poj2560

求最小生成树

 1 //Accepted    240 KB    0 ms

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 const int MAXN =105;

 6 const int inf = 100000000;

 7 double x[MAXN],y[MAXN];

 8 double a[MAXN][MAXN];

 9 bool vis[MAXN];

10 double dis[MAXN];

11 int n;

12 void pre()

13 {

14     for (int i=0;i<n;i++)

15     {

16         for (int j=0;j<n;j++)

17         {

18             a[i][j]=sqrt((double )(x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));

19         }

20     }

21 }

22 double prim()

23 {

24     double ans=0;

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

26     for (int i=0;i<n;i++)

27     dis[i]=(double )inf;

28     dis[0]=0;

29     for (int i=0;i<n;i++)

30     {

31         double temp=(double )inf;

32         int k=0;

33         for (int j=0;j<n;j++)

34         {

35             if (vis[j]) continue;

36             if (temp>dis[j])

37             {

38                 temp=dis[j];

39                 k=j;

40             }

41         }

42         vis[k]=1;

43         ans+=temp;

44         for (int j=0;j<n;j++)

45         {

46             if (vis[j]) continue;

47             if (dis[j]>a[k][j])

48             {

49                 dis[j]=a[k][j];

50             }

51         }

52     }

53     return ans;

54 }

55 int main()

56 {

57     while (scanf("%d",&n)!=EOF)

58     {

59         for (int i=0;i<n;i++)

60         {

61             scanf("%lf%lf",&x[i],&y[i]);

62         }

63         pre();

64         double ans=prim();

65         printf("%.2f\n",ans);

66     }

67     return 0;

68 }

poj2560