poj 2349 Arctic Network 【最小生成树-Kruskal】

Arctic Network

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 18709 Accepted: 5918

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

12 40 1000 3000 600150 750

Sample Output

212.13

AC:

#include<stdio.h>  #include<math.h>  #include<algorithm>  using namespace std;  int father[550], m, k;  double d[550];  struct post  {      double x, y;  }p[550];  struct edge  {      int u, v;      double w;  }e[500005];  bool comp(edge e1, edge e2)  {      return  e1.w < e2.w;  }  double get_dis(double x1, double y1, double x2, double y2)  {      return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));  }  void Init(int n)  {      for(int i = 1; i <= n; i++)          father[i] = i;  }  int Find(int x)  {      if(x != father[x])          father[x] = Find(father[x]);      return father[x];  }  void Merge(int a, int b)  {      int p = Find(a);      int q = Find(b);      if(p > q)          father[p] = q;      else          father[q] = p;  }  void Kruskal(int n)  {      k = 0;      double Max = 0;      for(int i = 0; i < m; i++)          if(Find(e[i].u) != Find(e[i].v))          {              Merge(e[i].u, e[i].v);              d[k++] = e[i].w;              n--;              if(n == 1) //当所有顶点都遍历完时，退出该函数                return;          }  }  int main()  {      int t, S, P, i, j;      double x, y;      scanf("%d",&t);      while(t--)      {          m = 0;          scanf("%d%d",&S,&P);          Init(P);          for(i = 1; i <= P; i++)              scanf("%lf%lf",&p[i].x, &p[i].y);          for(i = 1; i <= P; i++)              for(j = i + 1; j <= P; j++)              {                  e[m].u = i;                  e[m].v = j;                  e[m++].w = get_dis(p[i].x, p[i].y, p[j].x, p[j].y);                  e[m].u = j;                  e[m].v = i;                  e[m++].w = get_dis(p[i].x, p[i].y, p[j].x, p[j].y);              }          sort(e, e+m, comp);  //以边上的权值大小进行排序        Kruskal(P);          printf("%.2lf\n",d[P-S-1]);      }      return 0;  }