POJ Arctic Network 【最小生成树】

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 12253		Accepted: 4001

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

Source

Waterloo local 2002.09.28

直接读错题。。以为是建S个生成树。。。。

题目大意： 有一些炮台，如果这个炮台有卫星接收器，那么任意两个有卫星接收器的炮台可以通信，不受距离限制；否者，两个炮台之间只能通过对讲机通信，这是受距离限制的。要买一种对讲机，用在需要的炮台上，要求所有炮台两两之间可以直接或者间接通信，问要买通信距离至少为多少的对讲机可以满足要求。输入：S卫星接收器的数量，P炮台的数量，然后是P行，每行代表一个炮台的坐标。输出要求的对讲机的通信距离D。

因为是有S个卫星 贪心思想肯定是距离最远的S个点有这几个卫星  所以可以直接将这个几个卫星利用prim算法   然后用数组记录下  最小生成树的距离和原始距离的二者最小值。。  最后对该数组从大到小排序  然后输出第S个  即下标S-1的数组。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

bool used[510];
double cost[510][510];
double d[510];
int V;
int k;
double ans[510];
struct Point{
	double x,y;
}point[510];
bool cmp(double a,double b)
{
	return a>b;
}
void prim()
{
	
	for(int i=0;i<V;i++)
	{
		used[i]=false;
		d[i]=INF;
	}
	d[0]=0;
	double res=0;
	while(1)
	{
		int v=-1;
		for(int i=0;i<V;i++)
		{
			if(!used[i]&&(v==-1||d[v]>d[i]))v=i;
		}
		if(v==-1)break;
		used[v]=true;
		res+=d[v];
		ans[k++]=d[v];
		for(int i=0;i<V;i++)                                                                                                                                                                                                                                           
		{
			d[i]=min(d[i],cost[v][i]);
		}
	}
}
int main()
{
	int t;
	int N;
	scanf("%d",&t);
	while(t--)
	{
		k=0;
		memset(ans,0,sizeof(ans));
		scanf("%d%d",&N,&V);
		double a,b;
		for(int i=0;i<V;i++)
		{
			scanf("%lf%lf",&point[i].x,&point[i].y);
			for(int j=i;j<V;j++)
			{
				cost[i][j]=cost[j][i]=INF;
			}
		}
		for(int q=0;q<V;q++)
		{
			for(int w=q+1;w<V;w++)
			{
				double tmp1=(point[q].x-point[w].x)*(point[q].x-point[w].x);
				double tmp2=(point[q].y-point[w].y)*(point[q].y-point[w].y);
				double tmp=sqrt(tmp1+tmp2);
				cost[q][w]=cost[w][q]=tmp;
			}
		}
		prim();
		sort(ans,ans+k,cmp);
		//for(int i=0;i<k;i++)
		//printf("%lf**",ans[i]);
		printf("%.2lf\n",ans[N-1]);
	}
	return 0;
}