poj2031 Building a Space Station

题目链接：http://poj.org/problem?id=2031
题意：给你n个球，让你用最小的花费，把这n个球联通，如果两个球相交，就默认联通，否者连接这两个球的花费就是，球面之间的最短距离,输出最小花费
解析：最小生成树裸题

#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;}