思路：看出跟dfs的顺序有关就很好写了， 对于一棵树来说确定了起点那么访问点的顺序就是dfs序，每个点经过

其度数遍，每条边经过2边， 那么我们将边的权值×2加上两端点的权值跑最小生成树，最后加上一个最小的点的

权值最为dfs的起点。

#include
     
      #define LL long long#define fi first#define se second#define mk make_pair#define pii pair
      
       #define piii pair
       
         >using namespace std;const int N = 1e4 + 10;const int M = 1e5 + 7;const int inf = 0x3f3f3f3f;const LL INF = 0x3f3f3f3f3f3f3f3f;const int mod = 1e9 + 7;const double eps = 1e-6;struct Edge {    int u, v, cost;    bool operator < (const Edge &rhs) const {        return cost < rhs.cost;    }}edge[M];int n, m, c[N], fa[N];int getRoot(int x) {    return x == fa[x] ? x : getRoot(fa[x]);}LL kruscal() {    sort(edge + 1, edge + 1 + m);    for(int i = 1; i <= n; i++) fa[i] = i;    LL ans = 0, cnt = 0;    for(int i = 1; i <= m; i++) {        int u = edge[i].u, v = edge[i].v, cost = edge[i].cost;        int x = getRoot(u), y = getRoot(v);        if(x != y) {            cnt++;            ans += cost;            fa[x] = y;            if(cnt == n - 1) break;        }    }    return ans;}int main() {    scanf("%d%d", &n, &m);    int mn = inf;    for(int i = 1; i <= n; i++) {        scanf("%d", &c[i]);        mn = min(mn, c[i]);    }    for(int i = 1; i <= m; i++) {        int u, v, cost;        scanf("%d%d%d", &u, &v, &cost);        edge[i].u = u;        edge[i].v = v;        edge[i].cost = 2 * cost + c[u] + c[v];    }    LL ans = kruscal() + mn;    printf("%lld\n", ans);    return 0;}/**/