poj 2349 Arctic Network 最小生成树~~

浏览数：39 / 时间：2015年06月09日

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11429		Accepted: 3734

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

Sample Output

212.13

先把最小生成树的每个结点记住！然后排序即可，求出满足题意的最小值即可

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define MAX 550
#define INF 1000000000
using namespace std ;
struct Point{
	int x ,y ;
}point[MAX];
double lowCost[MAX] , graph[MAX][MAX] ;
bool visited[MAX] ;
double prim(int n)
{
	memset(visited,false,sizeof(visited)) ;
	for(int i = 0 ; i < n ; ++i)
	{
		lowCost[i] = graph[0][i] ;
	}
	lowCost[0] = INF ;
	visited[0] = false ;
	double sum = 0 ;
	for(int i = 0 ; i < n-1 ; ++i)
	{
		int index = 0 ;
		for(int j = 1 ; j < n ; ++j)
		{
			if(!visited[j] && lowCost[index]>lowCost[j])
			{
				index = j ;
			}
		}
		if(index == 0)
		{
			break ;
		}
		sum += lowCost[index] ;
		visited[index] = true ;
		for(int j = 1 ; j < n ; ++j)
		{
			if(!visited[j] && lowCost[j]>graph[index][j])
			{
				lowCost[j] = graph[index][j] ;
			}
		}
	}
	return sum ;
}
bool cmp(const double &a , const double &b)
{
	return a>b ;
}
int main()
{
	int n , s , p;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d",&s,&p) ;
		for(int i = 0 ; i < p ; ++i)
		{
			scanf("%d%d",&point[i].x,&point[i].y) ;
		}
		for(int i = 0 ;  i < p ; ++i)
		{
			for(int j = i+1 ; j < p ; ++j)
			{
				graph[i][j] = graph[j][i] = sqrt((point[i].x-point[j].x)*(point[i].x-point[j].x)*1.0+(point[i].y-point[j].y)*(point[i].y-point[j].y)*1.0) ;
			}
		}
		double sum = prim(p) ;
		sort(lowCost+1,lowCost+p,cmp) ;
		printf("%.2lf\n",lowCost[s]) ;
	}
	return 0 ;
}