POJ 1861 Network【最小生成樹kruskal，板子題】

題目鏈接

Network

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 18694 Accepted: 7540 Special Judge

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

思路：

首先說明一下，題目給的測試樣例結果錯了，我在上面改了一下；
板子題，輸入網絡，輸入最小生成樹中最大的邊、邊的總對數(必然是n-
1)、具體的邊(從小到大輸出)

kruskal代碼

#include <iostream>
#include <algorithm>
#include <memory.h>
using namespace std;

struct edge{
    int from;
    int to;
    int value;
}a[15005];

int n,m;
int f[1005];
int mn;
int way[1005];

bool cmp(edge a,edge b){
    return a.value<b.value;
}

int getf(int x){//並查集一部分，查找老大
    if(f[x]==x){
        return x;
    }
    else{
        f[x]=getf(f[x]);
        return f[x];
    }
}

void mrge(int x,int y){//並查集一部分，合併，路徑壓縮
    int t1=getf(x);
    int t2=getf(y);
    if(t1<t2){//這樣寫是爲了更嚴謹，雖然一般輸入都是先小後大
        f[t2]=t1;
    }
    if(t2<t1){
        f[t1]=t2;
    }
}

void kruskal(){
    mn=0x3f3f3f3f;
    int k=0;
    for(int i=1;i<=n;i++)
        f[i]=i;
    sort(a+1,a+m+1,cmp);
    int sum=0;
    int cnt=0;
    for(int i=1;i<=m;i++){
        int t1=getf(a[i].from);
        int t2=getf(a[i].to);
        if(t1!=t2){
            mrge(t1,t2);
            way[++k]=i;//記錄路徑
            sum+=a[i].value;
            cnt++;
        }
        if(cnt==n-1){
            cout<<a[i].value<<endl;//最小中的最大
            cout<<n-1<<endl;//總對數
            for(int i=1;i<=n-1;i++){//輸出路徑
                int idx=way[i];
                cout<<a[idx].from<<" "<<a[idx].to<<endl;
            }
            return;//退出
        }
    }
}

int main()
{
    while(cin>>n>>m){
        for(int i=1;i<=m;i++){
            cin>>a[i].from>>a[i].to>>a[i].value;
        }
        kruskal();
    }
    return 0;
}