Poj 2349 Arctic Network

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本文介绍了一个使用卫星和无线电建立北部地区无线网络的问题。任务是确定无线电设备最小有效距离D，以确保所有站点间的通信，考虑到有限数量的卫星通道。

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9557		Accepted: 3187

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

代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<ctime>
#include<stack>
using namespace std; 
double map[505][505];
int dx[505][2];
double ad[255005];
int n, m;

double cl(int x, int y,int k , int l )
{
	double a = x;
	double b = y;
	double c = k;
	double d = l;
	return sqrt((a-c)*(a-c)+(b-d)*(b-d));
}

int prim()
{
	int i, j, k, s = 0;
	double dis[505]={0};
	int vis[505]={0};
	dis[0] = 200000;
	vis[1] = 1;
	for(i=1;i<=m;i++)	dis[i]=map[1][i];
	for(i=1;i<m;i++)
	{
		k = 0;
		for(j=1;j<=m;j++)				
		{
			if(dis[j]<dis[k] && vis[j]==0 && dis[j])	k = j;
		}
		vis[k] = 1;
		ad[s++]=dis[k];
		for(j=1;j<=m;j++)				
		{
			if(vis[j])	continue;
			if((dis[j]>map[k][j] || dis[j]==0 ) && map[k][j] )	dis[j]=map[k][j];
		}
	}
	return s;
}

int main()
{
	int a, b, c, key;
//	freopen("in.txt","r+",stdin);/*如果in.txt不在连接后的exe的目录，需要指定路径如D:\\in.txt*/
//	freopen("out.txt","w+",stdout);/*同上*/
	scanf("%d",&key);
	while(key--)
	{
		memset(map,0,sizeof(map));
		memset(map,0,sizeof(dx));
		memset(map,0,sizeof(ad));
		scanf("%d%d",&n,&m);
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d",&a,&b);	
			dx[i][0] = a;
			dx[i][1] = b;
		}	

		for(int i=1;i<=m;i++)
		{
			for(int j=i;j<=m;j++)
			{
				map[i][j] = map[j][i] = cl(dx[i][0],dx[i][1],dx[j][0],dx[j][1]);
			}	
		}
		
//		for(int i=1;i<=m;i++)
//		{
//			for(int j=1;j<=m;j++)	printf("%lf ",map[i][j]);
//			printf("\n");
//		}
			
		prim();
		sort(ad,ad+m-1);
		printf("%.2lf\n",ad[m-n-1]);			
	}
//	fclose(stdin);
//	fclose(stdout);
	

	return 0;
}

请选择C++编译器，不要选择G++，这就是为什么你WA的主要原因

论G++与C++的区别