Traveling by Stagecoach(POJ 2686)

• 原题如下：
  Traveling by Stagecoach

  Time Limit: 2000MS		Memory Limit: 65536K
  Total Submissions: 4494		Accepted: 1852		Special Judge

  Description

  Once upon a time, there was a traveler.

  He plans to travel using stagecoaches (horse wagons). His starting point and destination are fixed, but he cannot determine his route. Your job in this problem is to write a program which determines the route for him.

  There are several cities in the country, and a road network connecting them. If there is a road between two cities, one can travel by a stagecoach from one of them to the other. A coach ticket is needed for a coach ride. The number of horses is specified in each of the tickets. Of course, with more horses, the coach runs faster.

  At the starting point, the traveler has a number of coach tickets. By considering these tickets and the information on the road network, you should find the best possible route that takes him to the destination in the shortest time. The usage of coach tickets should be taken into account.

  The following conditions are assumed.

  • A coach ride takes the traveler from one city to another directly connected by a road. In other words, on each arrival to a city, he must change the coach.
  • Only one ticket can be used for a coach ride between two cities directly connected by a road.
  • Each ticket can be used only once.
  • The time needed for a coach ride is the distance between two cities divided by the number of horses.
  • The time needed for the coach change should be ignored.

  Input

  The input consists of multiple datasets, each in the following format. The last dataset is followed by a line containing five zeros (separated by a space).

  n m p a b 
  t1 t2 ... tn 
  x1 y1 z1 
  x2 y2 z2 
  ... 
  xp yp zp

  Every input item in a dataset is a non-negative integer. If a line contains two or more input items, they are separated by a space.

  n is the number of coach tickets. You can assume that the number of tickets is between 1 and 8. m is the number of cities in the network. You can assume that the number of cities is between 2 and 30. p is the number of roads between cities, which may be zero.

  a is the city index of the starting city. b is the city index of the destination city. a is not equal to b. You can assume that all city indices in a dataset (including the above two) are between 1 and m.

  The second line of a dataset gives the details of coach tickets. ti is the number of horses specified in the i-th coach ticket (1<=i<=n). You can assume that the number of horses is between 1 and 10.

  The following p lines give the details of roads between cities. The i-th road connects two cities with city indices xi and yi, and has a distance zi (1<=i<=p). You can assume that the distance is between 1 and 100.

  No two roads connect the same pair of cities. A road never connects a city with itself. Each road can be traveled in both directions.

  Output

  For each dataset in the input, one line should be output as specified below. An output line should not contain extra characters such as spaces.

  If the traveler can reach the destination, the time needed for the best route (a route with the shortest time) should be printed. The answer should not have an error greater than 0.001. You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.

  If the traveler cannot reach the destination, the string "Impossible" should be printed. One cannot reach the destination either when there are no routes leading to the destination, or when the number of tickets is not sufficient. Note that the first letter of "Impossible" is in uppercase, while the other letters are in lowercase.

  Sample Input

  3 4 3 1 4
  3 1 2
  1 2 10
  2 3 30
  3 4 20
  2 4 4 2 1
  3 1
  2 3 3
  1 3 3
  4 1 2
  4 2 5
  2 4 3 4 1
  5 5
  1 2 10
  2 3 10
  3 4 10
  1 2 0 1 2
  1
  8 5 10 1 5
  2 7 1 8 4 5 6 3
  1 2 5
  2 3 4
  3 4 7
  4 5 3
  1 3 25
  2 4 23
  3 5 22
  1 4 45
  2 5 51
  1 5 99
  0 0 0 0 0

  Sample Output

  30.000
  3.667
  Impossible
  Impossible
  2.856
  

  Hint

  Since the number of digits after the decimal point is not specified, the above result is not the only solution. For example, the following result is also acceptable.

  30.0
  
  3.66667
  
  Impossible
  
  Impossible
  
  2.85595

• 题解：如果把城市看作顶点，道路看作边建图，由于有车票相关的限制，无法直接使用Dijkstra算法求解。不过，这种情况下只需要把状态作为顶点，而把状态的转移看成边建图就可以很好地避免这个问题。考虑"现在在城市v，此时还剩下的车票的集合为S"这样的状态，从这个状态出发，使用一张车票i∈S移动到相邻的城市u，就相当于转移到了"在城市u，此时还剩下的车票的集合为S\{i}"这个状态。把这个转移看成一条边，那么边上的花费是(v-u间道路的长度)/t_i。按照上述的方法所构的图就可以用Dijkstra算法求解了。集合S使用状态压缩的方法表示就可以了。由于剩余的车票的集合S随着移动元素个数不断变小，因此这个图实际上一个DAG，计算DAG的最短路不需要是用Dijkstra算法，可以简单地通过DP求解。
• 代码：

   #include <cstdio>
   #include <cstdio>
   #include <cctype>
   #include <cstring>
   #include <algorithm>
   #define number s-'0'
  
   using namespace std;
  
   const int INF=0x3f3f3f3f;
   const int MAX_N=;
   const int MAX_M=;
   int n,m,p,a,b;
   int t[MAX_N];
   int d[MAX_M][MAX_M];
   double dp[<<MAX_N][MAX_M];
  
   void read(int &x)
   {
       char s;
       x=;
       bool flag=;
       while(!isdigit(s=getchar()))
           (s=='-')&&(flag=true);
       for(x=number;isdigit(s=getchar());x=x*+number);
       (flag)&&(x=-x);
   }
  
   double min(double x, double y)
   {
       if (x<=y) return x;
       return y;
   }
  
   int main()
   {
       read(n);read(m);read(p);read(a);read(b);
       while (n+m+p+a+b)
       {
           for (int i=; i<n; i++) read(t[i]);
           memset(d, -, sizeof(d));
           for (int i=; i<p; i++)
           {
               int x, y, z;
               read(x);read(y);read(z);
               x--;y--;
               x[y[d]]=y[x[d]]=z;
           }
           for (int i=; i<<<n; i++)
           {
               fill(dp[i], dp[i]+m, INF);
           }
           dp[(<<n)-][a-]=;
           double res=INF;
           for (int S=(<<n)-; S>=; S--)
           {
               res=min(res, dp[S][b-]);
               for (int v=; v<m; v++)
               {
                   for (int i=; i<n; i++)
                    {
                        if ((S>>i)&)
                        {
                            for (int u=; u<m; u++)
                            {
                                if (d[v][u]>=)
                                {
                                    dp[S&~(<<i)][u]=min(dp[S&~(<<i)][u],dp[S][v]+d[v][u]/(double)t[i]);
                                }
                            }
                        }
                    }
                }
           }
           if (res==INF) puts("Impossible");
           else printf("%.3f\n", res);
           read(n);read(m);read(p);read(a);read(b);
       }
   }