Play Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

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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#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i <b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 22

int a[N],b[N];
int dp[N][N][N][N];
int n;
int suma[N],sumb[N];

int dfs(int le,int ri,int lle,int rri)
{
	if(dp[le][ri][lle][rri]!=-1) return dp[le][ri][lle][rri];
	int i,j;
	int temp=0;

    if(le>ri&&lle>rri)
			return dp[le][ri][lle][rri]=0;
	temp+=suma[ri]-suma[le-1];
	temp+=sumb[rri]-sumb[lle-1];
	int ans=0;

	if(le<=ri)
	{
		ans=max(ans,temp-dfs(le+1,ri,lle,rri));
		ans=max(ans,temp-dfs(le,ri-1,lle,rri));
	}
   if(lle<=rri)
	{
		ans=max(ans,temp-dfs(le,ri,lle+1,rri));
		ans=max(ans,temp-dfs(le,ri,lle,rri-1));
	}
   return dp[le][ri][lle][rri]=ans;
}

int main()
{
	int i,j;
	int t;
	sf(t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			suma[i]=suma[i-1]+a[i];
		}
        for(i=1;i<=n;i++)
		{
			sf(b[i]);
			sumb[i]=sumb[i-1]+b[i];
		}
	   mem(dp,-1);
       pf("%d\n",dfs(1,n,1,n));
	}
  return 0;
}