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

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <cmath>
# define LL long long
using namespace std ;

const int INF=0x3f3f3f3f;
const int MAXN=110;
bool vis[MAXN];
double lowc[MAXN];
int n ;
double cost[MAXN][MAXN] ;

struct poin
{
    int x ;
    int y ;
}p[110];

double Prim()//点是0~n-1
{
    double ans=0;
    memset(vis,false,sizeof(vis));
    vis[0]=true;
    for(int i=1;i<n;i++)lowc[i]=cost[0][i];
    for(int i=1;i<n;i++)
    {
        double minc=INF;
        int p=-1;
        for(int j=0;j<n;j++)
            if(!vis[j]&&minc>lowc[j])
            {
                minc=lowc[j];
                p=j;
            }
            if(minc==INF)return -1;//原图不连通
            ans+=minc;
            vis[p]=true;
            for(int j=0;j<n;j++)
                if(!vis[j]&&lowc[j]>cost[p][j])
                    lowc[j]=cost[p][j];
    }
    return ans;
}

int main()
{

   // freopen("in.txt","r",stdin) ;
    int T ;
    scanf("%d" , &T) ;
    while(T--)
    {
        scanf("%d" , &n) ;
        int i , j ;
        for (i = 0 ; i < n ; i++)
            for (j = 0 ; j < n ; j++)
               cost[i][j] = INF ;
        for (i = 0 ; i < n ; i++)
            scanf("%d %d" , &p[i].x , &p[i].y) ;
        for (i = 0 ; i < n ; i++)
            for (j = i+1 ; j < n ; j++)
            {
                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)) ;
                if (t <10.0 || t > 1000.0)
                    continue ;
                cost[i][j] = t ;
                cost[j][i] = t ;
            }
        double k = Prim() ;
        if (k == -1)
            printf("oh!
") ;
        else
            printf("%.1lf
" , k*100) ;

    }
    return 0 ;
}