Arctic Network

                                                 Arctic Network

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 3

Problem Description

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel.
Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts.

Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

 

Input

The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000).

 

Output

For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.

 

Sample Input

1
2 4
0 100
0 300
0 600
150 750

 

Sample Output

212.13

 

题意：第一个数字表示数据的组数，每组测试数据第一行，第二个数字表示点的个数，连成最小生成树之后，第一个数字就表示第多少长的边，裸地prim，需要记录每次挑选出来的min，然后排序，代码如下

#include<stdio.h>
#include<string.h>
#define INF 0xfffffff
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int cmp(double a,double b)
{
	if(a>b)
	return 1;
	return 0;
}
double x[1010],y[1010],map[1010][1010],ans[1010];
int v[1010];
double dis(double x1,double y1,double x2,double y2)
{
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int s,p,i,j,flag,cot=0;
		double min;
		scanf("%d%d",&s,&p);
		for(i=1;i<=p;i++)
		{
			for(j=1;j<=p;j++)
			{
				map[i][j]=INF;
			}
		}
		for(i=1;i<=p;i++)
		{
			scanf("%lf%lf",&x[i],&y[i]);
			for(j=1;j<i;j++)
			{
				map[i][j]=map[j][i]=dis(x[i],y[i],x[j],y[j]);
			}
		}
		memset(v,0,sizeof(v));
		v[1]=1;
		for(i=1;i<=p;i++)
		{
			flag=-1;
			min=INF;
			for(j=1;j<=p;j++)
			{
				if(!v[j]&&map[1][j]<min)
				{
					min=map[1][j];
					flag=j;
				}
			}
			if(flag==-1)
				break;
			v[flag]=1;
			ans[cot++]=min;
			for(j=1;j<=p;j++)
			{
				if(!v[j]&&map[1][j]>map[flag][j])
					map[1][j]=map[flag][j];
			}
		}
		sort(ans,ans+cot,cmp);
		printf("%.2lf\n",ans[p-s-1]);
	}
}