poj 2031Building a Space Station(几何判断+Kruskal最小生成树)

 1 /*
 2   最小生成树 + 几何判断
 3   Kruskal      球心之间的距离 - 两个球的半径 < 0 则说明是覆盖的！此时的距离按照0计算 
 4 */
 5 #include<iostream>
 6 #include<cstdio>
 7 #include<cstring>
 8 #include<cmath>
 9 #include<algorithm>
10 using namespace std;
11 int f[105];
12 struct ball{
13    double x, y, z, r;
14 };
15 
16 struct connect{
17    double dist;
18    int a, b;
19 };
20 
21 connect c[5005];
22 
23 ball b[105];
24 
25 bool cmp(connect a, connect b){
26    return a.dist < b.dist;
27 }
28 int n;
29 
30 int getFather(int x){
31    return x==f[x] ? x : f[x]=getFather(f[x]); 
32 } 
33 
34 int Union(int a, int b){
35     int fa=getFather(a), fb=getFather(b);
36     if(fa!=fb){
37         f[fa]=fb;
38         return 1;
39     }
40     return 0;
41 }
42 
43 int main(){
44    int i, j;
45    while(scanf("%d", &n) && n){
46       for(i=1; i<=n; ++i)
47          scanf("%lf%lf%lf%lf", &b[i].x, &b[i].y, &b[i].z, &b[i].r);
48       int cnt=0;
49       for(i=1; i<n; ++i)
50          for(j=i+1; j<=n; ++j){
51              double d = sqrt((b[i].x-b[j].x)*(b[i].x-b[j].x) + (b[i].y-b[j].y)*(b[i].y-b[j].y) + (b[i].z-b[j].z)*(b[i].z-b[j].z))
52                         - (b[i].r + b[j].r);
53                c[cnt].dist= d<0 ? 0: d;
54                c[cnt].a=i; 
55                c[cnt++].b=j;
56          }
57        sort(c, c+cnt, cmp); 
58        double minSum=0.0;
59        for(i=1; i<=n; ++i)
60           f[i]=i;
61        for(i=0; i<cnt; ++i){
62           if(Union(c[i].a, c[i].b))
63              minSum+=c[i].dist;
64        }
65        printf("%.3lf\n", minSum);
66    }
67    return 0;
68 }