uva 1048 最短路的建图 （巧，精品）

大白书 P341这题说的是给了NT种飞机票，给了价钱和整个途径，给了nI条要旅游的路线。使用飞机票都必须从头第一站开始坐，可以再这个路径上的任何一点下飞机一但下飞机了就不能再上飞机，只能重新买票，对于每张票使用的状态有经过了这个路途的前i个点使用它，那么点就有，n*len(n为城市编号，为旅途所经过的所有点的个数)，每张票在已经走完第一个点使用，第二个点，第三个点使用。。。最后从经过第一个点的旅途开始地方跑最长路，直到跑了len 旅途结束的点。

// LA3561 Low Cost Air Travel
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;

const int INF = ;
const int maxn =  + ;

struct Edge {
  int from, to, dist, val;
};

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

struct Dijkstra {
  int n, m;
  vector<Edge> edges;
  vector<int> G[maxn];
  bool done[maxn];    // 是否已永久标号
  int d[maxn];        // s到各个点的距离
  int p[maxn];        // 最短路中的上一条弧

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

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

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

  vector<int> GetShortestPath(int s, int t) {
    vector<int> path;
    while(t != s) {
      path.push_back(edges[p[t]].val);
      t = edges[p[t]].from;
    }
    reverse(path.begin(), path.end());
    return path;
  }
};

//////// 题目相关
#include<map>

int n_cities;
map<int,int> city_id;

int ID(int city) {
  if(city_id.count(city)) return city_id[city];
  city_id[city] = n_cities++;
  return n_cities-;
}

int ID(int visited, int cur) { return (visited-) * n_cities + cur; }

const int maxnt = ;
int cost[maxnt];
vector<int> cities[maxnt], iti;

Dijkstra solver;

int main() {
  int NT, NI, x, len, kase = ;
  while(scanf("%d", &NT) ==  && NT) {
    n_cities = ;
    city_id.clear();
    for(int i = ; i < NT; i++) {
      cities[i].clear();
      scanf("%d%d", &cost[i], &len);
      while(len--) { scanf("%d", &x); cities[i].push_back(ID(x)); }
    }
    scanf("%d", &NI);
    kase++;
    for(int trip = ; trip <= NI; trip++) {
      iti.clear();
      scanf("%d", &len);
      for(int i = ; i < len; i++) { scanf("%d", &x); iti.push_back(ID(x)); }

      solver.init(n_cities * len);
      for(int ticket = ; ticket < NT; ticket++)
        for(int visited = ; visited < len; visited++) {
          int cur = cities[ticket][]; // 当前状态为(visited, cur)
          int next = visited;          // 下一个需要访问的城市在iti中的下标
          for(int leg = ; leg < cities[ticket].size(); leg++) { // 使用前leg段
            if(cities[ticket][leg] == iti[next]) next++; // 行程上多经过一个城市
            solver.AddEdge(ID(visited, cur), ID(next, cities[ticket][leg]), cost[ticket], ticket+);
            if(next == len) break; // 行程单已经走完
          }
        }
      int src = ID(, iti[]), dest = ID(len, iti[len-]);
      solver.dijkstra(src);
      printf("Case %d, Trip %d: Cost = %d\n", kase, trip, solver.d[dest]);
      printf("  Tickets used:");
      vector<int> path = solver.GetShortestPath(src, dest);
      for(int i = ; i < path.size(); i++) printf(" %d", path[i]);
      printf("\n");
    }
  }
  return ;
}