我们看到最小生成树和输出该树的总权重以及所选边的索引序列，最合适的自然就是$Kruskal$算法，它可以模拟整颗树的构造过程，最适配这种题

这道题有点板，就不用过多解释了，但我就不用普通写法......

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+10;
struct node{int u,v,w,id;}G[N];
int n,m,fa[N];
long long ans;
vector<int>s;
bool cmp(node a,node b){return a.w<b.w;}
int find(int x){return (fa[x]==x)?x:fa[x]=find(fa[x]);}
bool pd(int x,int y)
{
	int rx=find(x),ry=find(y);
	if(rx==ry)return 0;
	fa[ry]=rx;
	return 1;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1,x,y,z;i<=m;i++)
	{
		scanf("%d%d%d",&x,&y,&z);
		G[i]={x,y,z,i-1};
	}
	sort(G+1,G+m+1,cmp);//保证贪心，优先小边权 
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<=m&&s.size()<n-1;i++)
	{
		if(pd(G[i].u,G[i].v))
		{
			ans+=G[i].w;
			s.push_back(G[i].id);
		}
	}
	printf("%lld\n",ans);
	for(int i=0;i<n-1;i++)printf("%d ",s[i]);
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N = 5e5 + 10;
#define int long long
typedef tuple<int, int, int, int> tp4;
priority_queue<tp4, vector<tp4>, greater<tp4> > edge;
int fa[N], n, m;
int findfa(int x){return (fa[x] == x ? fa[x] : fa[x] = findfa(fa[x]));}
bool merge(int x, int y)
{
    int xfa = findfa(x), yfa = findfa(y);
    if (xfa == yfa) return 0;
    fa[xfa] = yfa;
    return 1;
}
signed main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++) fa[i] = i;
    for (int i = 1; i <= m; i++)
    {
        int u, v, w; cin >> u >> v >> w;
        u ++; v ++; edge.push({w, u, v, i});
    }
    int t = n - 1, sum = 0;
    vector<int> vec;
    while (t)
    {
        auto [w, u, v, i] = edge.top(); edge.pop();
        if (merge(u, v)) t --, sum += w, vec.push_back(i);
    }
    cout << sum << endl;
    for (auto i : vec) cout << i - 1 << " ";
    return 0;
}