poj 2686 状态压缩DP

 

 

如题;http://poj.org/problem?id=2686

Traveling by Stagecoach

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 2206 Accepted: 755 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 43 1 21 2 102 3 303 4 202 4 4 2 13 12 3 31 3 34 1 24 2 52 4 3 4 15 51 2 102 3 103 4 101 2 0 1 218 5 10 1 52 7 1 8 4 5 6 31 2 52 3 43 4 74 5 31 3 252 4 233 5 221 4 452 5 511 5 990 0 0 0 0

Sample Output

30.0003.667ImpossibleImpossible2.856

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.

30.03.66667ImpossibleImpossible2.85595

Source

Japan 2005 Domestic

 

 

题目大意：有n张车票，m个城市，p条路，a是起点城市标号，b是终点城市标号，从一个城市到另一个城市的花费取决于使用哪一张车票，设使用的车票编号是i,花费就是2个城市间的距离/t[i]. t[i]是整数，两个城市之间的距离是整数。要求输出从a城市到b城市的最小花费。

 

思路：由于车票的缘故，这一题不能使用普通的最短路来解决。

对于顶点，我们将（当前可以使用的车票的集合，当前的顶点v）作为一种状态。

下面思考一下状态转移的过程。假设当前状态(S,v),我们用掉一张i车票，代表从S中剔除i。S变为S&~(i<<i).可以到达的是和V链接的所有顶点。

这样建立图，转移看成一条边，花费就是d(v,u)/t[i].顶点就是2个状态(S,v),(S&~(1<<i),u);这样使用dijsktra求最短路是可行的。

但是我们可以发现，由于S集合的不断减小，不可能出现圈，这个图是一个DGA（无圈的有向图）。可以使用DP解决。

因此dp[S&~(1<<i)][u]=min(dp[S&~(1<<i)][u],dp[S][v]+(double)d(v,u)/t[i]);

 

4月中旬前的最后一题新题了，接下来要复习之前的算法了。

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 15
#define MAXM 32
#define INF 0x0fffffff
#define min(a,b)(a<b?a:b)

int n,m,p,a,b;
int t[MAXN];
int d[MAXM][MAXM];
double dp[1<<MAXN][MAXM];   //dp[S][v]:当前可以使用的票的状态集合S，当前顶点v，到达当前状态S,V且的小的花费。

void solve()
{
 int i,j;
 for(i=0;i<1<<MAXN;i++)
  for(j=0;j<MAXM;j++)
   dp[i][j]=INF;
 dp[(1<<n)-1][a-1]=0;
 double res=INF;
 int S;
 int u,v;
 for(S=(1<<n)-1;S>=0;S--)
 {
  res=min(res,dp[S][b-1]);
  for(v=0;v<m;v++)
  {
   for(i=0;i<n;i++)
   {
    if(S>>i&1)  //如果i票在S中
    {
     for(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==INF)
  printf("Impossible\n");
 else
  printf("%.3lf\n",res);
}
int main()
{
// freopen("C:\\1.txt","r",stdin);
 while(~scanf("%d%d%d%d%d",&n,&m,&p,&a,&b))
 {
  if(n==0&&m==0&&p==0&&a==0&&b==0)
   break;
  memset(t,0,sizeof(t));
  memset(d,-1,sizeof(d));
  int i;
  for(i=0;i<n;i++)
   scanf("%d",&t[i]);
  for(i=0;i<p;i++)
  {
   int u,v,w;
   scanf("%d%d%d",&u,&v,&w);
   d[u-1][v-1]=w;
   d[v-1][u-1]=w;
  }
  solve();
 }
}