[NOIP2017] 宝藏

/*
宝藏的子集dp做法 看起来还是很靠谱的

 需要保存的状态就只有选择的状态和已选的最大深度
 令f[S][i] 表示当前选择的生成树集合为S, 树高为i的最小花费

 那么显然F[S][i] 可以从所有的 F[T subset S][i - 1]转移而来, 我们暴力考虑将其连接上的点深度都是最深的, 这样虽然不一定是最优解, 但是总体上不会漏掉最优解


*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#include<cmath>
#define ll long long
#define M 12
#define mmp make_pair
using namespace std;
int read() {
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
const int inf = 0x3e3e3e3e;
int dis[M][M], n, m, g[1 << M], f[1 << M][M + 1], note[M];

int main() {
	n = read(), m = read();
	memset(dis, 0x3e, sizeof(dis));
	memset(f, 0x3e, sizeof(f));
	for(int i = 1; i <= m; i++) {
		int vi = read() - 1, vj = read() - 1, v = read();
		dis[vi][vj] = min(dis[vi][vj], v);
		dis[vj][vi] = min(dis[vj][vi], v);
		note[vi] |= (1 << vj);
		note[vj] |= (1 << vi);
	}
	for(int i = 0; i < n; i++) f[1 << i][0] = 0, note[i] |= (1 << i);
	for(int s = 1; s < (1 << n); s++) {
		for(int i = 0; i < n; i++) {
			if(s & (1 << i)) {
				g[s] = g[s ^ (1 << i)] | note[i];
				break;
			}
		}
		for(int t = (s - 1) & s; t; t = (t - 1) & s) {
			if((g[t] | s) == g[t]) {
				int tot = 0, ss = s ^ t;
				for(int i = 0; i < n; i++) {
					if(ss & (1 << i)) {
						int tmp = inf;
						for(int j = 0; j < n; j++) {
							if(t & (1 << j)) {
								tmp = min(tmp, dis[j][i]);
							}
						}
						tot += tmp;
					}
				}
				for(int j = 1; j < n; j++) if(f[t][j - 1] != inf) f[s][j] = min(f[s][j], f[t][j - 1] + tot * j);
			}
		}
	}
	int ans = inf;
	for(int i = 0; i < n; i++) ans = min(ans, f[(1 << n) - 1][i]);
	cout << ans << "
";
	return 0;
}