本題是個次短路板子題，因為我太弱了所以我不會K短路。
考慮從1點跑一邊最短路，然後從N點再跑個最短路。之後遍歷這個圖，然後更新答案。
Code：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define ll long long
#define INF 0x3f3f3f3f
#define maxn 100001
using namespace std;
struct Edge{
    int nxt,to,dis;
}edge[maxn*2];
int
 
 head[maxn],num_edge,dis[maxn],vis[maxn],n,r,Dis[maxn];
inline void addedge(int from,int to,int dis){
    edge[++num_edge].nxt=head[from];
    edge[num_edge].to=to;
    edge[num_edge].dis=dis;
    head[from]=num_edge;
}
inline void spfa(int s){
    memset(vis,0,sizeof(vis));
    memset(dis,INF,sizeof
 
(dis));
    queue<int> q;
    q.push(s);
    vis[s]=1;
    dis[s]=0;
    while(!q.empty()){
        int x=q.front();
        q.pop();
        vis[x]=0;
        for(int i=head[x];i;i=edge[i].nxt){
            int y=edge[i].to;
            if(dis[y]>dis[x]+edge[i].dis){
                dis[y]=dis[x]+edge[i].dis;
                
 
if(vis[y]==0){
                    vis[y]=1;
                    q.push(y); 
                } 
            }
        }
    }
}
inline void spfa_(int s){
    memset(vis,0,sizeof(vis));
    memset(Dis,INF,sizeof(Dis));
    queue<int> q;
    q.push(s);
    vis[s]=1;
    Dis[s]=0;
    while(!q.empty()){
        int x=q.front();
        q.pop();
        vis[x]=0;
        for(int i=head[x];i;i=edge[i].nxt){
            int y=edge[i].to;
            if(Dis[y]>Dis[x]+edge[i].dis){
                Dis[y]=Dis[x]+edge[i].dis;
                if(vis[y]==0){
                    vis[y]=1;
                    q.push(y); 
                } 
            }
        }
    }
}
int main(){
    cin>>n>>r;
    for(int i=1;i<=r;i++){
        int u,v,w;
        cin>>u>>v>>w;
        addedge(u,v,w);
        addedge(v,u,w);
    }
    spfa(1);
    spfa_(n);
    int minn=dis[n];
    int ans=INF;
    for(int i=1;i<=n;i++){
        for(int j=head[i];j;j=edge[j].nxt){
            int v=edge[j].to,w=edge[j].dis;
            if(dis[i]+Dis[v]+w>minn&&dis[i]+Dis[v]+w<ans){
                ans=dis[i]+Dis[v]+w;
            }
        }
    }
    cout<<ans;
    return 0;
}