poj-2377模板题，最大生成树

版权声明：那个，最起码帮我加点人气吧，署个名总行吧 https://blog.csdn.net/qq_41670466/article/details/82720952

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：就是把最小生成树的那个sort排序给反过来就好了，那么既然这么简单，为什么还要写博客类？，当然是先给自己的分类+1，要不然一直是0多尴尬。不过还要注意一个细节，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;
}