主要思路就是通过dp数组记录状态进行剪枝，不明白为什么wa了，求条

#include<bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;

const int N = 15, inf = 200000000000;
int n, m;
int g[N][N];
bool in[N];
ll ans = inf;
int dep[N];
int dp[1 << N];
void dfs(int statue, int cnt, int sum)
{
	if(sum >= ans || dp[statue] <= sum)
		return ;
	dp[statue] = sum;
	if(cnt >= n)
	{
		ans = sum;
		return ;
	}
	for(int i = 0; i < n; i ++)//由集合里的哪个连出去
	{
		int x = i;
		if(!in[x])continue;
		for(int y = 0; y < n; y ++)
		{
			int w = g[x][y];
			if(w > inf || in[y] || y == x)continue;
			dep[y] = dep[x] + 1;
			in[y] = 1;
			dfs(statue ^ (1 << y), cnt + 1, sum + w * dep[x]);
			in[y] =0 ;
			dep[y] = 0;
			
		}
	}
}
signed main()
{
	cin >> n >> m;
	memset(g, 127, sizeof g);
	for(int i = 1; i <= m; i ++)
	{
		int u, v, w;
		cin >> u >> v >> w;
		u --;
		v --;
		g[u][v] = min(g[u][v], w);
		g[v][u] = min(g[v][u], w);
	}
	for(int i = 0; i < n; i ++)
		g[i][i] = 0;
	for(int i = 0; i < n; i ++)
	{
		memset(dp, 127, sizeof dp);
		in[i] = 1;
		dep[i] = 1;
		dfs(1 << i, 1, 0);
		dep[i] = 0;
		in[i] = 0;
	}
	cout << ans << endl;
	return 0;
	
}