1 #define _CRT_SECURE_NO_WARNINGS
2 /*
3 7 10
4 0 1 5
5 0 2 2
6 1 2 4
7 1 3 2
8 2 3 6
9 2 4 10
10 3 5 1
11 4 5 3
12 4 6 5
13 5 6 9
14
15 4 4
16 0 1 100
17 1 3 200
18 1 2 250
19 2 3 100
20 */
21 #include <iostream>
22 #include <functional>
23 #include <utility>
24 #include <queue>
25 #include <vector>
26 using namespace std;
27
28 const int maxn = 1010 + 20;
29 const int INF = 9999999;
30 typedef pair<int, int> P; //first-最短距离，second-顶点编号
31 struct edge
32 {
33 int to, cost;
34 };
35
36 //输入
37 int N, R; //N-点，R-边
38 vector<edge> G[maxn]; //图的邻接表表示
39
40 int dist[maxn]; //最短距离
41 int dist2[maxn]; //次短距离
42
43 void solve()
44 {
45 priority_queue<P, vector<P>, greater<P> > que;
46 fill(dist, dist + N, INF);
47 fill(dist2, dist2 + N, INF);
48 dist[0] = 0;
49 que.push(P(0, 0));
50
51 while (!que.empty())
52 {
53 P p = que.top(); que.pop(); //队列存放 （最短路和次短路）
54 int v = p.second, d = p.first;
55 if (dist2[v] < d) continue; //如果次小的值小，跳过该值
56 for (int i = 0; i < G[v].size(); i++) {
57 edge &e = G[v][i]; //得到v的临边
58 int d2 = d + e.cost; //d2 = 已知最短或次短 + 临边权值
59 if (dist[e.to] > d2) { //如果未确定的最短路 > d2
60 swap(dist[e.to], d2);
61 que.push(P(dist[e.to], e.to)); //添加最短路
62 }
63 if (dist2[e.to] > d2 && dist[e.to] < d2) {
64 dist2[e.to] = d2; //说明d2是次大值
65 que.push(P(dist2[e.to], e.to)); //添加次短路
66 }
67 }
68 }
69 printf("%d
", dist2[N - 1]); //N位置的dist2即为
70 }
71
72 void input()
73 {
74 int from, to, cost;
75 edge tmp;
76 for (int i = 0; i < R; i++) {
77 cin >> from >> to >> cost;
78 tmp.to = to; tmp.cost = cost;
79 G[from].push_back(tmp);
80 tmp.to = from;
81 G[to].push_back(tmp);
82 }
83 }
84
85 int main()
86 {
87 cin >> N >> R;
88 input();
89 solve();
90 return 0;
91 }