题意
给出2*n个数，分两列放置，每列n个，现在alice和bob两个人依次从任意一列的对头或队尾哪一个数，alice先拿，且两个人都想拿最多，问alice最后能拿到数字总和的最大值是多少。
思路
4月份通化邀请赛的题，当时竟然做不出来真是……记忆化搜索就OK了……
dp[a][b]{c}[d]=sum-min(dp[a+1][b]{c}[d],dp[a][b-1]{c}[d],dp[a][b]{c+1}[d],dp[a][b]{c}[d-1]);
其中dp[a][b]{c}[d]表示此状态下的最优情况，显然对方拿的越少自己就拿的越多，所以对于留给对手的4个状态应该让对方能拿的最大值最小，sum表示当前剩余的数字总和。然后直接搜索就行了。
代码
[cpp]
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#define MID(x,y) ((x+y)/2)
#define MEM(a,b) memset(a,b,sizeof(a))
#define REP(i, begin, end) for (int i = begin; i <= end; i ++)
using namespace std;
int dp[22][22][22][22];
int A[22], B[22];
int dfs(int a, int b, int c, int d, int sum){
if (~dp[a][b][c][d]) return dp[a][b][c][d];
if (a > b && c > d) return 0;
int res = 0;
if (a <= b){
res = max(res, sum - dfs(a+1, b, c, d, sum-A[a]));
res = max(res, sum - dfs(a, b-1, c, d, sum-A[b]));
}
if (c <= d){
res = max(res, sum - dfs(a, b, c+1, d, sum-B[c]));
res = max(res, sum - dfs(a, b, c, d-1, sum-B[d]));
}
return dp[a][b][c][d] = res;
}
int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int t;
scanf("%d", &t);
while(t --){
int n, sum = 0;
scanf("%d", &n);
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];
}
MEM(dp, -1);
printf("%d
", dfs(1, n, 1, n, sum));
}
return 0;
}
[/cpp]