hdu4597 区间dp

 1 //Accepted    1784 KB    78 ms
 2 //区间dp
 3 //dp[l1][r1][l2][r2] 表示a数列从l1到r1，b数列从l2到r2能得到的最大分值
 4 //
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <iostream>
 8 using namespace std;
 9 const int imax_n = 25;
10 int dp[imax_n][imax_n][imax_n][imax_n];
11 int n;
12 int a[imax_n],sa[imax_n],b[imax_n],sb[imax_n];
13 int max(int a,int b)
14 {
15     return a>b?a:b;
16 }
17 int dfs(int l1,int r1,int l2,int r2)
18 {
19     if (dp[l1][r1][l2][r2]!=-1) return dp[l1][r1][l2][r2];
20     dp[l1][r1][l2][r2]=0;
21     if (l1<=r1) dp[l1][r1][l2][r2]=sa[r1]-sa[l1-1]+sb[r2]-sb[l2-1]-dfs(l1+1,r1,l2,r2);
22     if (l1<=r1) dp[l1][r1][l2][r2]=max(dp[l1][r1][l2][r2],sa[r1]-sa[l1-1]+sb[r2]-sb[l2-1]-dfs(l1,r1-1,l2,r2));
23     if (l2<=r2)
24     {
25         dp[l1][r1][l2][r2]=max(dp[l1][r1][l2][r2],sa[r1]-sa[l1-1]+sb[r2]-sb[l2-1]-dfs(l1,r1,l2+1,r2));
26         dp[l1][r1][l2][r2]=max(dp[l1][r1][l2][r2],sa[r1]-sa[l1-1]+sb[r2]-sb[l2-1]-dfs(l1,r1,l2,r2-1));
27     }
28     return dp[l1][r1][l2][r2];
29 }
30 int main()
31 {
32     int T;
33     scanf("%d",&T);
34     while (T--)
35     {
36         scanf("%d",&n);
37         sa[0]=0;
38         for (int i=1;i<=n;i++)
39         {
40             scanf("%d",&a[i]);
41             sa[i]=sa[i-1]+a[i];
42         }
43         sb[0]=0;
44         for (int i=1;i<=n;i++)
45         {
46             scanf("%d",&b[i]);
47             sb[i]=sb[i-1]+b[i];
48         }
49         memset(dp,-1,sizeof(dp));
50         printf("%d\n",dfs(1,n,1,n));
51     }
52     return 0;
53 }