Network
Description
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 of the input contains two integer numbers: 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
4 6
1 2 1
1 3 1
1 4 2
2 3 1
3 4 1
2 4 1
Sample Output
1
3
1 2
1 3
3 4
题解：原题目数据输出有问题，该题就是典型的Krusal算法，这里使用并查集判断有没有环形成。
#include <iostream>
#include<algorithm>
#include<stdio.h>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=20000;
int pre[maxn],height[maxn];
void init_set(int n){
for (int i = 1; i <= n; i++){
pre[i]=i;
}
memset(height,0,sizeof(height));
}
int find_set(int x){
return x==pre[x]?x:pre[x]=find_set(pre[x]);
}
void union_set(int x,int y){
x= find_set(x);
y= find_set(y);
if(x==y)return;
if(height[x]==height[y]){
height[x]=height[x]+1;
pre[y]=x;
}else{
if(height[x]<height[y]) pre[x]=y;
else{
pre[y]=x;
}
}
}
struct edge{
int u,v,w;
bool operator <(const edge& a)const{
return w<a.w;
}
};
int main(){
int n,m;
while(~scanf("%d%d",&n,&m)){
init_set(n);
edge* edges = new edge[n+1];
bool *record=new bool[n+1];
memset(record,false,sizeof(record));
for (int i = 1; i <=m; i++){
scanf("%d%d%d",&edges[i].u,&edges[i].v,&edges[i].w);
}
sort(edges+1,edges+m+1);
int maxcable=edges[0].w,sums=0,num=0;
int u,v;
for (int i = 1; i <=m; i++){
u=edges[i].u;
v=edges[i].v;
if(find_set(u)!=find_set(v)){
union_set(u,v);
num++;
sums+=edges[i].w;
record[i]=true;
}
if(num>=n-1){
maxcable=edges[i].w;
break;
}
}
cout<<maxcable<<endl;
cout<<sums<<endl;
for (int i = 1; i <= m; i++){
if(record[i]){
cout<<edges[i].u<<" "<<edges[i].v<<endl;
}
}
}
}
View Code
POJ1861 Kruskal算法
原文：https://www.cnblogs.com/BlairGrowing/p/14288232.html
