poj 2686 Traveling by Stagecoach

Posted 2020-09-09 ZefengYao

                                                                                                                                                  Traveling by Stagecoach

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3293		Accepted: 1253		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

题意：旅行家要从城市a旅行到城市b，他的手里有n张马车票，使用马车票可以让旅行家在任意一条马路上通行，马车票上印有马匹的数量t[i]，并规定道路的长度除以马匹的数量就是通行该路所花的时间。那么旅行家从城市a到b至少需要花多少时间，如果无法通行，输出‘Impossible’
思路：状态压缩dp，设集合S为旅行家手里剩余的马车票，dp[S][v]含义:旅行家还剩余的马车票组成集合S，并且已到达城市v的时候所花的时间总和。若目前已到达城市v，且马车票组成的集合为S，此时使用集合S中的第i张马车票到达城市u,那么状态转移过程可以表示为:
dp[S&~(1<<i)][u]=min{dp[S&~(1<<i)][u],dp[s][v]+d[s][v]/t[i]};
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<functional>
using namespace std;
typedef long long ll;
const int M_MAX = 30,N_MAX=8+1;
int n, m, p,a, b, t[N_MAX];
int d[M_MAX][M_MAX];
double dp[1 << N_MAX][M_MAX];//dp[s][v]剩下的车票集合为S，当前在城市v,所花费的钱

int main() {
    while (scanf("%d%d%d%d%d", &n, &m, &p, &a, &b)&&n) {
        a--, b--;
        for (int i = 0; i < n; i++) {
            scanf("%d", &t[i]);
        }
        memset(d, -1, sizeof(d));
        for (int i = 0; i < p; i++) {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            x--, y--;
            d[x][y] = z;
            d[y][x] = z;
        }

        for (int i = 0; i < 1<<n; i++) {
            fill(dp[i], dp[i] + m,INT_MAX/2);
        }
        double res = INT_MAX / 2;
        dp[(1 << n) - 1][a] = 0;//车票都在，人在起点，花费为0
        for (int S = (1 << n) - 1; S >= 0; S--) {
            res = min(res,dp[S][b]);
            for (int v = 0; v < m;v++) {
                for (int i = 0; i < n; i++) {
                    if ((S >> i) & 1) {//这张票还没用的话就用掉
                        for (int u = 0; u < m; u++) {
                            if(d[v][u]>=0)//!!!!!!
                            dp[S&~(1 << i)][u] = min(dp[S&~(1 << i)][u],dp[S][v]+(double)d[v][u]/t[i]);
                            }
                        }
                    }
                }
            }
        if (res == INT_MAX / 2)
            printf("Impossible\n");
        else 
        printf("%.3f\n",res);
      }

return 0;
    }