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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 ai(1≤ai≤10000). The third line contains N integer bi (1≤bi≤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
题意：Alice和Bob轮流在两个数组中取数字，每次只能取走一个数组的第一个或最后一个数字，问Alice取走数字的最大和
思路： 表示第一个数组剩余
，第二个数组剩余
时，执行下一步的人可以获得的最大价值，它可以由上一步对方的最大值转移过来，4种转移方式：数组一的第一个、数组一的最后一个、数组二的第一个、数组二的最后一个。
for循环着实不好写，还是记忆化搜索8
#includeusing namespace std;typedef long long ll;const int inf = 0x3f3f3f3f;const int N = 25;int a[N], b[N], dp[N][N][N][N], n;int dfs(int i, int j, int p, int q, int sum) { if(i > j && p > q) return 0; if(dp[i][j][p][q]) return dp[i][j][p][q]; if(i <= j) { dp[i][j][p][q] = max(dp[i][j][p][q], sum - dfs(i + 1, j, p, q, sum - a[i])); dp[i][j][p][q] = max(dp[i][j][p][q], sum - dfs(i, j - 1, p, q, sum - a[j])); } if(p <= q) { dp[i][j][p][q] = max(dp[i][j][p][q], sum - dfs(i, j, p + 1, q, sum - b[p])); dp[i][j][p][q] = max(dp[i][j][p][q], sum - dfs(i, j, p, q - 1, sum - b[q])); } return dp[i][j][p][q];}int main(){// freopen("in.txt", "r", stdin); int t; scanf("%d", &t); while(t--) { scanf("%d", &n); int sum = 0; for(int i = 1; i <= n; ++i) { scanf("%d", &a[i]); sum += a[i]; } for(int i = 1; i <= n; ++i) { scanf("%d", &b[i]); sum += b[i]; } memset(dp, 0, sizeof(dp)); printf("%d\n", dfs(1, n, 1, n, sum)); } return 0;}
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