int t2 = getfa(res[i].v);
            if(t1!=t2)
            {
                ans += res[i].c;
                fa[t1] = t2;
            }
        }
        printf("%.3f\n",ans);
    }
    return 0;
}