UVa 11374 机场快线

https://vjudge.net/problem/UVA-11374

题意：

机场快线分为经济线和商业线两种，线路、速度和价格都不同。你有一张商业线车票，可以坐一站商业线，而其他时候只能乘坐经济线。你的任务是找一条去机场最快的线路。

思路：

因为商业线只能坐一站，所有可以枚举坐的是哪一站，用dijkstra算出起点到每个点的最短时间f(x)和终点到每个点的最短时间g(x)，则总时间为f(a)+T(a,b)+g(b)，其中T（a，b）为从a坐一站商业线到达b的时间。

#include <iostream>  
#include <cstring>  
#include <algorithm>   
#include <vector>
#include <queue>
using namespace std;

const int INF = 1000000000;
const int maxn = 500 + 5;

struct Edge
{
    int from, to, dist;
    Edge(int u, int v, int d) :from(u), to(v), dist(d){}
};

struct HeapNode
{
    int d, u;
    HeapNode(int x, int y) :d(x), u(y){}
    bool operator < (const HeapNode& rhs) const{
        return d > rhs.d;
    }
};

struct Dijkstra
{
    int n, m;                  //点数和边数
    vector<Edge> edges;        //边列表
    vector<int> G[maxn];       //每个结点出发的边编号（从0开始编号）
    bool done[maxn];           //是否已永久标号
    int d[maxn];               //s到各个点的距离
    int p[maxn];               //最短路中的上一条边

    void init(int n)
    {
        this->n = n;
        for (int i = 0; i < n; i++)    G[i].clear();
        edges.clear();
    }

    void AddEdges(int from, int to, int dist)
    {
        edges.push_back(Edge(from,to,dist));
        m = edges.size();
        G[from].push_back((m - 1));
    }

    void dijkstra(int s)
    {
        priority_queue<HeapNode> Q;
        for (int i = 0; i < n; i++)    d[i] = INF;
        d[s] = 0;
        memset(done, 0, sizeof(done));
        Q.push(HeapNode(0,s));
        while (!Q.empty())
        {
            HeapNode x = Q.top(); Q.pop();
            int u = x.u;
            if (done[u]) continue;
            done[u] = true;
            for (int i = 0; i < G[u].size(); i++)
            {
                Edge& e = edges[G[u][i]];
                if (d[e.to] > d[u] + e.dist)
                {
                    d[e.to] = d[u] + e.dist;
                    p[e.to] = e.from;
                    Q.push(HeapNode(d[e.to],e.to));
                }
            }
        }
    }

    void getpath(int s, int e, vector<int>& path)
    {
        int pos = e;
        while (true)
        {
            path.push_back(pos);
            if (pos == s)
                break;
            pos = p[pos];
        }
    }

}t[2];

int N, S, E;
vector<int> path;

int main()
{
    //freopen("D:\input.txt", "r", stdin);
    int time, kase = 0;
    while (scanf("%d%d%d", &N, &S, &E) != EOF)
    {
        S--; E--;
        if (kase != 0)   printf("
");
        kase++;
        t[0].init(N);
        t[1].init(N);
        path.clear();
        int M, K;
        int u, v, d;
        scanf("%d", &M);
        while (M--)
        {
            scanf("%d%d%d", &u, &v, &d); 
            u--; v--;
            t[0].AddEdges(u, v, d);
            t[1].AddEdges(u, v, d);
            t[0].AddEdges(v, u, d);
            t[1].AddEdges(v, u, d);
        }
        t[0].dijkstra(S);
        t[1].dijkstra(E);
        int ks = -1, ke = -1;
        time = t[0].d[E];
        scanf("%d", &K);
        while (K--)
        {
            scanf("%d%d%d", &u, &v, &d);
            u--;v--;
            if (d + t[0].d[u] + t[1].d[v] < time){
                time = d + t[0].d[u] + t[1].d[v];
                ks = u; ke = v;
            }
            if (d + t[0].d[v] + t[1].d[u] < time){
                time = d + t[0].d[v] + t[1].d[u];
                ks = v; ke = u;
            }
        }
        if (ks == -1)
        {
            t[0].getpath(S, E, path);
            reverse(path.begin(), path.end());
            for (int i = 0; i < path.size() - 1; i++)
                printf("%d ", path[i]+1);
            printf("%d
", E+1);
            printf("Ticket Not Used
");
            printf("%d
", time);
        }
        else
        {
            t[0].getpath(S, ks, path);
            reverse(path.begin(), path.end());
            t[1].getpath(E, ke, path);
            for (int i = 0; i < path.size() - 1; i++)
                printf("%d ", path[i]+1 );
            printf("%d
", E+1);
            printf("%d
", ks + 1);
            printf("%d
", time);
        }
    }
    return 0;
}