poj2349Arctic Network(最小生成树kruscal+第k长的边）

题目链接：

啊哈哈，点我点我

题意：
北极的某区域共有n座村庄( 1 ≤ n ≤ 500 )，每座村庄的坐标用一对整数(x, y)表示，其中 0 ≤ x, y ≤ 10000。为了加强联系，决定在村庄之间建立通讯网络。通讯工具可以是无线电收发机，也可以是卫星设备。所有的村庄都可以拥有一部无线电收发机， 且所有的无线电收发机型号相同。但卫星设备数量有限，只能给一部分村庄配备卫星设备。  不同型号的无线电收发机有一个不同的参数d，两座村庄之间的距离如果不超过d就可以用该型号的无线电收发机直接通讯，d值越大的型号价格越贵。拥有卫星设备的两座村庄无论相距多远都可以直接通讯。

这道题目主要是那卫星不好处理，但是如果正面不好想，那么可以从逆向来想，如果知道了最小距离D，那么去掉最小生成树里面大于D的边，那么图会存在k个连通支，这里给一个定理：如果去掉所有权大于d的边后最小生成树被分割成k个连通分支，图也被分割成k个连通分支，。那么给每个连通支分一个卫星电话即可，那么这个最短距离D就变成了去掉大于D后的最大的边，即第k长的边。。。。那么这个问题就解决了。。。

题目：

Arctic Network

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10211		Accepted: 3374

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

Source

Waterloo local 2002.09.28

代码：

#include
#include
#include
#include
#include
using namespace std;

const int maxn=500+10;
int s,p,root[maxn],cal;
double dis[maxn*maxn];

struct Point
{
    int x,y;
}point[maxn];

struct Edge
{
    int u,v;
    double distance;
}edge[maxn*maxn];

int findroot(int x)
{
    if(root[x]!=x)
        root[x]=findroot(root[x]);
    return root[x];
}

double caldistance(Point a,Point b)
{
    return sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y));
}


bool cmp(Edge a,Edge b)
{
    return a.distanceb;
}

void kruscal()
{
    memset(dis,0.0,sizeof(dis));
    int sum=0,i,k=0;
    for(i=1;i<=cal;i++)
    {
        int fx=findroot(edge[i].u);
        int fy=findroot(edge[i].v);
        if(fx==fy)
            continue;
        else
        {
            root[fx]=fy;
            sum++;
            dis[sum]=edge[i].distance;
            if(sum==p-1)
                break;
        }
    }
    sort(dis+1,dis+1+sum,cmp2);
    printf("%.2f\n",dis[s]);
}


void read_graph()
{
    for(int i=1;i<=p;i++)
        root[i]=i;
    for(int i=1;i<=p;i++)
       scanf("%d%d",&point[i].x,&point[i].y);
    for(int i=1;i