URAL 1160 Network(最小生成树)

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

题意：

给定几个需要链接的点以及能够利用的边，要求用这些变将所有的点都链接起来，且所用边长的最大值尽量小。

思路：

起初，看题意是最小生成树，但题目所给的样例，并不是最小生成树的结果，纠结半天，然后又看了几遍题目，还是不知所云，最后试着把模版敲上去，submit~~

然后便ac了，

后来问了才知道，这道题确实最小生成树，不过只要所用边的最大值不变，那些权值小的边可以任意加上去，因为题目只是对最大边最小有要求，对边的数量没有要求。

#include <stdio.h>
#include <string.h>
#include <cmath>
#include <iostream>
#include <stack>
#include <queue>
#include <algorithm>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = ;
const int M = ;
int  n,m,k,s,t,tot,sum=,maxn=;
int head[N],vis[N],dis[N],father[N];
int dfn[N],low[N],stack1[N],num[N],in[N],out[N];
struct man{
    int u,v,val,used;
}edg[M];
bool cmp(man f,man g){
    return f.val<g.val;
}
int find(int x){
    if(father[x]!=x)father[x]=find(father[x]);
    return father[x];
}
void Union(int x,int y){
    x=find(x);y=find(y);
    if(x!=y)father[y]=x;
    return;
}
void kruskal(){
    for(int i=;i<=m;i++){
        int u=edg[i].u,v=edg[i].v;
        if(find(u)==find(v))continue;
        Union(u,v);
        sum++;edg[i].used=;
        maxn=edg[i].val;
        if(sum==n-)return;
    }
}
int main() {
    int u,v,val;tot=;met(dfn,);met(vis,);met(head,-);
    for(int i=;i<N;i++)father[i]=i;
    scanf("%d%d",&n,&m);
    for(int i=;i<=m;i++){
        scanf("%d%d%d",&u,&v,&val);
        edg[i].u=u;edg[i].v=v;edg[i].val=val;edg[i].used=;
    }
    sort(edg+,edg+m+,cmp);
    kruskal();
    printf("%d\n%d\n",maxn,sum);
    for(int i=;i<=m;i++){
        if(edg[i].used){
            printf("%d %d\n",edg[i].u,edg[i].v);
        }
    }
    return ;
}