【POJ】Traveling by Stagecoach -最短路状压Dp

该问题要求确定一个旅行者从起点到终点的最佳路线,使用有限的马车票,并考虑每张票的速度。通过状态压缩和动态规划的方法,找到满足条件的最短时间路径。如果无法到达目的地,则输出'Impossible'。动态规划的状态转移方程为:f[i&(~1<< k)][j]=min(f[i&(~1<< k)][j],f[i][u]+w[u][j]/t[k])。" 37509577,1033215,Android API 深入解析:Notification 详解,"['Android开发', 'API使用', '通知管理']

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Description

A traveler plans to travel using stagecoaches=. His starting point and destination are fixed. Your job in this problem is to write a program which determines the route for him.

A road network connecting several cities in this country. The stagecoach can ride to another city from this city if there is a road between them. 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.


Solution

To solve this kind of problem, it would be better if we can think of Dp~、
And we must focus on the data limit infered in the Input:

The number of tickets is between 1 and 8.The number of cities is between 1 and 30.

So…State compression!
Now let’s consider the state transfer function,easily get the eqution below(assume that i is a binary non-negative intiger,if (i>>k)&1 == 1,it means we have already used the i-th ticket,j means now we are in the j-th city,w[u][j] means the distance of the road between the u-th city and the j-th city,and t[k] means the number of horses of the k-th ticket):

f[i&(~1<< k)][j]=min(f[i&(~1<< k)][j],f[i][u]+w[u][j]/t[k])

Maybe you can think more carefully again so as to understand it entirely.


Code

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#define eps 1e-3
#define db double
#define INF 1e30
const int N=1<<10;
using namespace std;
int n,m,a,t[202],tot,fr,to,path;
db f[N][35];db ans;
int w[35][35];
inline int read()
{
    char ch=getchar();int x=0;
    while(ch<'0' || ch>'9') ch=getchar();
    while(ch>='0' && ch<='9') {x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x;
}

inline db min(db a,db b){
    if(a>b) return b;
    return a;
}

inline void recover()
{
    memset(w,-1,sizeof(w));
    for(int i=0;i<(1<<n);i++){
        for(int j=0;j<=m;j++)
          f[i][j]=INF;
    }
    ans=INF;
    for(int i=0;i<n;i++) t[i]=read();
    while(path--){
        int x=read(),y=read(),z=read();
        if(w[x][y]<0){
            w[x][y]=z;w[y][x]=z;
        }else{
            w[x][y]=min(w[x][y],z);w[y][x]=min(w[y][x],z);
        }
    }
}

inline void deal()
{
    int s=(1<<n)-1;
    f[s][fr]=0;
    for(int i=s;i>=0;i--)
    {
      ans=min(ans,f[i][to]);
      for(int u=1;u<=m;u++){
        for(int k=0;k<n;k++)
            if(i>>k & 1){
            for(int v=1;v<=m;v++)
            if(w[u][v]>=0){
               f[i&(~(1<<k))][v]=min(f[i&(~(1<<k))][v],f[i][u]+w[u][v]*1.0/t[k]);
            }
        }
      }
    }    
}

int main(){
    n=read();m=read();path=read();fr=read();to=read();
    while(n || m || path || fr || to){
        recover();
        deal();
        if(ans>=INF)
          printf("Impossible\n");
        else printf("%.3f\n",ans);//lf->f ->ac qwq
        n=read();m=read();path=read();fr=read();to=read();
    }
    return 0;
}

皮这一下很开心~

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