2349 poj && 10369 uva Arctic Network

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13236		Accepted: 4295

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

1
2 4
0 100
0 300
0 600
150 750

Sample Output

212.13

题意：

给出所有的位置的坐标，，有两种通信方式，用卫星的话，无视距离，但是卫星的数量有限，另外就是直接传送消息，现给出所有的坐标和卫星的数量，让你求出，满足修建道路最短的情况下，需要修建的最长的道路长度...就这样理解吧...

题解：

按这个思路，也就是求最小生成树，但是需要求出的是最小生成树的第s+1大的边的长度.....

只要理解了题意，这个题就不难，下面完全就是克鲁斯卡尔算法的模板，不过注意处理方式...

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int per[505],n,s,a[505],b[505];
struct lu
{
	int a,b;
	double len;
}x[250005];
double dis(int i,int j)
{
	return sqrt((a[i]-a[j])*(a[i]-a[j])*1.0+(b[i]-b[j])*(b[i]-b[j]));
}
void init()
{
	for(int i=1;i<=n;++i)
	{
		per[i]=i;
	}
}
bool cmp(lu a,lu b)
{
	return a.len<b.len;
}
int find(int x)
{
	int r=x;
	while(r!=per[r])
	{
		r=per[r];
	}
	int i=x,j;
	while(i!=r)
	{
		j=per[i];per[i]=r;i=j;
	}
	return r;
}
int join(int x,int y)
{
	int fx=find(x),fy=find(y);
	if(fx!=fy)
	{
		per[fy]=fx;
		return 1;
	}
	return 0;
}
void kruskal()
{
	int cnt=0;double maxn=0;
	for(int i=0;cnt<n;++i)
	{
		if(join(x[i].a,x[i].b))
		{
			++cnt;//累加生成树的边数
			if(cnt==n-s)//找到需要的那条边之后
			{
				maxn=x[i].len;
				break;
			}
		}
	}
	printf("%.2lf\n",maxn);//对应输出
}
int main()
{
	int t,i,j;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&s,&n);
		init();
		for(i=0;i<n;++i)
		{
			scanf("%d%d",a+i,b+i);
		}
		int c=0;
		for(i=0;i<n-1;++i)
		{
			for(j=i+1;j<n;++j)
			{
				x[c].a=i+1;x[c].b=j+1;
				x[c].len=dis(i,j);
				++c;
			}
		}
		sort(x,x+c,cmp);
		kruskal();
	}
	return 0;
}