UvaOJ10369 - Arctic Network

 /*
     The first line of each test case contains 1 <= S <= 100, the number of satellite channels!
     注意：S表示一共有多少个卫星，那么就是有 最多有S-1个通道！ 然后将最小生成树中的后边的 S-1通道去掉就行了！
     思路：最小生成树中的第 k 个最小边！
 */
 //克鲁斯克尔算法.....
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;

 double x[], y[];

 struct node{
    int u, v;
    double d;
 };

 bool cmp(node a, node b){
     return a.d < b.d;
 }

 int f[];

 node nd[];
 double ret[];

 int getFather(int x){
    return x==f[x] ? x : f[x]=getFather(f[x]);
 }

 bool Union(int a, int b){
    int fa=getFather(a), fb=getFather(b);
    if(fa!=fb){
        f[fa]=fb;
        return true;
    }
    return false;
 }

 int main(){
    int n, m;
    int t;
    scanf("%d", &t);
    while(t--){
           scanf("%d%d", &m, &n);
        for(int i=; i<=n; ++i){
           scanf("%lf%lf", &x[i], &y[i]);
           f[i]=i;
        }
        int cnt=;
        for(int i=; i<n; ++i)
           for(int j=i+; j<=n; ++j){
               nd[cnt].u=i;
               nd[cnt].v=j;
               nd[cnt++].d=sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));
           }
        sort(nd, nd+cnt, cmp);
        int cc=;
        for(int i=; i<cnt; ++i)
           if(Union(nd[i].u, nd[i].v))
              ret[cc++]=nd[i].d;
         for(int i=; i<cc; ++i)
            cout<<ret[i]<<"fdsf"<<endl;
        printf("%.2lf\n", ret[n-m-]);
    }
    return ;
 }   

 //prim算法.......
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;
 const double INF = 0x3f3f3f3f*1.0;
 double x[], y[];

 int n, m;
 double map[][];
 int vis[];

 double ret[];

 void prim(){
     memset(vis, , sizeof(vis));
     vis[]=;
     for(int i=; i<=n; ++i)
        ret[i]=INF;
     int root=, p;
     for(int i=; i<n; ++i){
         double minLen=INF;
         for(int j=; j<=n; ++j){
            if(!vis[j] && ret[j]>map[root][j])
               ret[j]=map[root][j];
            if(!vis[j] && minLen>ret[j]){
               minLen=ret[j];
               p=j;
            }
         }
         root=p;
         vis[root]=;
     }
 }

 int main(){

    int t;
    scanf("%d", &t);
    while(t--){
           scanf("%d%d", &m, &n);
        for(int i=; i<=n; ++i)
           scanf("%lf%lf", &x[i], &y[i]);
        for(int i=; i<n; ++i)
           for(int j=i+; j<=n; ++j)
              map[i][j]=map[j][i]=sqrt( (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]));

        prim();
        sort(ret, ret+n+);

        printf("%.2lf\n", ret[n-m+]);
    }
    return ;
 }