Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1554    Accepted Submission(s): 677

本题非常简单，最小生成树问题，直接抄模板即可

代码：

 1 #include<stdio.h>
 2 #include<math.h>
 3  struct node
 4 {
 5     double x,y;
 6 }s[105];
 7  int n;
 8  double prim(int v0)
 9 {
10     int i,j,visit[105],u;
11     double sum=0,min,distance[105],s1;
12     for(i=1;i<=n;i++)
13     {
14         distance[i]=sqrt((s[v0].x-s[i].x)*(s[v0].x-s[i].x)+(s[v0].y-s[i].y)*(s[v0].y-s[i].y));
15         visit[i]=0;
16     }
17     visit[v0]=1;
18     for(i=1;i<n;i++)
19     {
20         min=0xfffffff;
21         for(j=1;j<=n;j++)
22         {
23             if(!visit[j] && distance[j]<min)
24             {
25                 u=j;
26                 min=distance[j];
27             }
28         }
29         sum+=min;
30         visit[u]=1;
31         for(j=1;j<=n;j++)
32         {
33             s1=sqrt((s[u].x-s[j].x)*(s[u].x-s[j].x)+(s[u].y-s[j].y)*(s[u].y-s[j].y));
34             if(!visit[j] && s1<distance[j])
35                 distance[j]=s1;
36         }
37     }
38     return sum;
39 }
40 int main()
41 {
42     int i;double x,y;
43     while(scanf("%d",&n)!=EOF)
44     {
45         for(i=1;i<=n;i++)
46         {
47             scanf("%lf%lf",&x,&y);
48             s[i].x=x;
49             s[i].y=y;
50         }
51         printf("%.2lf\n",prim(1));
52     }
53     return 0;
54 }