更新時變為3種情況：
1.最短路更新最短路；
2.最短路不能更新最短路但能更新次短路；
3.次短路更新次短路。

思考一下，就顯然了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <queue>
using namespace std;

#define MAXN (100100)
int V, E, tot = 0;
int first[MAXN], nxt[MAXN<<1], dis1[MAXN], dis2[MAXN];
bool
 
 used[MAXN];
struct edge{
    int from, to, cost;
}es[MAXN<<1];

void build(int ff, int tt, int dd)
{
    es[++tot] = (edge){ff,tt,dd};
    nxt[tot] = first[ff];
    first[ff] = tot;
}

#define INF (1e9)
queue <int> q;
void sub_spfa(int s)
{
    fill(dis1+1,dis1+V+1,INF); fill(dis2+1,dis2+V+1
 
,INF);
    dis1[s] = 0;
    q.push(s);
    used[s] = 1;
    while(q.size())
    {
        int x = q.front();
        q.pop();
        used[x] = 0;
        for(int i = first[x]; i != -1; i = nxt[i])
        {
            int v = es[i].to;
            if(dis1[v] > dis1[x] + es[i].cost)
            {
                dis2[v] = dis1[v];
                dis1[v] = dis1[x] + es[i].cost;
                if
 
(!used[v])
                {
                    q.push(v);
                    used[v] = 1;
                }
            }
            else if((dis1[v]<dis1[x]+es[i].cost) && (dis2[v]>dis1[x]+es[i].cost))
            {
                dis2[v] = dis1[x] + es[i].cost;
                if(!used[v])
                {
                    q.push(v);
                    used[v] = 1;
                }
            }
            else if(dis2[v] > dis2[x] + es[i].cost)
            {
                dis2[v] = dis2[x] + es[i].cost;
                if(!used[v])
                {
                    q.push(v);
                    used[v] = 1;
                }
            } 
        }
    }
}

int main()
{
    cin >> V >> E;
    memset(first,-1,sizeof(first));
    for(int i = 1; i <= E; ++ i)
    {
        int f,t,d;
        scanf("%d%d%d", &f, &t, &d);
        build(f,t,d);
    }
    sub_spfa(1);
    if(dis2[V] < INF) cout << dis2[V] << '\n';
    else puts("-1");
    return 0;
}