Traveling by Stagecoach POJ - 2686

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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

Output

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

30.0

3.66667

Impossible

Impossible

2.85595

题解:dp[S][v]:=到达剩下的车票集合为S并且现在城市在v的状态所需要的最小花费。DP好难啊！！！！

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 const double INF=10000000.0;
 8 
 9 int n,m,p,a,b;
10 int t[10],d[35][35];
11 double dp[1<<10][35];
12 
13 void Init(){
14     for(int i=0;i<(1<<n);i++)
15         fill(dp[i],dp[i]+m,INF);
16     for(int i=0;i<m;i++)
17         for(int j=0;j<m;j++) d[i][j]=-1;    
18 }
19 
20 void Read(){
21     for(int i=0;i<n;i++) scanf("%d",&t[i]);
22     int u,v,cost;
23     for(int i=0;i<p;i++){
24         scanf("%d%d%d",&u,&v,&cost);
25         d[u-1][v-1]=cost;
26         d[v-1][u-1]=cost;
27     }
28 }
29 
30 void solve(){
31     dp[(1<<n)-1][a-1]=0;
32     double ans=INF;
33     
34     for(int S=(1<<n)-1;S>=0;S--){
35         ans=min(ans,dp[S][b-1]);
36         for(int v=0;v<m;v++){
37             for(int i=0;i<n;i++){
38                 if(S>>i&1){
39                     for(int u=0;u<m;u++){
40                         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]);
41                     }
42                 }
43             }
44         }
45     }
46     if(ans==INF) cout<<"Impossible"<<endl;
47     else printf("%.3f\n",ans);
48 }
49 
50 int main()
51 {   while(scanf("%d%d%d%d%d",&n,&m,&p,&a,&b)){
52         if(!n&&!m&&!p&&!a&&!b) break;
53         
54         Init();
55         Read();
56         solve();
57     }
58     return 0;
59 }

Traveling by Stagecoach POJ - 2686

标签：lang   int   logs   eterm   eve   test   sage   country   cst   

原文地址：http://www.cnblogs.com/zgglj-com/p/7380527.html