hdu 1875 给出每个结点的坐标 权值为两点间的距离 （MST）

Sample Input
2
2
10 10 //坐标
20 20
3
1 1
2 2
1000 1000

Sample Output
1414.2   //最小权值和*100  保留1位小数
oh!       //不连通

prim

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <cmath>
 6 # define LL long long
 7 using namespace std ;
 8
 9 const int INF=0x3f3f3f3f;
10 const int MAXN=110;
11 bool vis[MAXN];
12 double lowc[MAXN];
13 int n ;
14 double cost[MAXN][MAXN] ;
15
16 struct poin
17 {
18     int x ;
19     int y ;
20 }p[110];
21
22 double Prim()//点是0~n-1
23 {
24     double ans=0;
25     memset(vis,false,sizeof(vis));
26     vis[0]=true;
27     for(int i=1;i<n;i++)lowc[i]=cost[0][i];
28     for(int i=1;i<n;i++)
29     {
30         double minc=INF;
31         int p=-1;
32         for(int j=0;j<n;j++)
33             if(!vis[j]&&minc>lowc[j])
34             {
35                 minc=lowc[j];
36                 p=j;
37             }
38             if(minc==INF)return -1;//原图不连通
39             ans+=minc;
40             vis[p]=true;
41             for(int j=0;j<n;j++)
42                 if(!vis[j]&&lowc[j]>cost[p][j])
43                     lowc[j]=cost[p][j];
44     }
45     return ans;
46 }
47
48 int main()
49 {
50
51    // freopen("in.txt","r",stdin) ;
52     int T ;
53     scanf("%d" , &T) ;
54     while(T--)
55     {
56         scanf("%d" , &n) ;
57         int i , j ;
58         for (i = 0 ; i < n ; i++)
59             for (j = 0 ; j < n ; j++)
60                cost[i][j] = INF ;
61         for (i = 0 ; i < n ; i++)
62             scanf("%d %d" , &p[i].x , &p[i].y) ;
63         for (i = 0 ; i < n ; i++)
64             for (j = i+1 ; j < n ; j++)
65             {
66                 double t = sqrt((double)(p[i].x - p[j].x) * (p[i].x - p[j].x) + (p[i].y - p[j].y) * (p[i].y - p[j].y)) ;
67                 if (t <10.0 || t > 1000.0)
68                     continue ;
69                 cost[i][j] = t ;
70                 cost[j][i] = t ;
71             }
72         double k = Prim() ;
73         if (k == -1)
74             printf("oh!\n") ;
75         else
76             printf("%.1lf\n" , k*100) ;
77
78     }
79     return 0 ;
80 }