思路：這道題的核心是對於狀態方程的設立，在這裡dp[i]表示的在已經堆好的磚塊下面再墊上第i塊石頭的高度，建立完dp後，再找出所有dp裡面的最大值。需要注意的是一個磚頭可以產生6種狀態。然後其中一個細節是需要用sort自定義排序，順序是長度最小的在前面，長度相同則寬度小的在前，這樣處理的好處是方便你在建立dp的過程中優化，具體的優化細節可以在程式碼中看；

程式碼：

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

using namespace std;

const int maxn = 200;
struct BLOCK
{
		int l, w, h;
}block[maxn];
int dp[maxn];

bool cmp(BLOCK a, BLOCK b)
{
		if (a.l == b.l)
				return a.w < b.w;
		return a.l < b.l;
}

int main()
{
		int n;
		int T = 0;
		while (scanf("%d", &n) && n)
		{
				memset(dp, 0, sizeof(dp));
				int len = 0;
				int p = 0;
				while (p != n)
				{
						p++;
						int a, b, c;
						scanf("%d %d %d", &a, &b, &c);
						block[len].l = a; block[len].w = b; block[len++].h = c;
						block[len].l = a; block[len].w = c; block[len++].h = b;
						block[len].l = b; block[len].w = a; block[len++].h = c;
						block[len].l = b; block[len].w = c; block[len++].h = a;
						block[len].l = c; block[len].w = a; block[len++].h = b;
						block[len].l = c; block[len].w = b; block[len++].h = a;
				}
				sort(block, block + len, cmp);
				dp[0] = block[0].h;
				int mx = 0;
				for (int i = 1; i < len; i++)
				{
						mx = 0;
						for (int j = 0; j < i; j++)
						{
								if (block[j].l < block[i].l&&block[j].w < block[i].w)
										mx = mx > dp[j] ? mx : dp[j];
						}
						dp[i] = mx + block[i].h;
				}
				mx = 0;
				for (int i = 0; i < len; i++)
				{
						if (mx < dp[i])
								mx = dp[i];
				}
				printf("Case %d: maximum height = %d\n", ++T,mx);
		}
	//	system("pause");
		return 0;
}