At the beginning, most people think that this is a minimum spanning tree problem, which can be solved by prim. However, in fact, due to the limitation of multiple and depth, this algorithm is not necessarily correct. So what should I do? We just need to make a judgment before finding the smallest point, and use random numbers to get modulus, so as to ensure that there is a very small probability to get the next small, smaller probability to get the next small Point, this process repeats 1000 times (almost), basically can pass.

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<iomanip>
#include<queue>
#include<cmath>
#include<ctime>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=20;
const int maxm=1010;
int e[maxn][maxn];
int dis[maxn],depth[maxn];
bool b[maxn],inq[maxn];
struct H
{
	int id;
};
int n,m,ans;
priority_queue<H> q;
queue<H> p;
bool operator < (const H &a,const H &b)
{
	return dis[a.id]>dis[b.id];
}
int f(int x)
{
	memset(dis,INF,sizeof(dis));
	memset(b,false,sizeof(b));
	memset(inq,false,sizeof(inq));
	memset(depth,0,sizeof(depth));
	while(!q.empty()) q.pop();
	while(!p.empty()) p.pop();
	
	dis[x]=0;
	depth[x]=1;
	q.push((H){x});
	inq[x]=true;
	int tot=0;
	for(int i=1;i<=n;i++)
	{
		H now=q.top();
		q.pop();
		while(!q.empty()&&(rand()%10==0))
		{
			p.push(now);
			now=q.top();
			q.pop();
		}
		while(!p.empty()) 
		{
			q.push(p.front());
			p.pop();
		}
		int tmp=now.id;
		b[tmp]=true;
		tot+=dis[tmp];
		if(tot>ans) return INF;
		for(int j=1;j<=n;j++)
		{
			if(!b[j]&&e[tmp][j]!=INF&&dis[j]>depth[tmp]*e[tmp][j])
			{
				dis[j]=depth[tmp]*e[tmp][j];
				depth[j]=depth[tmp]+1;
				if(!inq[j])
				{
					inq[j]=true;
					q.push((H){j});	
				}			
			}
		}
	}
	return tot;
}
int main()
{
	scanf("%d %d",&n,&m);
	memset(e,INF,sizeof(e));
	for(int i=1;i<=m;i++)
	{
		int x,y,z;
		scanf("%d %d %d",&x,&y,&z);
		if(z<e[x][y])
		{
			e[x][y]=z;
			e[y][x]=z;
		}
	}
	ans=INF;
	srand(1919);
	
	for(int i=1;i<=1000;i++)
	{
		for(int j=1;j<=n;j++)
		{
			ans=min(ans,f(j));
		}
	}
	
	printf("%d",ans);
	return 0;
}