poj 2031 Building a Space Station

题意：在三维坐标下给你n个球的坐标以及半径 ，覆盖或相邻的球不用建路  ，求最短路

解题思路：Kruskal在构建权值的时候判断是否相邻，如果相连，并查集合并 ，如果不相邻 则构建边，特别注意要用%f输出（%f可以输出double,且可移植性更强）

解题代码：

// File Name: d.cpp
// Author: darkdream
// Created Time: 2013年04月25日 星期四 00时16分20秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>

using namespace std;
struct point
{
  double x, y, z,r;
};
struct point points[105];
int p[105];
double w[10005];
int  v[10005];
int  u[10005];  
int  r[10005];
int cmp(const int i, const int j)
{
   return w[i] < w[j];
}
double dis(int i , int j )
{
    return sqrt(pow(points[i].x-points[j].x,2.0)+pow(points[i].z-points[j].z,2.0)+pow(points[i].y-points[j].y,2.0));
}
int find(int x)
{
  return p[x] == x? x:p[x] = find(p[x]);
}
int main(){
  int n ;
  //freopen("/home/plac/problem/input.txt","r",stdin);
  //freopen("output.txt","w",stdout);
  while(scanf("%d",&n) != EOF)
  {
    memset(p,0,sizeof(p));
    memset(v,0,sizeof(v));
    memset(u,0,sizeof(u));
    memset(r,0,sizeof(r));
    memset(w,0,sizeof(w));
    memset(points,0,sizeof(points));
    if(n == 0)
        break;
    for(int i =1 ;i <= n;i ++)
    {
      scanf("%lf %lf %lf %lf",&points[i].x,&points[i].y,&points[i].z,&points[i].r);
    }
    for(int i =1 ;i <= n;i ++)
        p[i] = i ;
    int m = 0 ;
    for(int i =1 ;i <= n; i++)
        for(int j = i+1 ;j <= n ;j ++ )
        {
        if(dis(i,j) > (points[i].r + points[j].r))
           {
              v[++m] = i ;
              u[m] = j;
              w[m] = dis(i,j) - points[i].r - points[j].r;
           }
           else
           {
        //    printf("%d %d\n",i,j);
             /*v[++m] = i;
             v[m] = j;
             w[m] = 0.0;*/
             int x = find(i);
             int y = find(j);
             if(x != j)
                 p[x] = y;
           }
              
        }
    //for(int i=1 ;i <= n;i ++)
    //    printf("%d ",p[i]);
    //printf("%lf\n",dis(3,4));
    //printf("%d %d\n",v[439],u[439]);
    for(int i =1;i <= m;i ++)
        r[i] = i;
    sort(r+1,r+m+1,cmp);
    double ans = 0.000 ;
    for(int i = 1;i <= m;i ++)
    {
      int e = r[i];
      int x = find(v[e]);
      int y = find(u[e]);
      if(x != y)
      {
    //     printf("%d %d %d\n",e,v[e],u[e]);
         ans += w[e];
         p[x] = y ;
      }
    }
    printf("%.3f\n",ans);
  }
return 0;
}