poj Building a Space Station

http://poj.org/problem?id=2031

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<cmath>

 4 #include<algorithm>

 5 using namespace std;

 6 

 7 const double eps=1e-8;

 8 const int inf=1<<23;

 9 int cmp(double x)

10 {

11     if(fabs(x)<eps) return 0;

12     if(x>0) return 1;

13     return -1;

14 }

15 

16 const double pi=acos(-1.0);

17 inline double sqr(double x)

18 {

19     return x*x;

20 }

21 

22 inline double Sqrt(double a)

23 {

24     return a<=0?0:sqrt(a);

25 }

26 

27 struct point

28 {

29    double x,y,z,r;

30    point(){}

31    point(double a,double b,double c,double d):x(a),y(b),z(c),r(d){}

32 };

33 

34 double dis(const point &a,const point &b)

35 {

36     return Sqrt(sqr(a.x-b.x)+sqr(a.y-b.y)+sqr(a.z-b.z));

37 }

38 

39 double dist[1000];

40 double diss[110][110],ans;

41 bool vis[1000];

42 bool prime(int n)

43 {

44     memset(vis,0,sizeof(vis));

45     for(int i=1; i<=n; i++)

46         dist[i]=inf;

47     ans=0;dist[1]=0;

48     for(int i=1; i<=n; i++){

49         double temp=inf;

50         int k=0;

51         for(int j=1; j<=n; j++)

52         {

53             if(!vis[j]&&dist[j]<temp)

54             {

55                 temp=dist[j];

56                 k=j;

57             }

58         }

59         if(temp==inf) return false;

60         vis[k]=true;

61         ans+=temp;

62         for(int j=1; j<=n; j++)

63         {

64             if(!vis[j]&&dist[j]>diss[k][j])

65             {

66                 dist[j]=diss[k][j];

67             }

68         }

69     }

70     return true;

71 }

72 int main()

73 {

74     int n;

75     while(scanf("%d",&n)&&n)

76     {

77         point a[1000];

78         memset(diss,0,sizeof(diss));

79         for(int i=1; i<=n; i++)

80         {

81             scanf("%lf%lf%lf%lf",&a[i].x,&a[i].y,&a[i].z,&a[i].r);

82         }

83         memset(diss,0,sizeof(diss));

84         for(int i=1; i<=n; i++)

85         {

86             for(int j=1; j<=n; j++)

87             {

88                 if(dis(a[i],a[j])-a[i].r-a[j].r<=0)

89                 diss[i][j]=0;

90                 else if(dis(a[i],a[j])-a[i].r-a[j].r>eps)

91                 diss[i][j]=dis(a[i],a[j])-a[i].r-a[j].r;

92             }

93         }

94         prime(n);

95         printf("%.3lf\n",ans);

96     }

97     return 0;

98 }

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