poj 1861（最小生成树）


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
4
1 2
1 3
2 3
3 4

Source

Northeastern Europe 2001, Northern Subregion

给出 n个点,m条边，输出最小生成树的最大边和边的条数，奇怪的是样例是错的，当初纠结好久(⊙o⊙)…看别人的才知道

记录生成树边得信息，输出就行，不多说，上代码

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;

#define maxe 15005

#define N 1005

int father[N];

int n,m;

struct stud{
int a,b,len;
}f[maxe];

int k,a[N],b[N];

int cmp(stud a,stud b)
{
    return a.len<b.len;
}

int cha(int x)
{
    if(x!=father[x])
        father[x]=cha(father[x]);
    return father[x];
}

int main()
{
    int i,j;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=0;i<m;i++)
            scanf("%d%d%d",&f[i].a,&f[i].b,&f[i].len);

        sort(f,f+m,cmp);

        for(i=0;i<=n;i++)
            father[i]=i;

        int num=1,ans=-maxe;
        k=0;
        for(i=0;i<m;i++)
        {
            int aa=cha(f[i].a);
            int bb=cha(f[i].b);
            if(aa!=bb)
            {
                father[aa]=bb;
                num++;
                a[k]=f[i].a;
                b[k]=f[i].b;
                k++;
                if(f[i].len>ans)
                    ans=f[i].len;

                if(num==n)
                    break;
            }
        }

        printf("%d\n%d\n",ans,k);

        for(i=0;i<k;i++)
            printf("%d %d\n",a[i],b[i]);
    }
    return 0;
}

poj 1861（最小生成树）

时间： 2024-10-18 12:35:04

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