思路：看出跟dfs的順序有關就很好寫了， 對於一棵樹來說確定了起點那麽訪問點的順序就是dfs序，每個點經過

其度數遍，每條邊經過2邊， 那麽我們將邊的權值×2加上兩端點的權值跑最小生成樹，最後加上一個最小的點的

權值最為dfs的起點。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> >

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;
}

/*
*/

bzoj 1232 [Usaco2008Nov]安慰奶牛cheer