（intermediate）最小生成树 UVA 10369 - Arctic Network-优快云博客

本文解决了一个涉及使用无线网络连接北部哨所的问题。采用卫星通信和无线电通信两种方式，旨在找到确保所有哨所能相互通信所需的最小无线电通信距离D。

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Problem C: 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.

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). 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

题意：给出一些output的坐标，有S个卫星频道，装了卫星频道的output，不管距离如何都能互相通信，但是没有这个频道的范围不能超过D才能相互通信。现在要求所有的output能互相通信，要求出这个D的最小值。

思路：如果我们有S个频道，那么只要图里面有S个连通块，我们一个连通块随意放一个频道，那么他们就能互相通信了，所以我们从小到大将边加进图中，当图中的连通块刚好等于S的时候就输出最后加入的边的长度。

代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<string.h>
#include<math.h>
#include<cstring>
#include<map>
#include<algorithm>
#include<vector>
using namespace std;
#define LL long long
const int maxn = 510;
int n , k , m;
int fa[maxn];
bool vis[maxn];

int find(int x)
{
	if (x==fa[x]) return x;
	return fa[x] = find(fa[x]);
}

struct Point
{
	int x;
	int y;
}p[maxn];

inline double sqr(LL x) { return x * x; }

inline double dist(const Point & p1,const Point & p2)
{
	return sqrt(sqr(p1.x-p2.x)+sqr(p1.y-p2.y));
}

struct Edge
{
	Edge(int uu,int vv,double ww) 
		: u(uu) , v(vv) , w(ww) { }
	Edge() { }
	int u , v;
	double w;
}edge[maxn*maxn];

bool operator < (const Edge&e1,const Edge&e2)
{
	return e1.w < e2.w;
}

void input()
{
	int x , y;
	for (int i = 0 ; i < n ; ++i) scanf("%d%d",&p[i].x,&p[i].y);
	m = 0;
	for (int i = 0 ; i < n ; ++i)
	{
		for (int j = i+1 ; j < n ; ++j)
		{
			edge[m].u = i;
			edge[m].v = j;
			edge[m].w = dist(p[i],p[j]);
			++m;
		}
	}
}

void solve()
{
	int rest = n;
	sort(edge,edge+m);
	for (int i = 0 ; i < n ; ++i) fa[i] = i;
	double ans = 0;
	memset(vis,0,sizeof(vis));
	for (int i = 0 ; i < m ; ++i)
	{
		int u = find(edge[i].u);
		int v = find(edge[i].v);
		if (u==v) continue;
		--rest;
		if (rest==k) { ans = edge[i].w; break; }
		fa[u] = v;
	}
	printf("%.2lf\n",ans);
}

int main()
{
	int T; cin>>T;
	while (T--)
	{
		scanf("%d%d",&k,&n);
		input();
		solve();
	}
}