[动态规划] Sum游戏 ( Game of Sum, Uva 10891 )

抓住状态转移方程即可   :  从子序列 i j 中取最大 =  i + 从子序列i+1,j中取最大        或         j +  从子序列i,j-1中取最大

#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;


const int maxn = 100+10;
int S[maxn], A[maxn], d[maxn][maxn], vis[maxn][maxn], n;

int dp(int i, int j){
	if (vis[i][j]) return d[i][j];
	vis[i][j] = 1;
	
	if (i == j) {
		d[i][j] = A[i];
		return d[i][j];
	}

	d[i][j] = max(A[i] + S[i+1,j] - dp(i + 1,j), A[j] + S[i,j - 1] - dp(i,j - 1));
	return d[i][j];
}

int main(){
	while (scanf("%d", &n) && n){
		S[0] = 0;
		for (int i = 1; i <= n;i++)
		{
			scanf("%d", &A[i]);
			S[i] = S[i - 1] + A[i];
		}
		memset(vis, 0, sizeof(vis));
		printf("%d\n", 2*dp(1, n)-S[n]);
	}
	return 0;
}

[动态规划] Sum游戏 ( Game of Sum, Uva 10891 )

原文：http://blog.csdn.net/qq_21970857/article/details/44024031