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。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <algorithm>
  4 #include <vector>
  5 #include <queue>
  6 using namespace std;
  7
  8 const int INF = 1000000000;
  9 const int maxn = 500 + 5;
 10
 11 struct Edge
 12 {
 13     int from, to, dist;
 14     Edge(int u, int v, int d) :from(u), to(v), dist(d){}
 15 };
 16
 17 struct HeapNode
 18 {
 19     int d, u;
 20     HeapNode(int x, int y) :d(x), u(y){}
 21     bool operator < (const HeapNode& rhs) const{
 22         return d > rhs.d;
 23     }
 24 };
 25
 26 struct Dijkstra
 27 {
 28     int n, m;                  //点数和边数
 29     vector<Edge> edges;        //边列表
 30     vector<int> G[maxn];       //每个结点出发的边编号（从0开始编号）
 31     bool done[maxn];           //是否已永久标号
 32     int d[maxn];               //s到各个点的距离
 33     int p[maxn];               //最短路中的上一条边
 34
 35     void init(int n)
 36     {
 37         this->n = n;
 38         for (int i = 0; i < n; i++)    G[i].clear();
 39         edges.clear();
 40     }
 41
 42     void AddEdges(int from, int to, int dist)
 43     {
 44         edges.push_back(Edge(from,to,dist));
 45         m = edges.size();
 46         G[from].push_back((m - 1));
 47     }
 48
 49     void dijkstra(int s)
 50     {
 51         priority_queue<HeapNode> Q;
 52         for (int i = 0; i < n; i++)    d[i] = INF;
 53         d[s] = 0;
 54         memset(done, 0, sizeof(done));
 55         Q.push(HeapNode(0,s));
 56         while (!Q.empty())
 57         {
 58             HeapNode x = Q.top(); Q.pop();
 59             int u = x.u;
 60             if (done[u]) continue;
 61             done[u] = true;
 62             for (int i = 0; i < G[u].size(); i++)
 63             {
 64                 Edge& e = edges[G[u][i]];
 65                 if (d[e.to] > d[u] + e.dist)
 66                 {
 67                     d[e.to] = d[u] + e.dist;
 68                     p[e.to] = e.from;
 69                     Q.push(HeapNode(d[e.to],e.to));
 70                 }
 71             }
 72         }
 73     }
 74
 75     void getpath(int s, int e, vector<int>& path)
 76     {
 77         int pos = e;
 78         while (true)
 79         {
 80             path.push_back(pos);
 81             if (pos == s)
 82                 break;
 83             pos = p[pos];
 84         }
 85     }
 86
 87 }t[2];
 88
 89 int N, S, E;
 90 vector<int> path;
 91
 92 int main()
 93 {
 94     //freopen("D:\\input.txt", "r", stdin);
 95     int time, kase = 0;
 96     while (scanf("%d%d%d", &N, &S, &E) != EOF)
 97     {
 98         S--; E--;
 99         if (kase != 0)   printf("\n");
100         kase++;
101         t[0].init(N);
102         t[1].init(N);
103         path.clear();
104         int M, K;
105         int u, v, d;
106         scanf("%d", &M);
107         while (M--)
108         {
109             scanf("%d%d%d", &u, &v, &d);
110             u--; v--;
111             t[0].AddEdges(u, v, d);
112             t[1].AddEdges(u, v, d);
113             t[0].AddEdges(v, u, d);
114             t[1].AddEdges(v, u, d);
115         }
116         t[0].dijkstra(S);
117         t[1].dijkstra(E);
118         int ks = -1, ke = -1;
119         time = t[0].d[E];
120         scanf("%d", &K);
121         while (K--)
122         {
123             scanf("%d%d%d", &u, &v, &d);
124             u--;v--;
125             if (d + t[0].d[u] + t[1].d[v] < time){
126                 time = d + t[0].d[u] + t[1].d[v];
127                 ks = u; ke = v;
128             }
129             if (d + t[0].d[v] + t[1].d[u] < time){
130                 time = d + t[0].d[v] + t[1].d[u];
131                 ks = v; ke = u;
132             }
133         }
134         if (ks == -1)
135         {
136             t[0].getpath(S, E, path);
137             reverse(path.begin(), path.end());
138             for (int i = 0; i < path.size() - 1; i++)
139                 printf("%d ", path[i]+1);
140             printf("%d\n", E+1);
141             printf("Ticket Not Used\n");
142             printf("%d\n", time);
143         }
144         else
145         {
146             t[0].getpath(S, ks, path);
147             reverse(path.begin(), path.end());
148             t[1].getpath(E, ke, path);
149             for (int i = 0; i < path.size() - 1; i++)
150                 printf("%d ", path[i]+1 );
151             printf("%d\n", E+1);
152             printf("%d\n", ks + 1);
153             printf("%d\n", time);
154         }
155     }
156     return 0;
157 }