題意：給你n個點，m條邊，然後問你從1到n的次短路是多少（比最短路長，比其他路都短的路）

題解：看書上用dij非常複雜，然後百度到了一個不錯的題解，用spfa，先算出每個點到1的最短路dis[i]，再算出每個點到n點的最短路disr[i]

然後列舉m條邊，計算次短路dis[u]+disr[v]+edge[u][v]

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <string>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <sstream>
#include <cstdlib>
#include <iostream>
#include <algorithm>

using namespace std;
#define   MAX       5005
#define   MAXN      205
#define   lson      l,m,rt<<1
#define   rson      m+1,r,rt<<1|1
#define   lrt       rt<<1
#define   rrt       rt<<1|1
#define   mid       int m=(r+l)>>1
#define   LL        long long
#define   ull       unsigned long long
#define   mem0(x)   memset(x,0,sizeof(x))
#define   mem1(x)   memset(x,-1,sizeof(x))
#define   meminf(x) memset(x,INF,sizeof(x))
#define   lowbit(x) (x&-x)

const int    mod   = 1000000007;
const int    prime = 999983;
const int    INF   = 0x3f3f3f3f;
const int    INFF  = 1e9;
const double pi    = 3.141592653589793;
const double inf   = 1e18;
const double eps   = 1e-10;

struct Edge{
    int v,cost,next;
}edge[200005];
int head[MAX];
int dis[MAX];
int disr[MAX];
int vis[MAX];
int tot;
int n,m;

void add_edge(int a,int b,int c){
    edge[tot]=(Edge){b,c,head[a]};
    head[a]=tot++;
}

void spfa(int u,int d[]){
    mem0(vis);
    queue<int> q;
    q.push(u);
    vis[u]=1;
    d[u]=0;
    while(!q.empty()){
        u=q.front();
        q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].v;
            int cost=edge[i].cost;
            if(d[v]>d[u]+cost){
                d[v]=d[u]+cost;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}
int main(){
    scanf("%d%d",&n,&m);
    mem1(head);
    meminf(dis);
    meminf(disr);
    int tot=0;
    for(int i=0;i<m;i++){
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        add_edge(a,b,c);
        add_edge(b,a,c);
    }
    spfa(1,dis);
    spfa(n,disr);
    int mini=dis[n];
    int ans=INF;
    for(int i=1;i<=n;i++){
        for(int j=head[i];j!=-1;j=edge[j].next){
            int v=edge[j].v;
            int cost=edge[j].cost;
            if(dis[i]+disr[v]+cost>mini&&dis[i]+disr[v]+cost<ans) ans=dis[i]+disr[v]+cost;
        }
    }
    printf("%d\n",ans);
    return 0;
}