#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define MAXN (5000 + 10)
#define INF (5000*5000*2)
using namespace std;

struct edge{
int to, cost;
edge(int tv = 0, int tc = 0):
to(tv), cost(tc){}
};
typedef pair<int ,int> P;
int N, R;
vector<edge> graph[MAXN];
int dist[MAXN]; //最短距離
int dist2[MAXN]; //次短距離

void solve(){
fill(dist, dist+N, INF);
fill(dist2, dist2+N, INF);
//從小到大的優先隊列
//使用pair而不用edge結構體
//是因為這樣我們不需要重載運算符
//pair是以first為主關鍵字進行排序
priority_queue<P, vector<P>, greater<P> > Q;
//初始化源點信息
dist[0] = 0;
Q.push(P(0, 0));
//同時求解最短路和次短路
while(!Q.empty()){
P p = Q.top(); Q.pop();
//first為s->to的距離，second為edge結構體的to
int v = p.second, d = p.first;
//當取出的值不是當前最短距離或次短距離，就舍棄他
if(dist2[v] < d) continue;
for(unsigned i = 0; i < graph[v].size(); i++){
edge &e = graph[v][i];
int d2 = d + e.cost;
、
if(dist[e.to] > d2){
swap(dist[e.to], d2);
Q.push(P(dist[e.to], e.to));
}
if(dist2[e.to] > d2 && dist[v] < d2){
dist2[e.to] = d2;
Q.push(P(dist2[e.to], e.to));

}
}
}
printf("%d\n", dist2[N-1]);
}

int main(){
int A, B, D;
scanf("%d%d", &N, &R);
for(int i = 0; i < R; i++){
scanf("%d%d%d", &A, &B, &D);
graph[A-1].push_back(edge(B-1, D));
graph[B-1].push_back(edge(A-1, D));
}
solve();
return 0;
}

次短路經（dijsktra）