uva 10034 Problem A: Freckles

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=975

最小生成树。

 1 #include <cstdio>

 2 #include <cstring>

 3 #include <cmath>

 4 #include <algorithm>

 5 #define maxn 1000

 6 using namespace std;

 7 const double inf=(double)(1<<29);

 8 

 9 int t,n;

10 struct node

11 {

12     double x,y;

13 }p[maxn];

14 bool vis[maxn];

15 double dis[maxn];

16 double g[maxn][maxn];

17 double ans;

18 

19 double sqr(double x)

20 {

21     return x*x;

22 }

23 

24 void prim()

25 {

26     memset(vis,false,sizeof(vis));

27     for(int i=1; i<=n; i++) dis[i]=g[1][i];

28     dis[1]=0;

29     vis[1]=true;

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

31     {

32         double m=inf;

33         int x;

34         for(int y=1; y<=n; y++) if(!vis[y]&&dis[y]<m) m=dis[x=y];

35         ans+=m;

36         vis[x]=true;

37         for(int y=1; y<=n; y++) if(!vis[y]&&dis[y]>g[x][y]) dis[y]=g[x][y];

38     }

39 }

40 

41 int main()

42 {

43     scanf("%d",&t);

44     while(t--)

45     {

46         scanf("%d",&n);

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

48         {

49             scanf("%lf%lf",&p[i].x,&p[i].y);

50         }

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

52         {

53             for(int j=i; j<=n; j++)

54             {

55                 if(i==j) g[i][j]=0;

56                 else

57                 g[i][j]=g[j][i]=sqrt(sqr(p[i].x-p[j].x)+sqr(p[i].y-p[j].y));

58             }

59         }

60         ans=0;

61         prim();

62         printf("%.2lf\n",ans);

63         if(t) printf("\n");

64     }

65     return 0;

66 }

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