POJ 2686-Traveling by Stagecoach(DAG图-状态压缩DP)

￡神魔★判官ぃ 2022-08-22 12:25 148阅读 0赞

# Traveling by Stagecoach #

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong> 2000MS</td> 
   <td>&nbsp;</td> 
   <td colspan="3"><strong>Memory Limit:</strong> 65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong> 2802</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong> 1010</td> 
   <td>&nbsp;</td> 
   <td style="font-weight:bold; color:red">Special Judge</td> 
  </tr> 
 </tbody> 
</table>

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

Source

[Japan 2005 Domestic][]

## 题目意思： ##

有n张车票，m个城市，给出p条路径，每条路径有x，y，z三个元素。表示从x到y的路径长度为z。 给出a，b。商人要从a城市到达b城市。 每张车票只能通过一条道路，且每张车票上有车子的马的匹数。从一个城市到达它相邻的城市所需的时间等于城市之间的道路长度除以马的匹数。每张车票只能用一次，且不考虑换乘所需的时间。求城市a到城市b所需要的最短时间。 如果无法到达输出“Impossible”。

## 解题思路： ##

在状态压缩dp中我们用二进制表示一个集合中的元素的选取与否。这样就可以用0 ~~ (1<<n)-1表示每一张票是否选取。 我们将dp\[S\]\[v\]表示“现在在城市v，此时剩下的车票的集合为S” 。从这个状态出发使用一张车票i∈S移动到相邻的城市u，即相当于转移到了“在城市u，此时剩下的车票的集合为S\\\{i\}”这个状态（dp\[S\\\{i\}\]\[u\]）。 根据题意我们可以得到状态转移方程 dp\[S\]\[u\] = min(dp\[S\]\[u\],dp\[S∪i\]\[v\]+map\[v\]\[u\]/t\[i\])

[From：][From]

状压DP：其本质就是把数想像为二进制的状态，因为每个数其二进制中0和1的位置是不一样的，所以出现0和1的情况拿来做讨论，就表示一个状态。比如出现1的位置为开灯，0的位置为关灯：则长度为4的二进制数数字6二进制为：0110，则0110第二第三个位置为1表示开灯，第一第四个位置表示关灯，则0110表示一个状态，即数字四表示一个状态；又如数字2：0010也表示一个状态（第三个位置表示开灯）。 根据这些像排列组合的二进制0与1的不同位置来做的动态规划就是状态压缩动态规划。

在状压DP中经常用的二进制计算符号为&与|。

经常用的计算式子为：i&(1<<j)表示判断1右移 j 位后与i中相应的位置取&，如果结果为真，则说明i中这个位置为1，否则为0；

（i|(1<<j)）!=i表示判断1右移j位后与i中相应的位置取|，如果结果为真，则说明为i中的这个位置为0，否则为1。

#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<set>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define MAXN 1010//不能开得太大，不然会爆
    #define INF 99999//用来初始化，开的比较大就够用了
    int n,m,a,b;//n:车票数量，m:城市数量，a b表示起点和终点两个城市
    int t[MAXN];//车票
    int d[MAXN][MAXN];//图的邻接矩阵（-1表示没有边）
    double dp[MAXN][MAXN];//状态转移数组
    //dp[S][v]:到达剩下的车票集合为S并且现在在城市v的状态所需要的最小花费
    void solve()
    {
        int i;
        for(i=0; i<1<<n; ++i)
            fill(dp[i],dp[i]+m,INF);//用足够大的值初始化
        dp[(1<<n)-1][a-1]=0;//在起点城市不需要花费
        double res=INF;
        int S,v,u;
        for(S=(1<<n)-1; S>=0; --S)//用状态压缩的方法表示
        {
            res=min(res,dp[S][b-1]);
            for(v=0; v<m; ++v)//城市v
                for(i=0; i<n; ++i)//车票
                    //if((S>>i)%2)//这种写法也能AC而且速度从提交状态看还快了一丢丢……
                    if((S>>i)&1)//当前编号为i的车票还没有被用到
                    {
                        for(u=0; u<m; ++u)//城市u
                            if(d[u][v]>=0)//用车票i从v移动到u
                                dp[S & ~(1<<i)][u]=min(dp[S & ~(1<<i)][u],dp[S][v]+(double)d[v][u]/t[i]);//加上的double值表示花费
                    }
        }
        if(res==INF) cout<<"Impossible"<<endl;//无法到达
        else printf("%.3f\n",res);
    }
    int main()
    {
        int p,u,v,l;
        while(cin>>n>>m>>p>>a>>b&&n!=0&&m!=0&&a!=0&&b!=0)//注意p!=0不是结束条件之一
        {
            for(int i=0; i<n; ++i)
                cin>>t[i];//车票
            memset(d,-1,sizeof(d));
            for(int i=0; i<p; ++i)
            {
                cin>>u>>v>>l;//读入两城市u和v及其距离l
                d[u-1][v-1]=d[v-1][u-1]=l;//注意无向图而且元素下标从1开始
            }
            solve();
        }
        return 0;
    }

[Japan 2005 Domestic]: http://poj.org/searchproblem?field=source&key=Japan+2005+Domestic
[From]: http://blog.csdn.net/u011466175/article/details/19046137