hdu1069 经典dp例题 monkey and Banana

版权声明：那个，最起码帮我加点人气吧，署个名总行吧 https://blog.csdn.net/qq_41670466/article/details/83476575

思路：这道题的核心是对于状态方程的设立，在这里dp[i]表示的在已经堆好的砖块下面再垫上第i块石头的高度，建立完dp后，再找出所有dp里面的最大值。需要注意的是一个砖头可以产生6种状态。然后其中一个细节是需要用sort自定义排序，顺序是长度最小的在前面，长度相同则宽度小的在前，这样处理的好处是方便你在建立dp的过程中优化，具体的优化细节可以在代码中看；

代码：

#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;
}