Arctic Network最小生成树

最新推荐文章于 2021-08-10 16:56:27 发布

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本文探讨了如何使用无线电和卫星通信技术连接北部多个前哨站的问题。具体介绍了如何通过确定最小无线电传输距离D来确保所有站点之间的有效通信。文章还提供了一个C++实现案例，用于计算使所有前哨站能够直接或间接通信所需的最小D值。

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

用c++提交是对的，别用G++，别问我为什么，因为我也不知道

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
struct path
{
    int a,b;
    double w;
}sh[500*500+50];
double x[505];
double y[505];
double ans[505];
int pre[505];
int n,m;
double dis(int i,int j)
{
    double d=sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
    return d;
}
double cmp(path a,path b)
{
    return a.w<b.w;
}
int find(int x)
{
    if(pre[x]!=x)
    pre[x]=find(pre[x]);
    return pre[x];
}
int merge(int x,int y)
{
    int xx=find(x);
    int yy=find(y);
    if(xx!=yy)
    {
        pre[xx]=yy;
        return 1;
    }
    return 0;
}
int main()
{
	int T;cin>>T;
	while(T--)
	{
		
		cin>>n>>m;
		for(int i=1;i<=m;i++)pre[i]=i;
		for(int i=1;i<=m;i++){
			cin>>x[i]>>y[i];
		}
		int cou=0;
		for(int i=1;i<=m;i++){
			for(int j=i+1;j<=m;j++)
			{
				sh[cou].a=i;
				sh[cou].b=j;
				sh[cou++].w=dis(i,j);
			}
		}
		double aa[510];int cou2=0;
		sort(sh,sh+cou,cmp);
		for(int i=0;i<cou;i++){
			if(merge(sh[i].a,sh[i].b))
			{
				ans[cou2++]=sh[i].w;
			}
		}
		printf("%.2lf\n",ans[cou2-n]);
	}
	return 0;
}