POJ2349 Arctic Network（kruskal+并查集）

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

其实这就是一道最小生成树的题，如果没有卫星的作用，而答案就是最大的两个哨所的距离，但这里还有s个卫星，其作用就是能使哨所之间的距离相当于0，所以当有s个卫星时，就是可以减少s-1条边。所以，可以用kruskal算法，先将任意两个哨所的距离求出来，然后排序，这样只需要求出第p-s条边就行了。

#include
#define ll long long
#define INF 0x3f3f3f3f
#define qx std::ios::sync_with_stdio(false)
using namespace std;

int s,p,q[510];
struct Node{
    int x,y;
}no[510];

struct No{
    int x,y;double d;
}n[510*505];

double len(Node a,Node b){
    double temp=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
    return sqrt(temp);
}
int cmp(No i,No j){return i.d>t;
    while(t--){
        cin>>s>>p;
        for(int i=0;i>no[i].x>>no[i].y;
        for(int i=0;i<=p;i++)q[i]=i;
        int k=0;
        for(int i=0;iy)
                q[x]=y;else q[y]=x;
                ans=n[i].d;
            }
        }
        printf ("%.2f\n",ans);//一定是.2f,2.lfwa了无数次才发现的
    }
    return 0;
}