soj 1176. Two Ends

思路：

正向递归：

1）如果还剩n个数（n>2）以上，那么，模拟自己取右边的数，贪心的人取剩下n-1个数左右两段更大的一端；同理模拟自己取左边的数。

2）如果还剩2个数，那么自己取较大的一个

代码：TLE

#include <cstdio>
#define N 1002
int a[N], n, ans, t, Case = 0;
void dfs(int s, int e, int d)
{
	if (e - s == 1)
	{
		if (a[e]>a[s] && ans<d+a[e]-a[s]) ans = d+a[e]-a[s];
		else if (a[e]<a[s] && ans<d+a[s]-a[e]) ans = d+a[s]-a[e];
		return;
	}
	t = a[s];
	if (a[s+1] >= a[e]) dfs(s+2, e, d+t-a[s+1]);
	else dfs(s+1, e-1, d+t-a[e]);
	t = a[e];
	if (a[s] >= a[e-1]) dfs(s+1, e-1, d+t-a[s]);
	else dfs(s, e-2, d+t-a[e-1]);
}
int main()
{
	while (scanf("%d", &n), n)
	{
		for (int i = 1; i <= n; ++ i)
			scanf("%d", &a[i]);
		ans = 0;
		dfs(1, n, 0);
		printf("In game %d, the greedy strategy might lose by as many as %d points.\n", ++Case, ans);
	}
}

但是，细想一下不难发现，这样会超时，递归深度1000。

于是可以进行反向递推，先把所有剩下的2个数时的最大差值算出来，再算4个数，6个数，...，n个数即可。

代码：

#include <cstdio>
#define N 1002
#define abs(x) (x)<0?-(x):(x)
#define max(a,b) (a)<(b)?(b):(a)
int n, a[N], f[N][N], t1, t2, Case = 0;
int main()
{
	while (scanf("%d", &n), n)
	{
		for (int i = 1; i <= n; ++ i)
			scanf("%d", &a[i]);
		for (int i = 1; i < n; ++ i)
			f[i][i+1] = abs(a[i]-a[i+1]);
		for (int i = 4; i <= n; i += 2)
		{
			for (int j = 1; j+i-1 <= n; ++ j)
			{
				// take right end
				t1 = a[j+i-1];
				if (a[j] >= a[j+i-2])
				{ // greedy takes left end
					t1 -= a[j];
					t1 += f[j+1][j+i-2];
				}
				else
				{ // greedy takes right end
					t1 -= a[j+i-2];
					t1 += f[j][j+i-3];
				}
				// take left end
				t2 = a[j];
				if (a[j+1] >= a[j+i-1])
				{ // greedy takes left end
					t2 -= a[j+1];
					t2 += f[j+2][j+i-1];
				}
				else
				{ // greedy takes right end
					t2 -= a[j+i-1];
					t2 += f[j+1][j+i-2];
				}
				f[j][j+i-1] = max(t1, t2);
			}
		}
		printf("In game %d, the greedy strategy might lose by as many as %d points.\n", ++Case, f[1][n]);
	}
}