題意

T組測試資料，每組先輸入n，m表示有一個n個點m條邊的無向圖，然後輸入n個點的點權，然後輸入m條邊（u，v，w），求一棵以1為根節點的數使得該樹花費最小，輸出最小花費，如果無法構建成一棵樹輸出“No Answer”。
樹的花費為：$\sum$（邊權*子樹的點權和）

思路

通過推樣例可以發現每個點的點權需要乘以這個點到根結點的所有邊的邊權，所以可以得到
$\sum$邊權*子樹的點權和 = $\sum$該點根節點的路徑*點權
所以直接跑一遍最短路統計答案即可。
注意開long long。

//#include<bits/stdc++.h>
#include <iostream>
 

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN = 1e5+5;
typedef long long LL;
const LL INF = 0x3f3f3f3f3f3f3f3f;

struct Edge
{
    int from, to; LL dist;       //起點,終點,距離
    Edge(int from, int to, LL dist):from(from), to(to), dist(dist) {}
 

};

struct Dijkstra
{
    int n, m;                 //結點數,邊數(包括反向弧)
    vector<Edge> edges;       //邊表。edges[e]和edges[e^1]互為反向弧
    vector<int> G[MAXN];      //鄰接表，G[i][j]表示結點i的第j條邊在edges陣列中的序號
    int vis[MAXN];            //標記陣列
    LL d[MAXN];              //s到各個點的最短路
    int p[MAXN];              //上一條弧
 


    void init(int n)
    {
        this->n = n;
        edges.clear();
        for (int i = 0; i <= n; i++) G[i].clear();
    }

    void AddEdge(int from, int to, int dist)
    {
        edges.push_back(Edge(from, to, dist));
        m = edges.size();
        G[from].push_back(m - 1);
    }

    struct HeapNode
    {
        int from; LL dist;
        bool operator < (const HeapNode& rhs) const
        {
            return rhs.dist < dist;
        }
        HeapNode(int u, LL w): from(u), dist(w) {}
    };

    void dijkstra(int s)
    {
        priority_queue<HeapNode> Q;
        for (int i = 0; i <= n; i++) d[i] = INF;
        memset(vis, 0, sizeof(vis));
        d[s] = 0;
        Q.push(HeapNode(s, 0));
        while (!Q.empty())
        {
            HeapNode x = Q.top(); Q.pop();
            int u = x.from;
            if (vis[u]) continue;
            vis[u] = true;
            for (int i = 0; i < G[u].size(); i++)
            {
                Edge& e = edges[G[u][i]];
                if (d[e.to] > d[u] + e.dist)
                {
                    d[e.to] = d[u] + e.dist;
                    p[e.to] = G[u][i];
                    Q.push(HeapNode(e.to, d[e.to]));
                }
            }
        }
    }
}gao;

int a[MAXN];
int main()
{
    int T; scanf("%d", &T);
    while (T--)
    {
        int n, m; scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        gao.init(n);
        while (m--)
        {
            int u, v, w; scanf("%d%d%d", &u, &v, &w);
            gao.AddEdge(u, v, w);
            gao.AddEdge(v, u, w);
        }
        gao.dijkstra(1);
        LL ans = 0;
        for (int i = 1; i <= n; i++)
        {
            if (gao.d[i] == INF) { ans = -1; break; }
            ans += a[i]*gao.d[i];
        }
        if (ans != -1) printf("%lld\n", ans);
        else printf("No Answer\n");
    }
    return 0;
}
/*
2
2 1
1 1
1 2 15
7 7
200 10 20 30 40 50 60
1 2 1
2 3 3
2 4 2
3 5 4
3 7 2
3 6 3
1 5 9
*/