撰写The space station is made up with a number of units, called cells. All cells are sphere-shaped, but their sizes are not necessarily uniform. Each cell is fixed at its predetermined position shortly after the station is successfully put into its orbit. It is quite strange that two cells may be touching each other, or even may be overlapping. In an extreme case, a cell may be totally enclosing another one. I do not know how such arrangements are possible.
All the cells must be connected, since crew members should be able to walk from any cell to any other cell. They can walk from a cell A to another cell B, if, (1) A and B are touching each other or overlapping, (2) A and B are connected by a `corridor’, or (3) there is a cell C such that walking from A to C, and also from B to C are both possible. Note that the condition (3) should be interpreted transitively.
You are expected to design a configuration, namely, which pairs of cells are to be connected with corridors. There is some freedom in the corridor configuration. For example, if there are three cells A, B and C, not touching nor overlapping each other, at least three plans are possible in order to connect all three cells. The first is to build corridors A-B and A-C, the second B-C and B-A, the third C-A and C-B. The cost of building a corridor is proportional to its length. Therefore, you should choose a plan with the shortest total length of the corridors.
You can ignore the width of a corridor. A corridor is built between points on two cells’ surfaces. It can be made arbitrarily long, but of course the shortest one is chosen. Even if two corridors A-B and C-D intersect in space, they are not considered to form a connection path between (for example) A and C. In other words, you may consider that two corridors never intersect.
Input
n
x1 y1 z1 r1
x2 y2 z2 r2
…
xn yn zn rn
The first line of a data set contains an integer n, which is the number of cells. n is positive, and does not exceed 100.
The following n lines are descriptions of cells. Four values in a line are x-, y- and z-coordinates of the center, and radius (called r in the rest of the problem) of the sphere, in this order. Each value is given by a decimal fraction, with 3 digits after the decimal point. Values are separated by a space character.
Each of x, y, z and r is positive and is less than 100.0.
The end of the input is indicated by a line containing a zero.
Output
Note that if no corridors are necessary, that is, if all the cells are connected without corridors, the shortest total length of the corridors is 0.000.
Sample Input
3 10.000 10.000 50.000 10.000 40.000 10.000 50.000 10.000 40.000 40.000 50.000 10.000 2 30.000 30.000 30.000 20.000 40.000 40.000 40.000 20.000 5 5.729 15.143 3.996 25.837 6.013 14.372 4.818 10.671 80.115 63.292 84.477 15.120 64.095 80.924 70.029 14.881 39.472 85.116 71.369 5.553 0
Sample Output
20.000 0.000 73.834
题意：三维空间内给你很多个圆心坐标与半径，相接触的代表有路到接触的两个圆，问加使其连通所加的最小边权和，这里加的道路相交不算相交。
题解：读入坐标，通过位置构造边权，相交为0，不相交则为两圆最短距离，跑一次K树即可。
代码：
#include<algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
const ll mod = 1000000007;
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
const int maxn=1e2+5;
const int maxm=1e6+10;
int F[maxn];
int tol;
int find(int x){
return F[x]==-1?x:F[x]=find(F[x]);
}
struct Edge{
int u;
int v;
double w;
}edge[maxm];
void addedge(int u,int v,double w){
edge[tol].u=u;
edge[tol].v=v;
edge[tol++].w=w;
}
bool cmp(Edge &a, Edge &b){
return a.w<b.w;
}
double Kruskal(int n)//传入点数，返回最小生成树的权值，如果不连通返回-1
{
memset(F,-1,sizeof(F));
sort(edge,edge+tol,cmp);
int cnt=0;//计算加入的边数
double ans=0;
for(int i=0;i<tol;i++) {
int u=edge[i].u;
int v=edge[i].v;
double w=edge[i].w;
int t1=find(u);
int t2=find(v);
if(t1!=t2) {
ans+=w;
F[t1]=t2;
cnt++;
}
if(cnt==n-1)break;
}
if(cnt<n-1)return -1;//不连通
else return ans;
}
struct circle{
double x;
double y;
double z;
double r;
}c[maxn];
double cont(int index1,int index2){
double t1=sqrt((c[index1].x-c[index2].x)*(c[index1].x-c[index2].x)+(c[index1].y-c[index2].y)*(c[index1].y-c[index2].y)+(c[index1].z-c[index2].z)*(c[index1].z-c[index2].z));
double t2=c[index1].r+c[index2].r;
if(t1>t2)return t1-t2;
else return 0;
}
int main(){
int n;
double x,y,z,r;
while(scanf("%d", &n)&&n){
tol=0;
rep(i,1,n){
scanf("%lf%lf%lf%lf", &c[i].x,&c[i].y,&c[i].z,&c[i].r);
}
rep(i,1,n-1){
rep(j,i+1,n){
double t=cont(i,j);
addedge(i,j,t);
}
}
printf("%.3f\n",Kruskal(n));
}
return 0;
}