http://poj.org/problem?id=2686                             
                    Traveling by Stagecoach
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 2276   Accepted: 787   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
//1代表没走,0代表走了。
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define inf 100000000
int n,m,a,b;
int t[],d[][];
double dp[<<][];
void solve()
{
//memset(dp,inf,sizeof(dp));
for(int i=;i<<<n;i++)
for(int j=;j<<<n;j++)
dp[i][j]=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 & ) {//第i个城市是否为1.
for (int u = ; u <m; u++) {
if (d[u][v] >= ) {
dp[S & ~( << i)][u] = min(dp[S & ~( << i)][u],
dp[S][v] + d[u][v] / (double)t[i]);//把第i位变为1.
}
}
}
}
}
}
if(res == inf)
printf("Impossible\n");
else
printf("%.3f\n", res);
}
int main()
{
int p;
for(;;)
{
scanf("%d%d%d%d%d",&n,&m,&p,&a,&b);
if(n==&&m==&&p==&&a==&&b==)
break;
int i,j;
memset(d, -, sizeof(d)); for(i=;i<n;i++)
scanf("%d",&t[i]);
int x,y,z;
for(i=;i<p;i++)
{
scanf("%d%d%d",&x,&y,&z);
d[x-][y-]=d[y-][x-]=z;
}
solve();
}
return ;
}

poj2686 Traveling by Stagecoach的更多相关文章

  1. POJ2686 Traveling by Stagecoach 状态压缩DP

    POJ2686 比较简单的 状态压缩DP 注意DP方程转移时,新的状态必然数值上小于当前状态,故最外层循环为状态从大到小即可. #include <cstdio> #include < ...

  2. POJ2686 Traveling by Stagecoach(状压DP+SPFA)

    题目大概是给一张有向图,有n张票,每张票只能使用一次,使用一张票就能用pi匹马拉着走过图上的一条边,走过去花的时间是边权/pi,问从a点走到b点的最少时间是多少. 用dp[u][S]表示当前在u点且用 ...

  3. [状压dp]POJ2686 Traveling by Stagecoach

    题意: m个城市, n张车票, 每张车票$t_i$匹马, 每张车票可以沿某条道路到相邻城市, 花费是路的长度除以马的数量. 求a到b的最小花费, 不能到达输出Impossible $1\le n\le ...

  4. 【状压DP】poj2686 Traveling by Stagecoach

    状压DP裸题,将({当前车票集合},当前顶点)这样一个二元组当成状态,然后 边权/马匹 当成边长,跑最短路或者DAG上的DP即可. #include<cstdio> #include< ...

  5. POJ2686 Traveling by Stagecoach(状压DP)

    题意: 有一个旅行家计划乘马车旅行.他所在的国家里共有m个城市,在城市之间有若干道路相连.从某个城市沿着某条道路到相邻的城市需要乘坐马车.而乘坐马车需要使用车票,每用一张车票只可以通过一条道路.每张车 ...

  6. POJ2686 Traveling by Stagecoach (状压DP)

    将车票的使用情况用二进制表示状态,对其进行转移即可. 但是我一开始写的代码是错误的(注释部分),看似思路是正确的,但是暗藏很大的问题. 枚举S,我们要求解的是dp[S][v],这个是从u转移过来的,不 ...

  7. Traveling by Stagecoach 状态压缩裸题

    Traveling by Stagecoach dp[s][v]  从源点到达  v,状态为s,v的最小值.  for循环枚举就行了. #include <iostream> #inclu ...

  8. POJ 2686 Traveling by Stagecoach(状压二维SPFA)

    Traveling by Stagecoach Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3407   Accepted ...

  9. Traveling by Stagecoach(POJ 2686)

    原题如下: Traveling by Stagecoach Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 4494   Ac ...

随机推荐

  1. keepalived+haproxy-部署高可用负载均衡

    环境: 准备两台机子,安装haproxy及keepalive都一样,只是配置不一样而已. 这里只说明一台机子上安装haproxy及keepalive. ======================== ...

  2. 《jQuery UI开发指南》勘误收集

    此书由罗晴明 (http://weibo.com/sunnylqm)和我合译完成,此篇博客作为勘误收集而用,若译文有误或者有任何疑问,欢迎留下评论,或者给我发邮件(地址:gzooler@gmail.c ...

  3. jsonp使用规范

    这两天花了很多时间弄研究jsonp这个东西, 可是无论我怎么弄..TMD就是不进入success函数,并且一直进入error函数...让我着实DT啊. 可以看下我之间的提问(这就是我遇到的烦恼).. ...

  4. iis7如何取消目录的可执行权限

    我们需要把IIs中某一个目录的可执行权限去掉.这在IIs6中是非常方便的,可是因为iis7的机制小编也找了不少资料才找到. 第一步:先选择需要取消权限的目录,然后在右边可以看到 “处理程序映射” 双击 ...

  5. C#表驱动法+一点反射实现“得到指定位数随机不重复字符串”三种方式的封装

    1.结构 第一个类 public class GetMethods{...}      类中的变量:                                                   ...

  6. Source Insight 显示中文乱码

    Source Insight 3.X utf8支持插件震撼发布 继上次SI多标签插件之后,因为公司内部编码改为utf8编码,因此特意做了这个Source Insight 3.X utf8插件. 下载地 ...

  7. MSTest不支持参数化测试的解决方案

    之前的项目中做单元测试一直用的是NUnit,这次做新项目,负责人要求统一用MsTest,理由是MsTest是Visual Studio内置的.用就用吧,我没什么意见.不过用了两天,我就发现一个大问题: ...

  8. scrollView的几个属性contentSize contentOffset contentInset

    01-  ontentSize是scrollview可以滚动的区域 比如frame = (0 ,0 ,320 ,480) contentSize = (320 ,960), 代表你的scrollvie ...

  9. linux相关办公软件汇总

    ubuntu pdf阅读器 FoxitReader_1.1.0_i386.deb ubuntu 下的PDF阅读器(超级好使) Ubuntu下的chm和PDF阅读器 ubuntu便签软件xpad sud ...

  10. 坚果云创业团队访谈:我们 DIY 云存储(不要过度关注竞争对手,尤其当我们还是小公司的时候)

    坚果云(http://jianguoyun.com/)是一款用于多平台文件同步.备份和交换的云存储工具,立志于提供“便捷,安全”的服务.坚果云自去年年初启动内测,至今年三月初刚刚正式发布.近日我们拜访 ...