Play Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 24    Accepted Submission(s): 18

Problem Description Alice and Bob are playing a game. There are two piles of cards. There are N cards in each pile, and each card has a score. They take turns to pick up the top or bottom card from either pile, and the score of the card will be added to his total score. Alice and Bob are both clever enough, and will pick up cards to get as many scores as possible. Do you know how many scores can Alice get if he picks up first?  

Input The first line contains an integer T (T≤100), indicating the number of cases. 

Each case contains 3 lines. The first line is the N (N≤20). The second line contains N integer a
_i (1≤a
_i≤10000). The third line contains N integer b
_i (1≤b
_i≤10000).  

Output For each case, output an integer, indicating the most score Alice can get.  

Sample Input 2 1 23 53 3 10 100 20 2 4 3  

Sample Output 53 105  

Source
2013 ACM-ICPC吉林通化全国邀请赛——题目重现  

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 1 /* ***********************************************
 2 Author        :kuangbin
 3 Created Time  :2013/8/24 12:37:04
 4 File Name     :F:\2013ACM练习\比赛练习\2013通化邀请赛\1008.cpp
 5 ************************************************ */
 6 
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <math.h>
17 #include <stdlib.h>
18 #include <time.h>
19 using namespace std;
20 int a[30],b[30];
21 int sum1[30];
22 int sum2[30];
23 int dp[30][30][30][30];
24 int solve(int l1,int r1,int l2,int r2)
25 {
26     if(dp[l1][r1][l2][r2] != -1)return dp[l1][r1][l2][r2];
27     if(l1 > r1 && l2 > r2)
28         return dp[l1][r1][l2][r2] = 0;
29     int ans = 0;
30     int sum = 0;
31     if(l1 <= r1)
32         sum += sum1[r1] - sum1[l1-1];
33     if(l2 <= r2)
34         sum += sum2[r2] - sum2[l2-1];
35     if(l1 <= r1)
36     {
37         ans = max(ans,sum - solve(l1+1,r1,l2,r2));
38         ans = max(ans,sum - solve(l1,r1-1,l2,r2));
39     }
40     if(l2 <= r2)
41     {
42         ans = max(ans,sum - solve(l1,r1,l2+1,r2));
43         ans = max(ans,sum - solve(l1,r1,l2,r2-1));
44     }
45     return dp[l1][r1][l2][r2] = ans;
46 }
47 int main()
48 {
49     //freopen("in.txt","r",stdin);
50     //freopen("out.txt","w",stdout);
51     int n;
52     int T;
53     scanf("%d",&T);
54     while(T--)
55     {
56         scanf("%d",&n);
57         sum1[0] = sum2[0] = 0;
58         for(int i = 1;i <= n;i++)
59         {
60             scanf("%d",&a[i]);
61             sum1[i] = sum1[i-1] + a[i];
62         }
63         for(int i = 1;i <= n;i++)
64         {
65             scanf("%d",&b[i]);
66             sum2[i] = sum2[i-1] + b[i]; 
67         }
68         memset(dp,-1,sizeof(dp));
69         printf("%d\n",solve(1,n,1,n));
70     }
71     return 0;
72 }