题目大意：

给出的案例结果得出步骤，如下图所示，从结点1开始查找，找出的一条路径如绿色部分所标注。（关键处在于连接每条路径所需要的适配器的价格得加上去）

 代码实现：

 1 #include<iostream>
 2 #include<cstdio>
 3 using  namespace std;
 4 #define MAX 1000
 5 //注意此处范围得按照题意设置为>=1000，否则会Segmentation Fault
 6 #define MAXCOST 0x7fffffff
 7 
 8 int graph[MAX][MAX];
 9 
10 int prim(int graph[][MAX], int n)
11 {
12     int lowcost[MAX];
13     int mst[MAX];
14     int i, j, min, minid, sum = 0;
15     for (i = 2; i <= n; i++)
16     {
17         lowcost[i] = graph[1][i];
18         mst[i] = 1;
19     }
20     mst[1] = 0;
21     for (i = 2; i <= n; i++)
22     {
23         min = MAXCOST;
24         minid = 0;
25         for (j = 2; j <= n; j++)
26         {
27             if (lowcost[j] < min && lowcost[j] != 0)
28             {
29                 min = lowcost[j];
30                 minid = j;
31             }
32         }
33         sum += min;
34         lowcost[minid] = 0;
35         for (j = 2; j <= n; j++)
36         {
37             if (graph[minid][j] < lowcost[j])
38             {
39                 lowcost[j] = graph[minid][j];
40                 mst[j] = minid;
41             }
42         }
43     }
44     return sum;
45 }
46 
47 int main()
48 {
49     int i, j, k, t, en, n,x, y, cost,pre[1001];
50     scanf("%d",&t);
51     while(t--){
52         scanf("%d",&n);
53         for(i=1;i<=n;i++)
54             scanf("%d",&pre[i]);
55         for (i = 1; i <= n; i++)
56         {
57             for (j = 1; j <= n; j++)
58             {
59                 graph[i][j] = MAXCOST;
60             }
61         }
62         //构建图G
63         for(i = 1; i <= n; i++){
64             for(k=1;k<=n;k++){
65                 scanf("%d",&graph[i][k]);
66                 graph[i][k]+=pre[i];//直接将每条路径上存在的适配器价格给加到权值里面去即可
67                 graph[i][k]+=pre[k];
68             }
69         }
70         cost = prim(graph, n);
71         cout <<cost << endl;
72     }
73     return 0;
74 }