最小生成树之Swordfish

题目链接：http://openoj.awaysoft.com:8080/judge/contest/view.action?cid=384#problem/A

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <math.h>
 4 double dis[110][110];
 5 int vist[110];
 6 
 7 struct node
 8 {
 9     double x,y;
10 }p[105];
11 
12 double map(node a,node b)
13 {
14     return ( sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) ) );
15 }
16 
17 int main()
18 {
19     int t=1,i,j,n,T;
20     double s;
21     while (~scanf("%d",&n) && n)
22     {
23         if (t!=1)
24         printf("\n");
25         for (i=0;i<n;i++)
26             scanf("%lf%lf",&p[i].x,&p[i].y);
27         memset(vist,0,sizeof(vist));
28         for (i=0;i<n-1;i++)
29             for (j=i+1;j<n;j++)
30                 dis[i][j]=dis[j][i]=map(p[i],p[j]);
31         vist[0]=1;
32         T=n-1;
33         s=0;
34         while (T--)
35         {
36             double m=10010;
37             int pos;
38             for (i=1;i<n;i++)
39                 if (!vist[i]&&dis[0][i]<m)
40                 {
41                     m=dis[0][i];
42                     pos=i;
43                 }
44             vist[pos]=1;
45             s+=m;
46             for (i=1;i<n;i++)
47                 if (!vist[i]&&dis[pos][i]<dis[0][i])
48                     dis[0][i]=dis[pos][i];
49         }        
50         printf("Case #%d:\n",t++);
51         printf("The minimal distance is: %0.2lf\n",s);
52     }
53     return 0;
54 }