标签：cee eterm follow cout location reg input bsp dep
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
Output
Sample Input
1 2 4 0 100 0 300 0 600 150 750
Sample Output
212.13
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 7 using namespace std; 8 9 int f[505]; 10 double dian[505][2]; 11 struct node 12 { 13 int u,v; 14 double w; 15 friend bool operator <(node x,node y) 16 { 17 return x.w<y.w; 18 } 19 }G[250005]; 20 21 int find(int x) 22 { 23 if(x!=f[x]) 24 return find(f[x]); 25 return x; 26 } 27 28 int main() 29 { 30 int n,m,T,S; 31 cin>>T; 32 while(T--) 33 { 34 scanf("%d%d",&S,&n); 35 memset(G,0,sizeof(G)); 36 m=0; 37 for(int i=0;i<=n;i++) 38 f[i]=i; 39 for(int i=0;i<n;i++) 40 cin>>dian[i][0]>>dian[i][1]; 41 for(int i=0;i<n;i++) 42 for(int j=i+1;j<n;j++) 43 { 44 double w1=sqrt((dian[i][0]-dian[j][0])*(dian[i][0]-dian[j][0])+(dian[i][1]-dian[j][1])*(dian[i][1]-dian[j][1])); 45 G[m].u=i; 46 G[m].v=j; 47 G[m++].w=w1; 48 } 49 sort(G,G+m); 50 double ans=0; 51 int s=S-1; 52 for(int i=0;i<m;i++) 53 { 54 int f1=find(G[i].u); 55 int f2=find(G[i].v); 56 if(f1!=f2) 57 { 58 f[f2]=f1; 59 ans=G[i].w; 60 s++; 61 // cout<<G[i].u<<" "<<G[i].v<<endl; 62 } 63 if(s==n-1) 64 break; 65 } 66 //cout<<"s "<<s<<endl; 67 printf("%.2f\n",ans); 68 } 69 70 return 0; 71 }
标签：cee eterm follow cout location reg input bsp dep
原文地址：http://www.cnblogs.com/xibeiw/p/7340565.html
