hdu1158 dp

 1 //Accepted    232 KB    0 ms
 2 //DP
 3 //dp[i][j]表示第i个月人数为a[j]时的最小花费
 4 //dp[i][j]=min(dp[i][j],dp[i-1][k]+hire*(a[j]-a[k])+pay*a[j]) a[j]>a[k]时
 5 //dp[i][j]=min(dp[i][j],dp[i-1][k]+fire*(a[k]-a[j])+pay*a[j]) a[k]>=a[j]时
 6 //a[i] 表示第i个月要用到的最少人数
 7 //
 8 #include <cstdio>
 9 #include <cstring>
10 #include <iostream>
11 using namespace std;
12 const int imax_n = 13;
13 const int inf = 100000000;
14 int a[imax_n];
15 int dp[imax_n][imax_n];
16 int hire,fire,pay;
17 int n;
18
19 int min(int a,int b)
20 {
21     return a>b?b:a;
22 }
23 int Dp()
24 {
25     for (int i=0;i<=n;i++)
26     for (int j=0;j<=n;j++)
27     dp[i][j]=inf;
28     dp[0][0]=0;
29     for (int i=1;i<=n;i++)
30     {
31         for (int j=0;j<=n;j++)
32         if (a[j]>=a[i])
33         {
34             for (int k=0;k<=n;k++)
35             {
36                 if (a[k]>a[j])
37                 dp[i][j]=min(dp[i][j],dp[i-1][k]+fire*(a[k]-a[j])+pay*a[j]);
38                 else
39                 dp[i][j]=min(dp[i][j],dp[i-1][k]+hire*(a[j]-a[k])+pay*a[j]);
40             }
41         }
42     }
43     int ans=inf;
44     for (int i=0;i<=n;i++)
45     ans=min(ans,dp[n][i]);
46     printf("%d\n",ans);
47     return ans;
48 }
49 int main()
50 {
51     while (scanf("%d",&n),n)
52     {
53         scanf("%d%d%d",&hire,&pay,&fire);
54         for (int i=1;i<=n;i++)
55         scanf("%d",&a[i]);
56         a[0]=0;
57         Dp();
58     }
59     return 0;
60 }

hdu1158 dp