1 /**\
2 https://ac.nowcoder.com/acm/contest/26077/1004
3 最短路dijstra,既然从a -> b -> c最长,等价求b ->a + b -> c
4 枚举中继点 然后求出最短路的和的最大值即可
5 \**/
6 #include <bits/stdc++.h>
7 using namespace std;
8
9 const int N = 1e3 + 10;
10 struct edge
11 {
12 int to, w, nextt;
13 } e[N << 1];
14 int cnt, head[N];
15
16 bool cmp(int a, int b)
17 {
18 return a > b;
19 }
20
21 void add_edge(int u, int v, int w)
22 {
23 e[++cnt].to = v;
24 e[cnt].w = w;
25 e[cnt].nextt = head[u];
26 head[u] = cnt;
27 }
28
29 struct v
30 {
31 int x, dis;
32 bool operator < (const v& rhs) const
33 {
34 return dis > rhs.dis;
35 }
36 };
37 int n, m, dis[N], vis[N];
38
39 int dijstra(int s)
40 {
41 memset(vis, 0, sizeof vis);
42 memset(dis, 0x3f, sizeof dis);
43
44 priority_queue<v> q;
45
46 dis[s] = 0;
47 q.push({s, 0});
48
49 while(!q.empty())
50 {
51 v now = q.top();
52 q.pop();
53
54 if(vis[now.x]) continue;
55 vis[now.x] = 1;
56
57 for(int i = head[now.x]; i != -1; i = e[i].nextt)
58 {
59 int y = e[i].to;
60 if(vis[e[i].to]) continue;
61 if(dis[y] > dis[now.x] + e[i].w)
62 {
63 dis[y] = dis[now.x] + e[i].w;
64 q.push({y, dis[y]});
65 }
66 }
67 }
68
69 sort(dis + 1, dis + 1 + n, cmp);
70
71 //dis[n] 就是dis[s] = 0 不用判断
72 for(int i = 1; i < n - 1; ++i)
73 {
74
75 if(dis[i] != 0x3f3f3f3f && dis[i + 1] != 0x3f3f3f3f)
76 {
77 return dis[i] + dis[i + 1];
78 }
79 }
80 return -1;
81 }
82
83 void accept()
84 {
85 cnt = 0;
86 memset(head, -1, sizeof head);
87
88 cin >> n >> m;
89 for(int x, y, z, i = 0; i < m; ++i)
90 {
91 cin >> x >> y >> z;
92 add_edge(x, y, z);
93 add_edge(y, x, z);
94 }
95
96 int ans = -1;
97 for(int i = 1; i <= n; ++i) ans = max(ans, dijstra(i));
98
99 cout << ans << "\n";
100 }
101
102 signed main()
103 {
104 ios::sync_with_stdio(false);
105 cin.tie(nullptr);
106 cout.tie(nullptr);
107
108 int _;
109 cin >> _;
110 for(int i = 0; i < _; ++i)
111 {
112 accept();
113 }
114
115 return 0;
116 }
