Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie. Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a "tree".

思路：就是把最小生成樹的那個sort排序給反過來就好了，那麼既然這麼簡單，為什麼還要寫部落格類？，當然是先給自己的分類+1，要不然一直是0多尷尬。不過還要注意一個細節，累加和要用long long

#include<cstdio>
#include<iostream>
#include<algorithm>

using namespace std;

const int maxn=2e4+10;

struct edge
{
	int from,val,to;
}edg[maxn];
int pre[maxn];
int n,m;

void ini()
{
	for(int i=0;i<=n;i++)
		pre[i]=i;
}

int find(int x)
{
	if(pre[x]==x)
		return x;
	return pre[x]=find(pre[x]);
}

void join(int x,int y )
{
	int fx=find(x),fy=find(y);
	pre[fx]=fy;
}

bool cmp(edge a,edge b)
{
	return a.val>b.val;
}

long long  solve()
{
	int eg=0;
	long long res=0;
	sort(edg+1,edg+1+m,cmp);
	for(int i=1;i<=m;i++)
	{
		if(find(edg[i].from)!=find(edg[i].to))
		{
			join(edg[i].from,edg[i].to);
			res+=edg[i].val;
			eg++;
		}
	}
	if(eg<n-1)
		return res=-1;
	return res;
}

int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d %d %d",&edg[i].from,&edg[i].to,&edg[i].val);
	}
	ini();
	long long res=solve();
	cout<<res<<endl;
	return 0;
}