hdu 4597 Play Game 区间dp

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题意：

有两排数，AB依次拿，每次只能从第一/二排最左边和最右边拿

问你A拿的和是多少，假设两个人都是很聪明的

思路：

http://blog.youkuaiyun.com/shuangde800/article/details/10277697

出现聪明这个词的时候，这种题不是博弈论就是dp吧

dp[x][y][i][j]表示当前玩家从a堆的x~y,b堆的i~j能获得的最大价值   

区间dp

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

ll a[25],b[25],dp[25][25][25][25];

ll solve(int l1,int r1,int l2,int r2){
	if(dp[l1][r1][l2][r2]!=-1) return dp[l1][r1][l2][r2];

	if(l1>r1) dp[l1][r1][l2][r2] = 0;
	if(l2>r2) dp[l1][r1][l2][r2] = 0;

	ll sum = 0,ans = 0;
	if(l1<=r1)
		sum += a[r1]-a[l1-1];
	if(l2<=r2)
		sum += b[r2]-b[l2-1];

	if(l1<=r1){
		ans = max(ans,sum-solve(l1+1,r1,l2,r2));
		ans = max(ans,sum-solve(l1,r1-1,l2,r2));
	}
	if(l2<=r2){
		ans = max(ans,sum-solve(l1,r1,l2+1,r2));
		ans = max(ans,sum-solve(l1,r1,l2,r2-1));
	}
	return dp[l1][r1][l2][r2] = ans;
}

int main(){
	int T; scanf("%d",&T);
	while(T--){
		memset(dp,-1,sizeof(dp));
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		int n; scanf("%d",&n);
		for(int i=1; i<=n; i++){
			scanf("%d",&a[i]);
			a[i] += a[i-1];
		}
		for(int i=1; i<=n; i++){
			scanf("%d",&b[i]);
			b[i] += b[i-1];
		}

		cout << solve(1,n,1,n) << endl;
	}
}