算法导论-装配线调度问题

 

全局变量

 1 /*
 2 There are 2 lines, each line has 5 station.
 3 Pass only one station per line, from left to right.
 4 Find the fastest way from enter to exit.
 5 Dynamic Programming
 6 */
 7 
 8 //enter price per line  
 9 int e[]={2,4};
10 
11 //exit price per line
12 int x[]={3,6};
13 
14 //price per station
15 int a[2][5]={7,9,3,4,8,
16                  8,5,6,4,5};
17 
18 //price when switch station crossing line
19 //switch station in one line need no price
20 int t[2][4]={2,3,1,3,
21                  2,1,2,2};
22 
23 //the lowest price when arriving a station
24 int f[2][5];
25 
26 //n stations each line
27 int n=5;
28 
29 //the lowest price
30 int _f;
31 
32 //when arriving at a station at the lowest price, which is the last station you pass, just store the line number
33 int l[2][4];
34 
35 //which station you will pass just before exit
36 int _l;

 

动态规划

 1 void Fastest_Way()
 2 {
 3     //there is only one way to get to the first station in each line
 4     f[0][0]=e[0]+a[0][0];
 5     f[1][0]=e[1]+a[1][0];
 6     
 7     //two way to get to the i's station, we will choose the lower one.
 8     for(int i=1;i<n;++i)
 9     {
10         if(f[0][i-1]+a[0][i]<f[1][i-1]+t[1][i-1]+a[0][i])
11         {
12             f[0][i]=f[0][i-1]+a[0][i];
13             l[0][i]=0;
14         }
15         else
16         {
17             f[0][i]=f[1][i-1]+t[1][i-1]+a[0][i];
18             l[0][i]=1;
19         }
20 
21         if(f[1][i-1]+a[1][i]<f[0][i-1]+t[0][i-1]+a[1][i])
22         {
23             f[1][i]=f[1][i-1]+a[1][i];
24             l[1][i]=1;
25         }
26         else
27         {
28             f[1][i]=f[0][i-1]+t[0][i-1]+a[1][i];
29             l[1][i]=0;
30         }
31     }
32     //when passing the last station, we should add the exit price, the lower will be the fastest
33     if(f[0][n-1]+x[0]<f[1][n-1]+x[1])
34     {
35         _f=f[0][n-1]+x[0];
36         _l=0;
37     }
38     else
39     {
40         _f=f[1][n-1]+x[1];
41         _l=1;
42     }
43     cout <<"The fastest time is "<<_f<<endl;
44     
45     //get the full path backward, using a stack is a good way
46     stack<int> s;
47     s.push(_l);
48     for(int i=n-1;i>=1;--i)
49     {
50         s.push(l[_l][i]);
51         _l=l[_l][i];
52     }
53     cout <<"The fastest path is: ";
54     for(int i=0;i<n;++i)
55     {
56         cout <<"S"<<"["<<s.top()+1<<"]"<<"["<<i+1<<"]";
57         if(i!=n-1)
58             cout <<"->";
59         s.pop();
60     }
61     cout <<endl;
62 }