次小生成树（kruskal）

In order to prepare the \The First National ACM School Contest" (in 20??) the major of the citydecided to provide all the schools with a reliable source of power. (The major is really afraid ofblackoutsJ). So, in order to do that, power station \Future" and one school (doesn't matter which one)must be connected; in addition, some schools must be connected as well.

You may assume that a school has a reliable source of power if it's connected directly to \Future",or to any other school that has a reliable source of power. You are given the cost of connection betweensome schools. The major has decided to pick out two the cheapest connection plans { the cost of theconnection is equal to the sum of the connections between the schools. Your task is to help the major| nd the cost of the two cheapest connection plans.

Input

The Input starts with the number of test cases,T(1<T<15) on a line. ThenTtest cases follow. The rst line of every test case contains two numbers, which are separated by a space,N(3<N<100)the number of schools in the city, andMthe number of possible connections among them. NextMlines contain three numbersAi,Bi,Ci, whereCiis the cost of the connection (1<Ci<300) betweenschoolsAiandBi. The schools are numbered with integers in the range 1 toN.

Output

For every test case print only one line of output. This line should contain two numbers separated by asingle space { the cost of two the cheapest connection plans. LetS1be the cheapest cost andS2thenext cheapest cost. It's important, thatS1=S2if and only if there are two cheapest plans, otherwiseS1<S2. You can assume that it is always possible to nd the costsS1andS2.

Sample Input

2

5 8

1 3 75

3 4 51

2 4 19

3 2 95

2 5 42

5 4 31

1 2 9

3 5 66

9 14

1 2 4

1 8 8

2 8 11

3 2 8

8 9 7

8 7 1

7 9 6

9 3 2

3 4 7

3 6 4

7 6 2

4 6 14

4 5 9

5 6 10

Sample Output

110 121

37 37

---

题目要求求出最小生成树和次小生成树。

首先要熟练最小生成树的求法，将最小生成树求出，再求次小生成树。

次小生成树是在最小生成树的基础上，遍历每一个没有入队的“一个”边入队，之后将最小生成树的队伍重置，在有一个之前没有入队的边入队的情况下，再次使用求最小生成树的方法求出次小生成树。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[10001],book[10001];

struct bu
{
	int i,j,k;
}e[110*100];

bool cmp(bu x,bu y)
{
	return x.k<y.k;
}

int getf(int x)
{
	while(x!=a[x])
	{
		a[x]=a[a[x]];
		x=a[x];
	}
	return x;
}

int merge(int x,int y)
{
	int tx=getf(x);
	int ty=getf(y);
	if(tx!=ty)
	{
		a[ty]=tx;
		return 1;
	}
	return 0;
}

void kruskal(int n,int m)
{
	memset(book,0,sizeof(book));
	//memset(a,0,sizeof(a));
	for(int i=1;i<=n;i++)
		a[i]=i;
	int ans=0,count=0;
	for(int i=1;i<=m;i++)
	{
		if(merge(e[i].i,e[i].j))
		{
			book[i]=1;
			ans+=e[i].k;
			count++;
		}
		if(count==n-1)
			break;
	}
	printf("%d ",ans);//求出最小生成树
	
	ans=0;count=0;
	int minn=99999999;
	
	for(int i=1;i<=m;i++)//遍历没有入队的边
	{
		if(book[i]==0)//将之前没有入队的边入队
		{
			//memset(a,0,sizeof(a));
			for(int i=1;i<=n;i++)
				a[i]=i;

			merge(e[i].i,e[i].j);
			ans=e[i].k;
			count=1;
			
			for(int j=1;j<=m;j++)
			{
				if(merge(e[j].i,e[j].j))
				{
					ans+=e[j].k;
					count++;
				}
				if(count==n-1)
					break;
			}
			minn=min(ans,minn);//遍历中最小的数即为次小生成树
		}
		
	}
	printf("%d\n",minn);
		
}

int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m;
		
		scanf("%d%d",&n,&m);
		
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d%d",&e[i].i,&e[i].j,&e[i].k);
		}
		sort(e+1,e+m+1,cmp);
		//for(int i=1;i<=m;i++)
		//	printf("%d %d %d\n",e[i].i,e[i].j,e[i].k);
		
		kruskal(n,m);
		
	}
return 0;
 }