UVa 10369 Arctic Network

Problem C: Arctic Network

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.

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). 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

 

也是最小生成樹問題，與UVa 10397(http://www.cnblogs.com/lzj-0218/p/3571562.html)類似。

這個題中說，有此前哨之間能夠經過衛星設施相互通訊而無需考慮兩點之間的距離，設有x部衛星通訊設施，則能夠在用Kruskal求完最小生成樹以後，將求出的邊從大到小減去(x-1)條邊，餘下的就是須要用radio transceiver通訊的了

 

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<queue>
  4 #include<cmath>
  5 #include<vector>
  6 #define MAX_V 505
  7 
  8 using namespace std;
  9 
 10 typedef struct
 11 {
 12     int x;
 13     int y;
 14 } POINT;
 15 
 16 typedef struct
 17 {
 18     int s;
 19     int e;
 20     double dis;
 21 } CABLE;
 22 
 23 struct cmp
 24 {
 25     bool operator()(CABLE a,CABLE b)
 26     {
 27         return a.dis>b.dis;
 28     }
 29 };
 30 
 31 int V,E;
 32 int par[MAX_V];
 33 POINT p[MAX_V];
 34 double G[MAX_V][MAX_V];
 35 
 36 double Distance(POINT a,POINT b)
 37 {
 38     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 39 }
 40 
 41 void Initial()
 42 {
 43     for(int i=1;i<=V;i++)
 44         par[i]=i;
 45 }
 46 
 47 int Find(int x)
 48 {
 49     if(par[x]==x)
 50         return x;
 51     return par[x]=Find(par[x]);
 52 }
 53 
 54 bool Unite(int x,int y)
 55 {
 56     int par_x=Find(x);
 57     int par_y=Find(y);
 58     if(par_x!=par_y)
 59     {
 60         par[par_y]=par_x;
 61         return true;
 62     }
 63     return false;
 64 }
 65 
 66 int main()
 67 {
 68     int kase;
 69 
 70     scanf("%d",&kase);
 71 
 72     while(kase--)
 73     {
 74         scanf("%d %d",&E,&V);
 75         for(int i=1;i<=V;i++)
 76             scanf("%d %d",&p[i].x,&p[i].y);
 77 
 78         priority_queue<CABLE,vector<CABLE>,cmp> q;
 79 
 80         for(int i=1;i<=V;i++)
 81             for(int j=i+1;j<=V;j++)
 82             {
 83                 G[i][j]=Distance(p[i],p[j]);
 84                 CABLE temp;
 85                 temp.s=i;
 86                 temp.e=j;
 87                 temp.dis=G[i][j];
 88                 q.push(temp);
 89             }
 90 
 91         int total=0;
 92         Initial();
 93         double ans[MAX_V];
 94 
 95         while(total<V-1)
 96         {
 97             CABLE temp=q.top();
 98             q.pop();
 99             if(Unite(temp.s,temp.e))
100                 ans[total++]=temp.dis;
101         }
102 
103         total-=E;
104 
105         printf("%.2lf\n",ans[total]);
106     }
107 
108     return 0;
109 }

[C++]