1160. Network
1160. Network
Time limit: 1.0 second Memory limit: 64 MB
Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network, each hub must be accessible by cables from any other hub (with possibly some intermediate hubs).
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem - not each hub can be connected to any other one because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.
Input
The first line contains two integer: N - the number of hubs in the network (2 ≤ N ≤ 1000) and M — the number of possible hub connections (1 ≤ M ≤ 15000). All hubs are numbered from 1 to N. The following M lines contain information about possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot be connected to itself. There will always be at least one way to connect all hubs.
Output
Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding cable. Separate numbers by spaces and/or line breaks.
Sample
input | output |
---|---|
4 6 1 2 1 1 3 1 1 4 2 2 3 1 3 4 1 2 4 1 |
1 4 1 2 1 3 2 3 3 4 |
Problem Author: Andrew Stankevich Problem Source: ACM ICPC 2001. Northeastern European Region, Northern Subregion
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最小生成树(好像样例错啦!!!!!!!!!!!!!)krustra
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1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 #include<stack> 7 #include<algorithm> 8 using namespace std; 9 struct edge 10 { 11 int s,t; 12 bool bo; 13 int w; 14 edge() 15 { 16 bo=false; 17 } 18 }ed[15006]; 19 int fa[1003]; 20 int n,m,s,t,I; 21 int ans[1003]; 22 int k; 23 int sum; 24 void add(int x,int y,int w) 25 { 26 ed[I].s=x; 27 ed[I].t=y; 28 ed[I].w=w; 29 ++I; 30 } 31 32 bool cmp(edge a,edge b) 33 { 34 if(a.w<b.w) 35 return true; 36 if(a.w>b.w) 37 return false; 38 if(a.w==b.w) 39 { 40 if(a.s<b.s) 41 return false; 42 if(a.s>b.s) 43 return true; 44 if(a.t<b.t) 45 return true; 46 return false; 47 } 48 } 49 void make_set() 50 { 51 for(int it=1;it<=n;it++) 52 fa[it]=it; 53 } 54 int find(int x) 55 { 56 if(fa[x]!=x) 57 return fa[x]=find(fa[x]); 58 return x; 59 } 60 bool Uon(int x,int y) 61 { 62 x=find(x); 63 y=find(y); 64 if(x==y) 65 return true; 66 else 67 { 68 fa[x]=y; 69 } 70 return false; 71 } 72 73 void kru() 74 { 75 int i,j; 76 sort(ed+1,ed+m+1,cmp); 77 for(i=1;i<=m;i++) 78 { 79 if(k==n-1)break; 80 if(!Uon(ed[i].s,ed[i].t)) 81 { 82 ed[i].bo=true; 83 sum=ed[i].w; 84 ++k; 85 ans[k]=i; 86 } 87 else 88 continue; 89 } 90 91 } 92 int main() 93 { 94 int i,j; 95 k=0;sum=0; 96 cin>>n>>m; 97 int h1,h2,we; 98 make_set(); 99 I=1; 100 for(i=1;i<=m;i++) 101 { 102 cin>>h1>>h2>>we; 103 add(h1,h2,we); 104 } 105 kru(); 106 cout<<sum<<endl<<k<<endl; 107 for(i=1;i<=k;i++) 108 { 109 cout<<ed[ans[i]].s<<' '<<ed[ans[i]].t<<endl; 110 111 } 112 return 0; 113 }