題目意思就是有n個站點，要從s前往e，有經濟路線和商務路線，只能坐一次商務路線，求花費的最優解。

只要列舉選擇哪一條商務路線就好了，所以兩遍Dijkstra，一遍是從起點開始，一遍是從終點開始。如果需要使用由a站到b站的商務車票，則總開銷就是起點到a的開銷+商務路線開銷+b到終點的開銷。經過的站點就注意哪裡需要倒序輸出就行。格式卡了好幾次。。每組資料後要換行，行末不能有多餘空格。

要去青島了，也沒個底。。最近也沒寫幾道題，唉

/*
* @Author: SamsonHo
* @Date:   2018-10-20-23.06.40
* @URL:https://vjudge.net/problem/UVA-11374
*/
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;
int g[505][505],dis1[505],dis2[505],pre1[505],pre2[505],vis[505];
int n,kase = 0;
stack<int> st;
void Dijkstra(int s,int dis[],int pre[])
{
    for(int i = 0; i <= n; ++i)
    {
        dis[i] = INF;
        vis[i] = 0;
        pre[i] = -1;
    }
    dis[s] = 0;
    pre[s] = -1;
    for(int i = 1; i <= n; ++i)
    {
        int Min = INF,pos = -1;
        for(int j = 1; j <= n; ++j)
        {
            if(!vis[j] && dis[j] < Min)
            {
                Min = dis[j];
                pos = j;
            }
        }
        if(pos == -1)   break;
        vis[pos] = 1;
        for(int j = 1; j <= n; ++j)
        {
            if(!vis[j] && dis[j]>dis[pos]+g[pos][j])
            {
                dis[j] = g[pos][j]+dis[pos];
                pre[j] = pos;
            }
        }
    }
}
int main(void)
{
    int u,v,c,eco,com,s,e;
    while(~scanf("%d%d%d",&n,&s,&e))
    {
        for(int i = 0; i <= n; ++i)
            for(int j = 0; j <= n; ++j)
                g[i][j] = i==j?0:INF;
        scanf("%d",&eco);
        for(int i = 1; i <= eco; ++i)
        {
            scanf("%d%d%d",&u,&v,&c);
            g[u][v] = g[v][u] = c;
        }
        Dijkstra(s,dis1,pre1);
        Dijkstra(e,dis2,pre2);
        scanf("%d",&com);
        int pos = -1,tmp,a,b,ans = dis1[e]; //先將ans初始化為不需要使用車票的花費
        for(int i = 1;i <= com; ++i)
        {
            scanf("%d%d%d",&u,&v,&c);
            tmp = dis1[u]+dis2[v]+c;
            if(ans > tmp)
            {
                ans = tmp;
                pos = u;
                a = u,b = v;
            }
            tmp = dis1[v]+dis2[u]+c;
            if(ans > tmp)
            {
                pos = v;
                ans = tmp;
                a = v,b = u;
            }
        }
        if(kase++)    puts("");
        if(pos == -1) //不需要使用商務路線
        {
            while(!st.empty())  st.pop();
            for(int i = e; i != -1; i = pre1[i])
                st.push(i);
            while(!st.empty())
            {
                printf("%d",st.top());
                st.pop();
                if(st.empty())
                    printf("\n");
                else
                    printf(" ");
            }
            puts("Ticket Not Used");
        }
        else
        {
            while(!st.empty())  st.pop();
            for(int i = a; i!=-1; i = pre1[i])
                st.push(i);
            int flag = 1;
            while(!st.empty())
            {
                if(flag)    flag = 0;
                else    printf(" ");
                printf("%d",st.top());
                st.pop();
            }
            for(int i = b; i != -1; i = pre2[i])
                printf(" %d",i);
            puts("");
            printf("%d\n",pos);
        }
        printf("%d\n",ans);
    }
}