http://acm.hdu.edu.cn/showproblem.php?pid=1875

最小生成树，由于是稠密图，prim:O(T*n^2),kruskal:O(T*n^2logn^2)，kruskal好写，勉强可以过。

枚举任两点，符合条件时加边即可。

#include<bits/stdc++.h>
using namespace std;
#define dist2(i,j) ((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]))

int t,c,m,x[105],y[105],tot;
double ans;
int p[105];
struct Edge{
	int u,v;
	double w;
	bool operator < (Edge x){
		return w<x.w;
	}
}edge[10005];

int findset(int x){return p[x]==x?x:p[x]=findset(p[x]);}

int main()
{
	freopen("input.in","r",stdin);
	scanf("%d",&t);
	while(t--)
	{
		m=0;ans=0;tot=0;
		scanf("%d",&c);
		for(int i=1;i<=c;i++)scanf("%d%d",&x[i],&y[i]);
		for(int i=1;i<=c;i++)for(int j=i+1;j<=c;j++)
		{
			if(dist2(i,j)>=100&&dist2(i,j)<=1000000)
			{
				edge[++m].u=i;
				edge[m].v=j;
				edge[m].w=sqrt(dist2(i,j));
			}
		}
		sort(edge+1,edge+1+m);
		for(int i=1;i<=c;i++)p[i]=i;
		for(int i=1;i<=m&&tot<c-1;i++)
		{
			int x=findset(edge[i].u),y=findset(edge[i].v);
			if(x!=y)
			{
				tot++;
				p[x]=y;
				ans+=edge[i].w;
			}
		}
		if(tot<c-1)puts("oh!");
		else printf("%.1f\n",ans*100);
	}
	return 0;
}