POJ 2349 Arctic Network

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：

要在一群前哨站建立通讯，有卫星的前哨站之间可以无视距离通讯，其他的需要通过接收器，所有站用的接收器是一样的。要求找到可以把所有前哨站连起来的最小的接收距离

，题目给出卫星数S和站点数P，以及每个站的坐标。

思路：

这个题用prim的话还需要记录距离， 所以直接用克鲁斯卡尔算法。找P-S个边，则最后一个边即是结果。

代码：

#include<stdio.h>
#include<iostream>
#include<queue>
#include<cstring>
#include<cmath>
#include<vector>

#define INF 0x3f3f3f3f

using namespace std;

int pre[505];
int S,P;
int map[505][2];
struct D
{
	int a,b;
	double c;
	D(int A,int B,double C)
	{
		a = A;
		b = B;
		c = C;
	}
};

struct cmp
{
	bool operator()(D a,D b)
	{
		return a.c > b.c;
	}
};

int find(int a)
{
	if(pre[a] == a)return a;
	return pre[a] = find(pre[a]);
}

int judge(int a,int b)
{
	int A = find(a);
	int B = find(b);
	if(A!=B)
	{
		pre[A] = B;
		return 1;
	}
	return 0;
}

int main()
{
	int N;
	scanf("%d",&N);
	while(N--)
	{
		priority_queue<D,vector<D>,cmp>Q;
		scanf("%d %d",&S,&P);
		for(int i=1 ; i<=P ; i++)pre[i] = i;

		for(int i=1 ; i<=P ; i++)
		{
			scanf("%d %d",&map[i][0],&map[i][1]);
		}
		for(int i=1; i<=P ; i++)
		{
			for(int j=i+1 ; j<=P ; j++)
			{
				double mid = sqrt((map[i][0] - map[j][0])*(map[i][0] - map[j][0]) + (map[i][1] - map[j][1])*(map[i][1] - map[j][1]));
				Q.push(D(i,j,mid));
			}
		}
		int flag = 0;
		while(!Q.empty())
		{
			D mid = Q.top();
			Q.pop();
			if(judge(mid.a,mid.b))flag++;
			if(flag == P-S)
			{
				printf("%.2f\n",mid.c);
				break;
			}
		}
	}
	return 0;
}