POJ-2031-Building a Space Station

链接：https://vjudge.net/problem/POJ-2031#author=0

题意:

三维空间，给n个圆心x,y,z,半径r的圆，求最短的连线。

接触不需要连。

思路：

求距离，接触权值为0，不接触为权值长度减半径。

代码：

#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
using namespace std;
typedef long long LL;
const int MAXM = 10000+10;
const int MAXN = 100+10;

struct Node
{
    double _x,_y,_z,_r;
}node[MAXN];

struct Path
{
    int _l,_r;
    double _value;
    bool operator < (const Path & that)const{
        return this->_value < that._value;
    }
}path[MAXM];

double Get_Len(Node a,Node b)
{
    return sqrt((a._x-b._x)*(a._x-b._x) + (a._y-b._y)*(a._y-b._y) + (a._z-b._z)*(a._z-b._z));
}

int Father[MAXN];

int Get_F(int x)
{
    return Father[x] = (Father[x] == x) ? x : Get_F(Father[x]);
}

int main()
{
    int n;
    while (cin >> n && n)
    {
        for (int i = 1;i<=n;i++)
            Father[i] = i;
        for (int i = 1;i<=n;i++)
            cin >> node[i]._x >> node[i]._y >> node[i]._z >> node[i]._r;
        int pos = 0;
        for (int i = 1;i<=n;i++)
        {
            for (int j = 1;j<=n;j++)
            {
                double len = Get_Len(node[i],node[j]);
                double r = node[i]._r + node[j]._r;
                path[++pos]._l = i;
                path[pos]._r = j;
                if (len <= r)
                    path[pos]._value = 0;
                else
                    path[pos]._value = len - r;
            }
        }
        sort(path+1,path+1+pos);
        double sum = 0;
        for (int i = 1;i <= pos;i++)
        {
            int tl = Get_F(path[i]._l);
            int tr = Get_F(path[i]._r);
            if (tl != tr)
            {
                Father[tl] = tr;
                sum += path[i]._value;
            }
        }
        printf("%.3lf
",sum);
    }

    return 0;
}